SN Name Figure Showing Properties
6. Parallelogram
7. Rectangle AD Opposite sides are equal in lengths and diagonals are
8. Rhombus not equal.
9. Square BC i.e., AB = CD, AC ≠ BD and BC = AD.
PS
10. Kite Opposite sides are equal and diagonals are equal.
11. Arrow Head QR i.e., PQ = SR, QR = PS and PR = QS.
12. Circle XW ‘OR’
Y Z Opposite sides are equal and square of diagonal is
D C equal to the sum of the square of adjacent sides.
A B i.e., PQ = SR, QR = PS and PR2 = PQ2 + QR2. COORDINATE
A GEOMETRY
All sides are equal and diagonals are not equal.
BD i.e., WX = XY = YZ = ZW and WY ≠ XZ.
C All sides are equal and diagonals are equal.
BD i.e., AB = BC = CD = DA and AC = BD.
A ‘OR’
All sides are equal and satisfies Pythagoras theorem.
C B i.e., AB = BC = CD = DA and AC2 = AB2 + BC2.
A
Any two pairs of opposite adjacent side are equal
O and the main (vertical) diagonal is longer than other
diagonal.
i.e., AB = AD, BC = CD and AC > BD.
Any two pairs of opposite adjacent side are equal
and the main (vertical diagonal is shorter than next
diagonal.
i.e., AB = AD, BC = CD and AC < BD.
All radii are equal.
i.e., OA = OB, where O is the center of the circle.
Example 1 : Prove that the points (–2, 1), (2, –1) and (2, 3) are the vertices of an isosceles triangle.
Solution : Here, the given three points are A(– 2, 1), B(2, – 1) and C(2, 3). A(–2, 1)
Now, we have
d = (x2 – x1)2 + (y2 – y1)2 B(2, –1) C(2, 3)
∴ AB = (2 + 2)2 + (– 1 – 1)2 = 16 + 4 = 20 = 2 5 units.
BC = (2 – 2)2 + (3 + 1)2 = 0 + 16 = 16 = 4 units.
CA = (– 2 – 2)2 + (1 – 3)2 = 16 + 4 = 20 = 2 5 units.
Here, AB = CA. Since two sides of a triangle are equal so the given points are the vertices
of the isosceles triangle.
Distance Formula and its Applications 101
Example 2 : Show that points A(1, a) and B(–a, 1) subtend a right angle at the origin.
Solution : Here, A(1, a), B(– a, 1) and O(0, 0) are the vertices of ΔABC. Then
OA = 12 + a2 OB = (– a)2 + 12 = a2 + 1 Y A (1, a)
AB = (– a – 1)2 + (1 – a)2 = a2 + 2a + 1 + 1 – 2a + a2 = 2a2 + 2 B (– a, 1)
∴ AB2 = 2a2 + 2
Now, OA2 + OB2 = a2 + 1 + a2 + 1 = 2a2 + 2 X' O(0, 0) X
∴ OA2 + OB2 = AB2 and ∠AOB = 90o.
Hence, the points A and B subtend a right angle at the origin. Y'
Example 3 : Show that pointsA(3, 0), B(7, 1), C(5, 3) and D(1, 2) are the vertices of the parallelogram.
Solution : Here, the given points are A(3, 0), B(7, 1), C(5, 3) and D(1, 2). D(1, 2) C(5, 3)
Now, we know that the distance formula,
d = (x2 – x1)2 + (y2 – y1)2 A(3, 0) B(7, 1)
∴ AB = (7 – 3)2 + (1 – 0)2 = 16 + 1 = 17 units.
BC = (5 – 7)2 + (3 – 1)2 = 4 + 4 = 8 = 2 2 units.
COORDINATE CD = (1 – 5)2 + (2 – 3)2 = 16 + 1 = 17 units.
GEOMETRY
DA = (3 – 1)2 + (0 – 2)2 = 4 + 4 = 8 = 2 2 units.
Here, AB = CD and BC = DA. Since the opposite sides of quadrilateral are equal so the
given points are the vertices of the parallelogram.
Example 4 : Prove that the quadrilateral PQRS with vertices P(2, – 1), Q(3, 4), R(– 2, 3) and
S(– 3, – 2) is a rhombus.
Solution : Here, the vertices of the quadrilateral are P(2, – 1), Q(3, 4), R(– 2, 3) and S(– 3, – 2).
Now, we know that the distance formula, S(–3, –2) R(–2, 3)
d = (x2 – x1)2 + (y2 – y1)2
∴ PQ = (3 – 2)2 + (4 + 1)2 = 1 + 25 = 26 units. P(2, 1) Q(3, 4)
QR = (– 2 – 3)2 + (3 – 4)2 = 25 + 1 = 26 units.
RS = (3 + 2)2 + (– 2 – 3)2 = 1 + 25 = 26 units.
PS = (– 3 – 2)2 + (– 2 + 1)2 = 25 + 1 = 26 units.
PR = (– 2 – 2)2 + (3 + 1)2 = 16 + 16 = 32 = 2 2units.
QS = (–3 – 3)2 + (–2 – 4)2 = 36 + 36 = 72 = 6 2units.
Here, PQ = QR = RS = PS and PR ≠ QS. Since all sides are equal and the diagonals are not
equal in the quadrilateral PQRS, so it is rhombus.
Example 5 : Prove that the points (–4, 2) and (4, –2) lie on the same circle having center (2, 4).
Solution : Let A(– 4, 2) and B(4, – 2) by given two points and C(2, 4) is the center of the circle.
Then, AC = (2 + 4)2 + (4 – 2)2 = 62 + 22 = 36 + 4 = 40 = 2 10 units.
BC = (2 – 4)2 + (4 + 2)2 = (– 2)2 + 62 = 4 + 36 = 40 = 2 10 units.
Here, AC = BC
i.e, the radii of the circle are equal. Hence, the given two points lie on the same circle.
102 Illustrated Optional Mathematics-9
Example 6 : Find the co-ordinates of a point which is equidistant from the point (7, – 3), (0, 2) and
(8, – 2). A (7, –3)
Solution : Let the given three points be A(7, – 3), B(0, – 2) and C(8, – 2)
and O(x, y) be the required point, then,
OA = (7 – x)2 + (–3 – y)2 O (x, y)
= 49 + 14x + x2 + 9 + 6y + y2 C (8, –2) B (0, –2)
= x2 + y2 – 14x + 6y + 58
OB = (0 – x) + (–2 – y) = x2 + 4 + 4y + y2 = x2 + y2 + 4y + 4
OC = (8 – x)2 + (–2 – y)2 = 64 – 8x + x2 + 4 + xy + y2 = x2 + y2 – 16x + 4y + 68
Since O(x, y) is equidistant from the given points,
So, OA = OB = OC
or, x2 + y2 – 14x + 6y + 58 = x2 + y2 + 4y + 4 = x2 + y2 – 16x + 4y + 68
or, x2 + y2 –14x + 6y + 58 = x2 + y2 + 4y + 4 = x2 + y2 – 16x + 4y + 68
or, – 14x + 6y + 58 = 4y + 4 = – 16x + 4y + 68 COORDINATE
GEOMETRY
Taking the first and the second expressions, Taking the second and the third expression,
– 14x + 6y + 58 = 4y + 4 4y + 4 = – 16x + 4y + 68
or, 14x + 4y – 6y + 4 – 58 = 0 or, 16x = 68 – 4
or, 14x – 2y – 54 = 0 or, x = 64
16
or, 7x – y – 27 = 0 ………. (i) ∴ x = 4
Putting the value of x in equation (i), we get
7 × 4 – y – 27 = 0
or, 28 – y – 27 = 0
or, y = 1
∴ The coordinates of the required point is (4, 1).
EXERCISE – 4.1 (B)
1. (a) Write any one property of a parallelogram.
(b) Why is equilateral triangle different from the isosceles triangle?
(c) Write any one different property between rectangle and square.
(d) In which condition do the given points lie on the circumference of a circle?
2. (a) Prove that the points P(– 1, 1), Q(2, 3) and R(2, 1) are the vertices of a scalene triangle.
(b) Prove that the points A(4, 5), B(4, 4) and C(2, – 1) form a scalene triangle.
(c) Show that the points P(2, 4), Q(5, –2) and R(– 3, 4) are the vertices of a scalene triangle.
3. (a) Show that the points (2, –1), (2, 5) and (– 2, 2) are the vertices of an isosceles triangle.
(b) Prove that the points (3, 1), (4, 2) and (2, 0) are the vertices of an isosceles triangle.
(c) Verify that: A(a, a), B(–a, a) and C(– a, – a) are the vertices of an isosceles triangle.
Distance Formula and its Applications 103
COORDINATE 4. (a) Prove that the triangle with vertices (– 1, 1), (1, – 1) and ( 3 , 3 ) is an equilateral triangle.
GEOMETRY (b) Prove that the points (2, 2), (– 2, – 2) and (– 2 3 , 2 3 ) are the vertices of an equilateral
triangle.
(c) If L(4, – 4), M(– 4, 4) and N(– 4 3 , 4 3 ) are three given points, prove that they form an equilateral
triangle.
5. (a) Prove that the points (5, 3), (1, 2) and (2, – 2) are the vertices of a right-angled triangle.
(b) Show that P(3, 0), Q(4, 7) and R(1, 6) are the vertices of a right-angled triangle.
(c) Prove that the triangle with vertices (2, – 1), (– 1, 2) and (4, 1) is a right triangle.
(d) Prove that the points (– 2, 1) and (– 2, – 4) subtend a right-angled triangle at the origin.
6. (a) Show that the points (0, – 1), (2, 1) and (0, 3) are the vertices of a right-angled isosceles triangle.
(b) Prove that the triangle with vertices (1, 1), (–1, 3) and (–3, 1) forms an isosceles right triangle.
(c) It is given that the points (2, 1), (2, – 4) and (– 3, 1) are on the same coordinate plane. Prove
that they form a right –angled isosceles triangle.
(d) Prove that the points (2, 5) and (–2, 5) subtend a right-angled triangle at the origin and the
triangle formed by these points is also isosceles.
7. (a) Prove that the points (1, 1), (3, 3), (7, 3) and (5, 1) are the vertices of a parallelogram.
(b) Show that the points (1, 1), (2, – 2), (5, 2) and (4, 5) form a parallelogram.
8. (a) Show that the points (3, 1), (4, 6), (– 1, 5) and (– 2, 0) are the vertices of a rhombus.
(b) Show that the quadrilateral with vertices (– 1, 1), (– 2, 0), (– 5, – 1) and (– 2, – 2) is a rhombus.
9. (a) Prove that the points (– 2, 1), (0, – 1), (3, 2) and (1, 4) are the vertices of a rectangle.
(b) Prove that the points P(0, 3), Q(– 3, 2), R(– 1, – 4) and S(2, – 3) are the vertices of a rectangle.
10. (a) Show that the points A(1, 3), B(– 1, 5), C(– 3, 3) and D (– 1, 1) are the vertices of the square.
(b) Prove that the quadrilateral ABCD with vertices A(1, 1), B(4, 2), C(5, – 1) and D (2, – 2) is a
square.
11. (a) Prove that the points A(4, 3), B(2, 1), C(4, – 2) and D(6, 1) are the vertices of the kite.
(b) Show that the points P(6, – 1), Q(3, 1), R(6, – 2) and S(9, 1) are the vertices of an arrow head.
12. (a) P(1, – 2), Q(3, –1), R(0, 2) and S(– 2, 1) are the corners of a geometric figure. What figure does it make?
(b) Which shape is made by joining the points (3, 1), (–2 , 2) and (– 1, – 3) ?
13. (a) Show that the circle with center C(2, 3) passes through the points A(5, 7) and B(6, 6).
(b) Prove that the points O(0, 0), A(3, 3), B(6, 0) and C(3, – 3) lie on the circle having center (3, 0).
14. (a) Find the coordinates of a point which is equidistant from the points (3, – 4), (3, 1) and (– 5, 4).
(b) Find the center and radius of the circum-circle circumscribing the triangle with vertices (1, 2),
(3, – 4) and (5, – 6).
15. (a) A pair of opposite vertices of a square is (1, 2) and (3, 4). Find the coordinates of the remaining
vertices of the square.
(b) The opposite angular points of a square are (1, – 1) and (5, 3). Find the coordinates of its
remaining angular points.
ANSWERS (b) (11, 2) 12. (a) Parallelogram (b) Scalene triangle
15. (a) (1, 4), (3, 2) (b) (1, 3), (5, – 1)
1.-11. Show to your teachers.
14. (a) – 3, 7
26
104 Illustrated Optional Mathematics-9
41.21 LOCUS
Learning Objectives
At the end of this topic, the students will be able to:
• identify the fundamental principle of locus. • find the equation of a locus under given condition.
4.2 (A) Introduction to Locus
In the figure (i), take a moving point P such that its distance from the y-axis is constant as a. The dot line
made by P is straight line which is parallel to the y-axis. In the figure (ii) every point Q is equidistant from
the both axes. The dot line made by Q is also a straight line. In the figure (iii) every point R is constant
distance from the fixed point as origin O. The dot line made by R is a path of circle.
YY
P x=y Y COORDINATE
NQ GEOMETRY
R
a X X' OM X OX
X' O
X' x2 + y2 = r2
Y'
x=a
fig: (iii)
Y' Y'
fig:(i) fig: (ii)
The path of moving point under certain condition is called the locus of a point.
Equation of a Locus
A locus may be any line, straight or curve. In the co–ordinate plane, the locus of a moving point P(x, y)
represents an algebraic relation between x and y satisfying the given condition, under which moves P.
In the above figure (i), the locus of the point P(x, y) is equidistant from the y-axis. Suppose the distance
from the y-axis is a unit i.e., OP = a, where OP is x-intercept or x–coordinate. Hence, x = a, which is
required equation of the locus (or straight line) of the point P(x, y) equidistant from the y-axis. In the above
figure (ii), the locus of the point Q(x, y) is equidistant from the x-axis and the y-axis i.e., QM = QN. Thus,
x-intercept and y-intercept are equal in length. Hence, x = y, which is the required equation of the locus of
the point Q(x, y) equidistant from both axes. In the above figure (iii), the locus of a moving point R(x, y)
is the circle. Let its distance from the origin O(0, 0) be the radius r. We have the distance formula,
d = (x2 – x1)2 + (y2 – y1)2
or, OP2 = (x – 0)2 + (y – 0)2
or, r2 = x2 + y2
∴ x2 + y2 = r2, which is the required equation of the locus of a point R which is constant distance from
the origin O.
Locus 105
Fundamental Principle of Locus Y
Consider the equation 2x – y = 3 ……… (i)
Take a point (2, 1) and put x = 2 and y = 1 in the equation (i) then we get
2 × 2 –1 = 3 or, 3 = 3, which is true. (2, 1)
X' O X
(2, –1)
Hence, the point (2, 1) lies on the locus of the equation. But, take another
point (2, – 1), and put x = 2 and y = – 1 in the equation (i), then we get 2x – y = 3
2 × 2 – (– 1) = 3
or, 4 + 1 = 3 or, 5 = 3, which is false. Y'
Y
Hence, the point (2, – 1) does not lie on the locus of the equation.
Conversely, the point (4, 3) lies on the straight line having equation (1, 3) (4, 3)
3x – 2y = 6……………(ii) 3x – 2y = 6
Then, put x = 4 and y = 3 in equation (ii), we get X' O X
3×4–2×3=6
COORDINATE or, 12 – 6 = 6 or, 6 = 6, which is true.
GEOMETRY
Hence, the point (4, 3) satisfies the equation of the locus. Y'
Take another point (1, 3) which does not lie on the equation of the straight line then put x = 1 and y = 3
in equation (ii), we get
3×1–2×3=6
or, 3 – 6 = 6 – 3 or, – 3 = 3, which is false.
Hence, the point does not lie on the locus of an equation and the coordinates of the point does not satisfy
the equation.
Thus, from the above discussion, we arrive the following fundamental principle of locus:
(i) If a point satisfies the equation or condition or rule, it lies on the locus.
(ii) If a point lies on the locus, it satisfy the given equation or condition of the locus.
(iii) If any point does not satisfy the given equation, it does not lie on the locus and vice versa.
(iv) The locus of a moving point in the same plane under certain condition or rule is always straight
or curved line.
(v) If three or more straight lines intersect at a point or if any point lies on the all straight lines,
they are said to be concurrent.
Example 1 : For what value of k, the point (1, – 2) lies on the locus whose equation is x2 + y2 + 2x – ky = 9
Solution : The equation of the locus is x2 + y2 + 2x – ky = 9 ……………..(i)
Since the point (1, – 2) lies on the equation (i), then the equation must satisfy that point
i.e., put x = 1 and y = – 2, we get
(1)2 + (– 2)2 + 2 × 1 – k(– 2) = 9 or, 2k = 9 – 7 or, k= 2 ∴ k =1.
or, 1 + 4 + 2 + 2k = 9 or, 7 + 2k = 9 2
106 Illustrated Optional Mathematics-9
Example 2 : If the straight lines 2x + y = 2, 3x – y = 3 and 4x + 3y + k + 2 = 0 are concurrent, find
the value of k.
Solution : Here, the equations of the given straight lines are 2x + y = 2
(x, y)
2x + y = 2 ................... (i)
3x – y = 3 ................... (ii)
4x + 3y + k + 2 = 0 ................... (iii) 3x – y = 3
Adding equations (i) and (ii), we get 4x + 3y + k + 2 = 0
5x = 5 or, x = 1
Putting the value of x in eqn (i), we get
2 × 1 + y = 2 or, y = 0
∴ The coordinates of the intersecting point of the equations (i) and (ii) is (1, 0). Since the
given three lines are concurrent, so the point (1, 0) also lies on the line (iii). Then,
Put x = 1 and y = 0 in eqn (iii), we get
4 × 1 + 3 × 0 + k + 2 = 0 or, k = –6.
EXERCISE - 4.2 (A) COORDINATE
GEOMETRY
1 (a) What is locus? (b) Write any one fundamental principle of locus.
(c) What is the equation of a locus of moving point P which moves equidistant from the x-axis?
2. (a) Which of the points (2, 1) and (– 4, – 1) lie on the line with equation 3x – 2y = 4 ?
(b) Do the points (– 4, 5) and (0, 4) lie on the locus whose equation is x2 + y2 + 6x – 8y + 16 = 0?
3. (a) For what value of a, the point (–2, – 4) lies on the locus of ax – 3y = 14?
(b) Find the value of k so that the point (2, 3) lies on the locus of x2 + y2 – kx + 2y – 17 = 0.
(c) Find the value of p if a point (p, 2p) lies on the locus of x2 + y2 – 5y + 5 = 0.
4. (a) The point (2, – 1) lies on the locus whose equation is x – ky = 4. Prove that (8, 2) is another
point which lies on the same locus.
(b) If the point (1, – 2) lies on the locus defined by the equation 2x2 + 2y2 – ax = 8, then show that
another point (1, 2) also lies on the same locus.
5. (a) If three locii with the equations – 2x + 4y + 6 = 0, x – 3 = 0 and 3x + 4y + k = 6 are intersected
at a point, find the value of k.
(b) If the straight lines 2x – 3y = 4, x – 2y = 1 and 3x + 2ky = 3 are concurrent, find the value of k.
6. (a) In which points do the locii with the equations 2x – y = 1 and x2 + y2 = 2 intersect?
(b) Where do the locii with the equations 3y + x = 2 and 2x2 + 2y2 + 4x – 6y + 6 = 0 intersect?
7. (a) If the locii x – y = 0, x2 + y2 – 6y = 0 and x2 + y2 – 6x + p + 2 = 0 are concurrent, find the value of p.
(b) If the locii x2 – y = 0, x – y + 2 = 0 and x2 + y2 – 4x – 2y + k = 0 are intersected at a point, find
the value of k.
ANSWERS
2. (a) lies, does not lie (b) does not lie, lies 3. (a) –1 (b) 1 (c) 1
(b) – 4
4. (a) 2 (b) 2 5. (a) – 3 (b) – 3
6. (a) (1, 1), – 1, – 7 (b) (– 1, 1), 13, – 11 7. (a) –2
55 10 10
Locus 107
4.2 (B) Method of Finding Equation of a Locus
Learning Objectives
At the end of this topic, the students will be able to:
• find the equation of the locus under given condition.
The general procedure for finding the equation of a locus is as follows: Y
Step-1 : Draw a figure showing the given condition or data and consider
P4
a point P(x, y) on the locus. For example; a point P which moves
its distance from x-axis is twice of its distance from y-axis. a P3
Step-2 : Write down the given condition under which it moves. For example; N(0, y) P2
the point P(x, y) which moves its distance from x-axis (i.e., PM) is
twice of its distance from y-axis (i.e., PN). Then PM = 2PN. P1 2a X
X' O M(x, 0)
Step-3 : Express the condition in terms of the coordinates
(x, y), i.e., horizontal and vertical displacement of Y'
P and simplify, generally, by using distance formula
COORDINATE d = (x2 – x1)2 + (y2 – y1)2 . This simplest form of the given condition is the required equation of
GEOMETRY the locus. For example; PM = y-intercept (y) and PN = x-intercept (x). Therefore, y = 2x, which
is required equation of the locus.
Example 1 : Find the equation of the locus of a point so that it is twice as far from (6, 3) as from (– 4, 5).
Solution : Let a point P(x, y) which moves twice as far from A (6, 3) P(x, y)
as far from (– 4, 5).
Then PA = 2PB
or, (6 – x)2 + (3 – y)2 = 2 (–4 – x)2 + (5 – y)2 A(6, 3) B(– 4, 5)
or, 36 – 12x + x2 + 9 – 6y + y2 = 4(16 + 8x + x2 + 25 – 10y + y2)
[ squaring on both sides]
or, x2 + y2 –12x – 6y + 45 = 64 + 32x + 4x2 + 100 – 40y + 4y2
or, 0 = 4x2 + 4y2 + 32x – 40y + 164 – x2 – y2 + 12x + 6y – 45
or, 3x2 + 3y2 + 44x – 34y + 119 = 0,
which is the required equation of the locus of the moving point.
Example 2 : Find the equation of a locus of a point whose distance 5 units P(x, y)
from the point (2, 2) is 5 units and prove that (2, – 3) lies A(2, 2)
on the locus.
Solution : Let P(x, y) be the locus of a moving point. (2, –3)
By question, AP = 4
or, (x – 2)2 + (y – 2)2 = 5 [ by using the distance formula]
or, x2 – 4x + 4 + y2 – 4y + 4 = 25 [ Squaring both sides]
or, x2 + y2 – 4x – 4y – 17 = 0, which is required equation of the locus.
108 Illustrated Optional Mathematics-9
Now, for (2, –3), let's put x = 2 and y = – 3 in the above equation, we get
22 + (– 3)2 – 4 × 2 – 4 (– 3) – 17 = 0
or, 4 + 9 – 8 + 12 – 17 = 0
or, 0 = 0, which is true.
Therefore, (2, – 3) lies on the locus.
Example 3 : Find the equation of the locus of a moving point such that the sum of the square of its
distances from the points A(– 1, 2) and B(– 2, 1) is unity. P(x, y)
Solution : Let P(x, y) be the locus of a moving point.
Then by question, PA2 + PB2 = 1
or, (–1 – x)2 + (2 – y)2 + (–2 – x)2 + (1 – y)2 = 1 B(–2, 1)
or, 1 + 2x + x2 + 4 – 4y + y2 + 4 + 4x + x2 + 1 – 2y + y2 = 1 A(–1, 2)
or, 2x2 + 2y2 + 6x – 6y + 9 = 0, which is the required equation of the locus.
Example 4 : Find the equation of the locus of a moving point which divides the line segment joining
the points A(a, 0) and B(0, b) in the ratio 2:3 and making a + b = 20.
Solution : Suppose, P(x, y) divide the line segment AB joining the points A(a, 0) and B(0, b) in the ratio 2:3. COORDINATE
GEOMETRY
Then, by using section formula, we get Y
B(0, b)
(x, y) = 2 × 0 + 3 × a, 2 × b + 3 × 0 = 35a, 2b
2 + 3 2 + 3 5
∴ x = 3a and y = 2b 3
5 5
b P(x, y)
or, a= 53x, b = 5y X' O P2
2
Now, by equation, A(a, 0)
aX
a + b = 20
or, 53x + 52y = 20 Y'
or, 5(2x6+ 3y) = 20
20 × 6
or, 2x + 3y = 5
or, 2x + 3y = 24, which is the required equation of the locus.
EXERCISE - 4.2 (B)
1. Write the equation of locus of a point which moves such that its distance from.
(a) x-axis is always 4 units. (b) y-axis is always 5units.
(c) x-axis and y-axis are always equal.
2. What is the locus of a point moves so that,
(a) Its abscissa is always 2. (b) its ordinate is always –5.
3. (a) What is the equation of the locus of a point so that it is parallel to x-axis and passes through (4, –1)?
(b) Write down the equation of the locus of a point so that it is parallel to y-axis and passes through
(3, –5).
Locus 109
4. Find the equation of a locus of a point which moves so that,
(a) its ordinate is twice as its abscissa. (b) its abscissa is twice of its ordinate.
5. Find the equation of a locus of a point which always moves,
(a) at a distance of 3 units from the origin. (b) at a distance of 4 units from the point (0, 0).
6. Find the equation of a locus of a moving point which moves such that its distance from,
(a) the point (2, 3) is 8 units. (b) the point (– 1, 2) is 3 units.
7. Find the equation of a locus of a point which moves so that,
(a) its distance from y-axis is twice of its distance from x-axis.
(b) its distance from x-axis is triple of its distance from y-axis.
8. Find the equation of a locus of a point which is always at equidistance from the points,
(a) (4, 2) and (3, 5) (b) (2, – 1) and (– 3, 2)
9. Find the equation of a locus of a point which moves such that the distance from,
(a) the point (1, – 2) is twice the distance from the point (0, 3).
(b) the point (0, 2) is half of the point (– 2, 0).
COORDINATE 10. Find the locus of a point which moves such that its distance from,
GEOMETRY
(a) x-axis is double the distance from the point (0, b).
(b) the point (–a, 0) is twice its distance from the y-axis.
11. Find the equation of the locus of a point which moves such that its distance from,
(a) the point (2, 3) is double of its distance from x-axis.
(b) the point (– 2, 3) is half of its distance from y-axis.
12. (a) A(3, 0) and B(0, –3) are the fixed points. Find the equation of the locus of the moving point
P (x, y) such that
(i) 2PA = 3PB (ii) PA2 = 2PB2
(iii) PA2 + PB2 = 12 (iv) PA2 + PB2 = AB2
(b) M(p, 0) and N(0, q) are two points. Find the locus of the point P which moves such that,
(i) PM = PN, (ii) PM2 = 1 PN2
2
(iii) PM2 + PN2 = p2 (iv) PM2 – PN2 = q2
13. (a) Find the locus of a point moving at the equal distance from the points (2, –1) and (0, –3).
Identify the point (2, 2) lies on the locus or not.
(b) Verify whether the point (8, 2) lies on the equation of a point which moves such that its distance
2 more than 3 times the distance from the x-axis is equal to its distance from the y-axis.
14. If A(5, 0) and B(–5, 0) are two fixed points, find the equation of the locus of a point P such that
(a) PA = PB (b) PA:PB = 3:4.
15. (a) Find the equation of a locus of a point P on the reference of two points A(3, 0) and B(– 2, – 1)
such that PA is perpendicular to PB.
(b) A(– 1, 5) and B(4, – 1) are fixed points. Find the locus of the moving points P at which A and
B subtend a right angle.
110 Illustrated Optional Mathematics-9
ANSWERS
1. (a) y = 4 (b) x = 5 (c) y = x 2. (a) x = 2 (b) y + 5 = 0
3. (a) y + 1 = 0 (b) x – 3 = 0 4. (a) y = 2x (b) x = 2y
5. (a) x2 + y2 = 9 (b) x2 + y2 = 16
6. (a) x2 + y2 – 4x – 6y – 51 = 0 (b) x2 + y2 + 2x – 4y – 4 = 0
7. (a) x = 2y (b) y = 3x 8. (a) x – 3y + 7 = 0 (b) 5x – 3y + 4 = 0
9. (a) 3x2 + 3y2 + 2x – 28y + 31 = 0 (b) 3x2 + 3y2 – 4x – 16y + 12 = 0
10. (a) 4x2 – 3y2 – 8by + 4b2 = 0 (b) 3x2 – y2 – 2ax – a2 = 0
11. (a) x2 – 3y2 – 4x – 6y + 13 = 0 (b) 3x2 + 4y2 + 16x – 24y + 52 = 0
12. (a) (i) 5x2 + 5y2 + 24x + 54y + 45 = 0 (ii) x2 + y2 + 6x + 12y + 9 = 0
(iii) x2 + y2 – 3x + 3y + 3 = 0 (iv) x2 + y2 – 3x + 3y = 0
(b) (i) 2px – 2qy – p2 + q2 = 0 (ii) x2 + y2 – 4px + 2py + 2p2 – q2 = 0
(iii) 2x2 + 2y2 – 2px – 2qy + q2 = 0 (iv) 2px – 2qy – p2 + 2q2 = 0
13. (a) x + y + 1 = 0, Not (b) x – 3y – 2 = 0, Yes COORDINATE
GEOMETRY
14. (a) x2 + y2 – 10x + 25 = 0 (b) 7x2 + 7y2 – 190x + 225 = 0
15. (a) x2 + y2 – x + y – 6 = 0 (b) x2 + y2 – 3x – 4y – 9 = 0
Project Work
Plot any two points A and B in the graph. Derive the equations of the locus of a moving point P(x, y) for
the following conditions:
(a) 3PA = 4AB (b) PA2 + PB2 = AB2 (c) PA2 – AB2 = 2 × PB2
Locus 111
41.31 SECTION FORMULA AND ITS
APPLICATIONS
4.3 (A) Section Formula
Learning Objectives
At the end of this topic, the students will be able to:
• derive the section formula, mid point of two points and centroid of triangle.
• find the coordinates point by using section, mid point and centroid formulae.
• solve the problems related to section, mid point and centroid formulae.
In this section, we discuss the division point which divides the line segment in the given ratio. There are
two types of division point according to the betweenness of the given two points. i.e., internal division and
external division.
COORDINATE Section Formula for Internal Division Y
GEOMETRY
Consider a line segment AB with two end points A(1, 1) and C(8, 8)
B(8, 1). Take a point P(4, 1) between A and B on it. The point
P(4, 1) divides internally AB in the ratio 3:4.
Similarly, the point Q(8, 4) divides internally the line segment R(4, 4) Q(8, 4)
joining the points B(8, 1) and C(8, 8) in the ratio of 3:4. Also,
the point R(4, 4) divides the line segment AC joining the points
A(1, 1) and C(8, 8) in the ratio 3:4.
But, which point divides internally the line segment PQ joining X' A(1, 1) P(4, 1) B(8, 1) X
the point P(– 6, – 2) and Q(4, 3) in the ratio 3:4 ? For this, we O
find the following section formula for internal division.
Let A(x1, y1) and B(x2, y2) be the points and C(x, y) Y' Q(4, 3)
divide the line segment AB internally in the ratio m1 : m2.
i.e., AC : CB = m1:m2 and AB = m1 + m2. 4
3
Draw AL ⊥ OX, BM ⊥ OX, CN ⊥ OX, AP ⊥ CN and
A(x, y)
CQ ⊥ BM. Then OL = x1, OM = x2, ON = x AL = y1,
BM = y2, CN = y P(– 6, – 2)
∴ AP = LN = ON – OL = x – x1 Y B(x2, y2)
CQ = NM = OM – ON = x2 – x C(x, y) m2
CP = CN – PN = CN – AL = y – y1 A(x1, y1) m1 Q
BQ = BM – QM = BM – CN = y2 – y P
Here, ΔACP and ΔCBQ are similar to each other, so
AP = CP = AC or, x – x1 = y – y1 = m1 X
CQ BQ CB x2 – x y2 – y m2 OL N M
112 Illustrated Optional Mathematics-9
Taking the first and last ratio, Taking the second third ratios,
x – x1 = mm21 or, y – y1 = m1
x2 – x y2 – y m2
or, m2x – m2x1 = m1 x2 – m1x or, m2y – m2y1 = m1 y2 – m1y
or, m1x + m2x = m1x2 + m2 x1 or, m1y + m2y = m1y2 + m2y1
or, x(m1 + m2) = m1 x2 + m2 x1 or, y(m1 + m2) = m1y2 + m2y1
or, x = m1x2 + m2x1 or, y = m1y2 + m2y1
m1 + m2 m1 + m2
Hence, the co-ordinates of the dividing point C(x, y) are m1x2 + mm22x1, m1y2 + m2y1 .
m1 + m1 + m2
Note: If C divides the line segment AB in the ratio k : 1, then the co-ordinates of C(x, y) is kxk2++1x1, ky2 + y1 .
k+1
Section Formula for External Division
Y
In the adjoining figure, AB is the line segment having two end
points A(1, 1) and B(5, 5). A point C (8, 8) divides the line C(8, 8)
segment AB in the ratio 7:3 from outside of AB.
3
7
Consider a line segment AB joins two points A(x1, y1) and B(x2, B(5, 5) COORDINATE
GEOMETRY
y2). The point C(x, y) divides externally. Then AB = AC – BC
= m1 – m2 = m1 + (–m2). So, the point C divides AB in the ratio
m1:(–m2). X' A(1, 1) X
Now, we have O
The co–ordinate of C is given by, Y'
(x, y) = m1x2 + (– mm22))x1, m1y2 + (– m2)y1 m2 m1 C(x, y)
m1 + (– m1 + (– m2)
∴ (x, y) = m1x2 – mm22x1, m1y2 – m2y1 B(x2, y2)
m1 – m1 – m2 A(x1, y1)
Mid Point Formula
In the given figure, the line segment AB joining the points A(1, 2) and B(7, 2) is divided by the point
C(4, 2) in the ratio 3:3 i.e 1:1 That means the point C divides AB in two equal segments.
Hence, C is the mid point of AB. Y
1 + 7, 2 + 2 C(4, 2)
2 2
Here, C(4, 2) = A(1, 2) B(7, 2)
Consider a line segment AB joining the points A(x1, y1) and X' O X
B(x2, y2). The C (x, y) is the mid point of AB. Y'
Then (x, y) divides the line segment AB internally in the ratio m1:m1.
Now, we have m1 B(x2, y2)
C(x, y)
The co–ordinates of the mid point C is given by, m2
A(x1, y1)
(x, y) = m1x2 + mm11x1, m1y2 + m1y1 = m1(x2 + x1), m1(y2 + y1)
m1 + m1 + m1 2m1 2m1
= x1 + x2, y1 + y2
2 2
Section Formula and its ApplicationS 113
Centroid of Triangle Y
In the figure, the points A(5, 7), B(1, 1) and C(9, 1) are the A(5, 7)
vertices of ΔABC and P(5, 1), Q(7, 4) and R(3, 4) are the mid
points of the sides BC, CA and AB of ΔABC respectively. R2 Q
Then, the medians AP, BQ and CR intersect at a point G. That X' B(1, 1) 1G C(9, 1)
means the medians of a triangle are concurrent. This intersecting O P X
point of the medians is called a centroid of the triangle. The
centroid G divides the median in the ratio 2:1. Here, the Y'
co-ordinates of the centroid G of ΔABC are (5, 3).
Let the points A(x1, y1), B(x2, y2) and C(x3, y3) the vertices of ΔABC and G(x, y) be the centroid of ΔABC.
Now,
Since P is the mid point of BC, so the co–ordinates of P are x2 + x3, y2 + y3 .
2 2
COORDINATE Also, the centroid G divides the median AP internally in the ratio 2:1. So, the co–ordinate of G is given by,
GEOMETRY
(x, y) = m1x2 + mm22x1, m1y2 + m2y1 A(x1, y1)
m1 + m1 + m2
= 2. x2 + x3 + 1.x1 2. y2 + y3 + 1.y1 2
2 +1 , 2 +1
2 2 1 G (x, y)
P
= x1 + x2 + x3, y1 + y2 + y3 B(x2, y2) C(x3, y3)
3 3
Example 1 : Find the coordinates of the point that divides the line segment joining the points (2, 3)
and (8, 4) internally in the ratio 3:5.
Solution : Let (x, y) divide the line segment joining the points (2, 3) and (8, 4) in the ratio 3:5. Then, we
have,
(x, y) = m1x2 + mm22x1, m1y2 + m2y1 5 (8, 4)
m1 + m1 + m2
3 (x, y)
= (2, 3)
3 × 8 + 5 × 2, 3 × 4 + 5 × 3
3 + 5 3 + 5
= 24 + 10, 12 + 15
8 8
= 384, 27
8
= 147, 27
8
( ) 17 27 .
Hence, the coordinate of the point is 4 , 8
114 Illustrated Optional Mathematics-9
Example 2 : In what ratio does the point (5, 4) divide the line segment joining the points (2, 1) and
(7, 6) ?
Solution : Let the point (5, 4) divide the line joining the points (2, 1) and (7, 6) in the ratio m1 : m2
Then we have,
(x, y) = m1x2 + mm22x1, m1y2 + m2y1 m2 (7, 6)
or, m1 + m1 + m2
m1 (5, 4)
m1 ×7 + m2 × 2, m1 ×6 + m2 × 1 (2, 1)
(5, 4) = m1 + m2 m1 + m2
or, (5, 4) = 7m1 + m2m2 2, 6m1 + m2
m1 + m1 + m2
∴ 5 = 7m1 + 2m2 [ Taking the first part only]
m1 + m2
or, 7m1 + 2m2 = 5m1 + 5m2
or, 7m1 – 5m1 = 5m2 – 2m2
or, 2m1 = 3m2
or, m1 = 3
m2 2
COORDINATE
Hence, the required ratio is 3:2. GEOMETRY
Example 3 : In what ratio does the x-axis intersect the line segment joining the points A(– 4, – 3)
and B(6, 3).
Solution : Let the point on the x-axis be (x, 0) which divides the line segment joining the points
A(– 4, – 3) and B(0, 3) in the ratio m:n. Then, we have
(x, y) = mx2 + nnx1, my2 + ny1 A(6, 3)
or, m + m + n x-axis X'
m × 6 + n (– 4), m × 3 + n (– 3) X
m+n m+n
(x, 0) =
or, (x, 0) = 6m – 4n ), 3m – 3n B(–4, –3)
m+n m + n
or, 0 = 3m – 3n [ Taking the second part only]
m + n
or, 3m – 3n = 0
or, 3m = 3n ∴ m:n = 1:1
or, mn = 33
Hence, the required ratio is 1:1.
Example 4 : If P(2, b) is the mid point of the line segment AB joining the points A(4, 3) and (0, 2),
find the value of b.
Solution : Here, P(2, b) is the mid point of the line segment AB joining the points A(4, 3) and B(0, 2).
Then we have, x1 + x2, y1 + y2
2 2
(x, y) = B(0, 2)
or, (2, b) = 4 + 0, 3 + 2 P(2, b)
2 2
or, (2, b) = 2, 5 ∴ b = 5 A(4, 3)
2 2
Section Formula and its ApplicationS 115
Example 5 : Find the coordinates of the centroid of ΔABC of the vertices A(1, 2) , B(– 1, – 5) and
C(– 3, 3).
Solution : Here, the vertices of ΔABC are A(1, 2), B(– 1, – 5) and C(– 3, 3). Let G (x, y) be the
centroid of ΔABC. Then we have. A(1, 2)
(x, y) = x1 + x2 + x3, y1 + y2 + y3
3 3
= 1 – 1 – 3, 2 – 5 + 3 G(x, y)
3 3
= –33, 0 = (–1, 0) B(–1, –5) C(–3, 3)
3
Hence, the required coordinates of the centroid of ΔABC is (–1, 0).
Example 6 : Find the coordinates of two points which trisect the line segment joining the points
(5, 3) and (8, 6).
Solution: Let P and Q divide the line segment AB joining the points (5, 3) and (8, 6). So, P divides AB
in the ratio 1:2 and the coordinate of P is given by
= m1x2 + mm22x1, m1y2 + m2y1 1P 1Q1 B(8, 6)
m1 + m1 + m2 A(5, 3)
COORDINATE
GEOMETRY = 1 × 8 + 2 × 5, 1 × 6 + 2 × 3
1 + 2 1 + 2
= 8 +310, 6 + 6
3
= 138, 12 = (6, 4)
3
Example 7 : If the mid points of the sides of triangle are (– 2, – 1), (6, – 3) and (4, 5), then find its
vertices. A(x1, y1)
Solution : Suppose, the mid points of the sides BC, CA and AB of
ΔABC of the vertices A(x1, y1), B(x2, y2) and C(x3, y3) are
P(– 2, – 1), Q(6, – 3) and R(4, 5) respectively. Then by using R(4, 5) Q(6, –3)
mid-point formula, we get
For P(– 2, – 1) B(x2, y2) P(–2, –1) C(x3, y3)
(– 2, – 1) = x2 + x3, y2 + y3 or, x2 + x3 = – 4 …….. (i) and
2 2 or, y2 + y3 = – 2 ….….. (ii)
∴ – 2 = x2 + x3 or, x3 + x1 = 12 ……….. (iii) and
2 or, y3 + y1 = – 6 ……….. (iv)
∴ – 1 = y2 + y3
2
For Q(6, – 3),
(6, –3) = x3 + x1, y3 + y1
2 2
∴ 6 = x3 + x1
2
∴ – 3 = y3 + y1
2
116 Illustrated Optional Mathematics-9
For R(4, 5),
(4, 5) = x1 + x2, y1 + y2
2 2
∴ 4 = x1 + x2 or, x1 + x2 = 8 ……………...….. (v)
2
∴ 5 = y1 + y2 or, y1 + y2 = 10 ……………….. (vi)
2
Now, subtracting eqn (i) from eqn (iii), we get
x1 – x2 = 16 ………….. (vii)
Adding equation (v) and equation (vii), we get
2x1 = 24, or, x1 = 12
Putting the value of x1 in equations (v) and (iii), we get
12 + x2 = 8, or, x2 = – 4 and
x3 + 12 = 12 or, x3 = 0
Similarly, solving (ii), (iv) and (vi), we get
y1 = 3, y2 = 7 and y3 = – 9,
Hence, the vertices of the triangle are (12, 3), (– 4, 7) and (0, – 9).
Example 8 : Show that the line 2x + y = 2 bisects the line segment joining the points (2, – 2) and (0, 2). COORDINATE
GEOMETRY
Solution: Here, the mid point of the line segment joining the points
(2, – 2) and (0, 2) is 2 + 0, – 2+ 2 = (1, 0).
2 2
(2, –2) (0, 2)
Now, put x = 1 and y = 0 in the given line 2x + y = 2, we get
2×1+0= 2
or, 2 = 2, which is true. 2x + y = 2
Hence, the given line bisects the line segment joining the given two points.
EXERCISE - 4.3 (A)
1. (a) Write the section formula for internal division and external division.
(b) What is the mid point of the line segment joining the points A(a1, b1) and B(a2, b2)?
(c) If the points A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of ∆ABC then how can you find
the centroid of ∆ABC?
2. (a) Find the coordinates of a point which divides the line segment joining the points (4, 3) and
(5, 6) internally in the ratio 3:2.
(b) Find the coordinates of a point which divides internally the line joining the points A(– 3, – 4)
and B(2, 5) in the ratio 3:5.
3. (a) Find the coordinates of a point which divides the line segment joining the points (3, 0) and
(– 2, – 4) externally in the ratio 1:4.
(b) Which point divides the line joining the points (1, – 2) and (4, 7) externally in the ratio 2:3 ?
4. (a) In what ratio does the point (0, 7) divide the line segment joining the points (2, 5) and (5, 2)?
(b) Find the ratio in which the point (1, 1) divides the segment joining the point (– 6, 8) and (8, – 6).
5. (a) In what ratio does the x-axis intersect the line segment joining the point (2, 4) and (–3 , – 2)?
(b) In what ratio is the line joining the points (– 2, 4) and (4, – 5) divided by the y-axis?
6. (a) In what ratio is the line segment (4, 5) and (1, 1) divided by the line x = 2 ?
Section Formula and its ApplicationS 117
(b) In what ratio does the line y = – 2 divide line segment joining the points (1, 2) and (– 5, – 6) ?
Also, find the coordinates of the point of division.
7. (a) The point (x, 1) divides the line joining the points (2, – 2) and (9, 5) in the ratio 3:4. Find the
value of x.
(b) If the point P 232, y divides the line segment joining the points A(2, 3) and B(10, 12) in the
ratio 2:1, then find the value of y.
8. (a) Find the point on the line joining the points A(2, 3) and B(8, 9) whose abscissa is 3.
(b) Find a point on the line joining the points P(– 3, 0) and Q(2, –5) whose ordinate is – 2.
9. (a) The point P(0, a) divides the line segment joining the points 8 units Y
A(– 2, 2) and B( 4, 8). If 2AP = BP, find the value of a. B
(b) From the adjoining figure, if 3AC = BC, find the values of p and q. C (3, q)
10. (a) Find the mid point of a line segment joining the points (– 3, 4) and (3, – 8). O p units A X
(b) If B(2, 0) and D(– 6, 4) are the opposite vertices of a quadrilateral ABCD, find the mid point
of the diagonal BD.
11. (a) If (2, y) is the mid point of the line segment joining the points (– 1, 2) and (5, 0), then find the value
of y.
(– 2, 4)
COORDINATE
GEOMETRY
(b) If the mid-point of the line segment joining the points (2, b) and (4, – 1) is (a, 5), find the values
of a and b.
12. (a) Find the coordinates of two points which divide the line joining the points (1, – 3) and (4, 3) into
three equal parts.
(b) Find the coordinates of two points which trisect line segment AB joining the points A(3, – 4) and
B(– 3, – 4).
13. (a) Find the centroid of ΔABC having the vertices A(4, 5), B(1, 2) and C(4, – 1).
(b) Find the coordinates centroid of a triangle having with vertices (1, – 2), P
(– 4, – 2) and (0, 1).
(c) In the adjoining figure, G is the centroid of ΔPQR. Find the coordinates of G. G
S(7,7)
14. (a) If (0, 1), (1, 2) and (–1, 2) are the mid points of the sides of a triangle, find the Q R
coordinates of the vertices of the triangle.
(b) If A(– 1, – 1) , B(– 1, – 3) and C(– 3, – 1) are the mid-points of the sides of ΔPQR, find the
coordinates of its vertices.
(c) The mid points of two sides of a triangle are (3, 2) and 72, 0 . If the centroid lying on the median
drawn on the third side of the triangle is (3, 1), find the remaining mid point and the vertices of the
triangle.
ANSWERS
) 2.
(a) 23, 24 (b) – 9, – 5 (3. (a) 14 , 4 (b) (– 5, – 20) 4. (a) – 2:5 (b) 1:1
55 88 3 3
5. (a) 2:1 (b) 1:2 6. (a) 2:1 (b) 1:1,(–2, –2) 7. (a) 5 (b) 9
8. (a) (3, 4) (b) (– 1, – 2) 9. (a) 4 (b) 4, 2 10. (a) (0, – 2) (b) (–2, 2)
11. (a) 1 (b) 3, 11
12. (a) (2, – 1), (3, 1) (b) (1, – 4), (– 1 – 4) 13. (a) (3, 2) (b) (– 1, – 1) (c) (4, 6)
14. (a) (– 2, 1), (2, 1), (0, 3) (b) (–3, 1), (1, –3), (–3, –3) (c) 5, 1 ; (2, 3), (4, 1), (3, – 1)
2
118 Illustrated Optional Mathematics-9
4.3 (B) Applications of Section Formula
Learning Objectives
At the end of this topic, the students will be able to:
• prove the geometric properties of triangles and quadrilaterals using section, mid–point and centroid formulae.
We can use section formula to prove some theorems or properties on geometrical figure such as isosceles
triangle, equilateral triangle, right-angled triangle, parallelogram, rectangle, square, rhombus, circle etc.
The methods of verification use the properties given below:
SN Name Figures Showing Properties
A Perpendicular drawn from the vertices to base bisects the
1. Isosceles base.
Triangle Bisector of vertical angle bisects the base. For both cases,
if AB = AC and AD ⊥ BC or ∠BAD = ∠CAD, then
BD C BD = DC.
P Perpendiculars drawn from each vertex or bisectors of
2. Equilateral CB vertical angles bisect the opposite sides. COORDINATE
Triangle GEOMETRY
i.e., PA = QB = RC.
QAR
A
3. Right-angled D Mid point of hypotense is equidistant from each vertex.
Triangle i.e., if AD = CD, then AD = CD = BD.
BC
A Line segment joining the mid points of any two sides of a
4. Triangle P Q triangle is half of its remaining side and parallel to it. i.e., if
C
AP = BP of AQ = CQ, then PQ = 1 BC or BC = 2PQ and
2
PQ // BC.
B
DC
5. Parallelogram O Mid-points of the diagonals are the same. i.e. O is mid point
of AC and BD.
AB
6. Rectangle S R Mid-point of the diagonal are the same.
O Diagonals are equal in length. i.e. O is the mid point of AC
P Q and BD. Also AC = BD.
Z Y Mid points are same.
7. Square O Diagonals are equal.
Any pair of adjacent sides are equal. i.e., O is the mid-point
WX of WY and XZ, WY = XZ and ZW = WX.
D C Mid–points are the same.
O Diagonals are not equal.
8. Rhombus Any pair of adjacent sides are equal.
AB i.e., O is the mid–point of AC and BD; AC ≠ BD and
AD = AB.
Section Formula and its ApplicationS 119
Illustrated Examples
Example 1 : The coordinates of the vertices of an isosceles triangle are (2, 0), (0, 2) and (3, 3). Prove
that the lengths of the medians drawn from the base points are equal.
Solution: Let P and Q be the mid points of the sides AB and AC A(3, 3)
of the isosceles triangle ABC respectively. Then,
P= 3 + 2, 3 + 0 = 25, 3 . PQ
2 2 2
Q= 3 + 0, 3 + 2 = 32, 5 .
2 2 2
B(2, 0) C(0, 2)
∴ CP = 5 – 0 2 + 3 – 2 2 = 24 + 1 = 1 26 units.
2 2 4 4 2
BQ = 3 – 2 2 + 5 – 0 2 = 1 + 25 = 1 26 units.
2 2 4 4 2
COORDINATE Hence, the length of the medians drawn from the base points are equal. Proved.
GEOMETRY
Example 2 : ∆ABC is an isosceles triangle with vertices A(2, 4), B(3, 4) and C(5, 2). Find the length of
perpendicular drawn from the vertex to the base.
Solution: Since the perpendicular AD drawn from the vertex A to the base BC bisects the base, so,
The coordinate of D = 3 + 5, 4 + 2 = (4, 3). A(2, 4)
2 2
∴ AD = (4 – 2)2 + (3 – 4)2
= 4 + 1
= 5 units. B(3, 4) D C(5, 2)
Example 3 : P(4, 0), Q(8, 2), R(5, 3) and S(1, 1) are four points. Prove that PR and QS bisect each
other. S(1, 1)
Solution: Given, the given four points are P(4, 0), Q(8, 2), R(5, 3) and S(1, 1).
Now, the mid point of PR = 5 + 4, 3 + 0 = 92, 3 P(4, 0) R(5, 3)
2 2 2 Q(8, 2)
The mid point of QS = 8 + 1, 2 + 1 = 29, 3
2 2 2
Since the mid points of PR and QS are the same, so they bisect each other.
Example 4 : Show that the points A(2, – 2), B(8, 4), C(5, 7) and D(– 1, 1) are the vertices of a rectangle.
Solution: Here, the given four points are A(2, – 2), B(8, 4), C(5, 7) and D(– 1, 1).
Now, we have, the mid– point of the diagonal AC is 2 + 5, – 2+ 7 D(–1, 1) C(5, 7)
2 2 A(2, –2) B(8, 4)
= 27, 5 .
2
The mid–point of the diagonal BD is 8 – 1, 4 + 1 = 72, 5
2 2 2
120 Illustrated Optional Mathematics-9
Again,
The length of the AC = (5 – 2)2 + (7 + 2)2 = 32 + 92 = 9 + 81 = 90 units.
The length of the BD = (–1 – 8)2 + (1 – 4)2 = (–9)2 + (–3)2 = 81 + 9 = 90 units.
Here, the mid-points of the diagonals AC and BD are the same and the length of the
diagonals AC and BD are equal. Hence, the given points are the vertices of a rectangle.
Example 5 : If the three vertices of a square ABCD are A(–1, –2), B(5, 2) and C(1, 8), find its fourth
vertex D and the length of BD.
Solution: Here, the three vertices of the square ABCD are A(–1, –2), B(5, 2) and C(1, 8).
Now, we have D(x1, y1) C(1, 8)
The mid-point of AC is – 1+ 1, – 2+ 8 = 20, 6 = (0, 3)
2 2 2
Let the co–ordinates of the fourth vertex be D(x1, y1),
then, the mid–point of BD is 5 + x1, 2 + y1 A(–1, –2) B(5, 2)
2 2
5 + x1, 2 + y1
Since the diagonals AC and BD have the same mid point, so 2 2 = (0, 3)
∴ 5 + x1 = 0 and 2 + y1 = 3 x1 = – 5 and y1 = 4 COORDINATE
2 2 GEOMETRY
Hence, the co–ordinates of the fourth vertex of the square is (–5, 4).
Example 6 : The four points A(– 1, 0), B(3, b), C(a, 2) and D(– 2, 1) are the vertices of a parallelogram
ABCD. Find the values of a and b. D(–2, 1) C(a, 2)
Solution: Here, the vertices of parallelogram ABCD are:
A(– 1, 0), B(3, b), C(a, 2) and D(– 2, 1).
Now, we have,
The mid-point of AC is – 1+ a, 0 + 2 = – 1+ a, 1 A(–1, 0) B(3, b)
2 2 2
The mid-point of BD is 3 – 2, b + 1 = 21, b + 1 .
2 2 2
Since, the diagonals AC and BD bisect each other, so, their mid-points are the same.
i.e., –1 + a, 1 = 12, b + 1
∴ 2 2
–1 + a = 1 and 1 = b + 1 ⇒ a = 1 + 1 and b + 1 = 2 ⇒ a = 2 and b = 1.
2 2 2
EXERCISE - 4.3 (B)
1. (a) Write any one property of an isosceles triangle.
(b) Write any one difference property of square from rhombus.
(c) In which conditions does a quadrilateral become a parallelogram?
(d) How can a parallelogram be a rectangle?
2. (a) If A(3, 3), B(– 2, 1) and C(1, – 2) are the vertices of an isosceles triangle ABC, find the length
of the medians drawn from the base points. Are they equal?
(b) Prove that the median drawn from the base points of an isosceles triangle with vertices (5, 3),
(0, 1) and (3, – 2) are equal in length.
Section Formula and its ApplicationS 121
COORDINATE 3. (a) Find the length of perpendicular drawn from the vertex A(5, 3) of the isosceles triangle ABC
GEOMETRY with vertices A(5, 3), B(0, 1) and C(3, –2).
(b) The points (8, 0), (0, 4) and (4, 8) are the vertices of an isosceles triangle. Find the length of
the bisector of vertical angle at (8, 0) up to the base of the isosceles triangle.
4. (a) Prove that the medians of an equilateral triangle having vertices (1, –1), (–1, 1) and ( 3, 3)
are equidistant.
(b) Show that the perpendicular drawn from each vertex two opposite side of an equilateral triangle
PQR with vertices (4, 4), (– 4, – 4) and (– 4 3, 4 3) are equal in length.
5. (a) The points (3, – 3), (7, 1) and (5, 3) are the vertices of right-angled triangle in which right angle
is at (7, 1). Prove that the mid point of the hypotenuse of the right-angled triangle is equidistant
from its vertices.
(b) Prove that the mid point of the hypotenuse AB of the right-angled triangle ABO with right
angle at origin O and other two vertices as A(4, 4) and B(– 2, 2) is equidistant from each
vertex.
6. (a) Prove that the points (1, 1), (3, 4), (3, 8) and (1, 5) are the vertices of the parallelogram.
(b) Prove that the quadrilateral ABCD with vertices A(1, – 3), B(– 3, – 4), C(– 5, – 2) and
D(– 1, – 1) is a parallelogram.
7. (a) Show that the points A(2, – 2), B(8, 4), C(5, 7) and D(– 1, 1) are the vertices of a rectangle.
(b) Prove that the points (0, – 2), (– 2, 0), (1, 3) and (3, 1) form a rectangle.
8. (a) Prove that the points A(0, – 1) B(2, 1) C(0, 3 ) and D (– 2,1) are the vertices of a square.
(b) Show that the quadrilateral PQRS with vertices P(2, 0), Q(0, 2), R(– 2, 0) and S(0, – 2) is a square.
9. (a) Show that the four points (2, 3), (5, 8), (0, 5) and (–3, 0) are the vertices of a rhombus.
(b) Prove that the points (– 1, – 1), (– 3, 2), (– 5, – 1) and (– 3, – 4) form a rhombus.
10. (a) The two consecutive vertices of a parallelogram are (– 2, 2) and (4, 8). If the point of the
25, 7
intersection of the diagonals of the parallelogram is 2 , then find the coordinates of the
remaining two vertices.
(b) If three consecutive vertices of a parallelogram are (3, 1), (2, 2) and D(8, b) C(8, 4)
(– 2, 1), find the co-ordinates of the fourth vertex.
(c) From the adjoining rectangle ABCD, find the values of a and b.
11. (a) P and Q are the mid points of AC and BC of ∆ABC with vertices
A(– 2, 3), B(2, 1) and C(4, 7). Prove that: AB = 2PQ. A(a, 2) B(4, 4)
(b) Prove that the line segment joining the mid points of any two sides of a triangle with vertices
(1, 0), (4, –1) and (3, 3) is half of its remaining side. Also, show that their slopes are equal.
12. (a) Prove that the line 2x + 3y – 5 = 0 bisects the line segment joining the points (4, – 3) and (– 2, 5).
(b) Verify that the straight line x + 2y = 4 bisects the line segment joining the points (– 2, – 2) and (6, 4).
ANSWERS
2. (a) 1 65 units, 1 65 units; yes.
2 2
3. (a) 7 units (b) 6 2 units
2
10. (a) (7, 5), (1, – 1) (b) (– 1, 0) (c) 4, 2
122 Illustrated Optional Mathematics-9
41.41 EQUATION OF STRAIGHT LINE
4.4 (A) Basic on Straight Line
Learning Objectives
At the end of this topic, the students will be able to:
• find the slope and inclination of a straight line. • find the intercepts made by a line on the axes.
X
A straight line is the locus of linear equation. The general form of Y
linear equation that represents a straight line is ax + by + c = 0, where (–2, 0)
O
a, b and c are ordinary constants. For example; if the equation of the X'
locus of a moving point is 2x + 3y + 4 = 0, we see its nature by taking (1, –2)
a set of at least two points which satisfies the given equation. COORDINATE
GEOMETRY
i.e. x –2 –5 1 (–5, 2)
y 0 2 –2 Y'
Draw the graph of the equation 2x + 3y + 4 = 0 by plotting the above points in the table. The nature of the
equation 2x + 3y + 4 = 0 represents the locus of the moving point as a straight line. So, the linear equation
in x and y gives always a straight line.
To find the equation of straight line, on the basis of the given condition.
a. Inclination of straight Line YB Y
Study the following figures: B
Y
B
θ θθ
A
O Fig: (i) O AX O X
A
Fig: (ii) Fig: (iii)
Y YY
θB AB AB
OQ X X OX
Fig: (vi)
Fig: (iv) A O
Fig: (v) Equation of Straight Line 123
COORDINATE In all the figures above, the straight line AB makes an angle of θ with x-axis in different positions.
GEOMETRY The angle θ made by the straight line AB with x-axis in the anti clockwise direction is the inclination
of the line AB. It is always measured positively.
The angle made by a straight line with x-axis in the anti clockwise direction is called the
inclination (or angle of inclination) of the line. It is usually denoted by θ.
Note:
(1) If the straight line is parallel to the y-axis, then it should be perpendicular to the x-axis. So, the
angle of inclination of the line will be a right angle. In the figure (ii), θ = 90o.
(2) If the straight line is perpendicular to the y-axis, then it should be parallel to the x-axis. Here, two
cases arise.
(3) If the straight line and the x-axis are parallel in the positive direction, then the inclination of
the straight line will be 0o i.e., θ = 0o, in the figure (v).
(4) If the straight line is parallel with the x-axis in negative (or opposite ) direction, then the
inclination of the line will be 180o, i.e. θ = 180o in the figure (vi).
There are four types of lines on the basis of inclination: They are as follows:
(i) Horizontal Line: A line parallel to the x-axis or perpendicular to the y-axis, i.e., the angle of
inclination of a line is zero angle (or null angle) then the line is called a horizontal line.
(ii) Vertical line: A line perpendicular to the x-axis or parallel to the y-axis, i.e., the angle of
inclination of a line is right angle, the line is called a vertical line.
(iii) Oblique line: A line which is neither parallel to x-axis nor parallel to y-axis i.e., the angle of
inclination of a line is other than 0o and 90o, then the line is called an oblique.
(iv) Negative line: A line which is parallel to x-axis but in opposite (or negative) direction i.e., the
angle of inclination of a line is 180o is called a negative line.
b. Slope (or Gradient) of a Straight Line YA
If θ is the angle of inclination of the line AB then the tangent of angle θ
θ (i.e. tan θ) is called the slope (or gradient) of the line AB. It is usually OB
denoted by m. i.e., m = tan θ when the value of θ is greater, then the slop
of the line is gradually increased. In the adjoining figure, AC⊥OX, then
tan θ = AC i.e., m = tan θ = y-intercept (Rise) CX
BC x-intercept (Rise)
Slope of line using coordinates
Consider a line passing through the two points A(x1, y1) and B(x2, y2). Y B(x2, y2)
Draw AL and BM perpendicular to the x-axis and draw AN⊥BM. A(x1, y1) θN
Then OL = x1, OM = x2, AL = y1, BM = y2
∴ AN = LM = OM – OL = x2 – x1
BN = BM – MN = BM – AL = y2 – y1 X' θ X
OL M
If θ is the angle of inclination of AB when produced up to C of the
Y'
x-axis, then the slope of AB (m) = tan θ = BN = y2 – yx11.
AN x2 –
Note: The point A, B and C are collinear if slope of AB = slope of BC.
124 Illustrated Optional Mathematics-9
c. Intercept Made by Line on the Axes
In the adjoining figure, the straight line intersects the x-axis at A and the y-axis at B then,
(i) OA is called x-intercept or the intercept made by the Y
line on x-axis is called x-intercept. It is usually denoted B
by a. The x-intercept a is considered positive if it is
measured to the right of the origin and negative, if it is b
measured to the left of the origin. In the adjoining figure,
OA = a and OD = – a. So, the coordinates of A and D are O a X
A(a, 0) and D(– a, 0) respectively. A
(ii) OB is called y-intercept or the intercept made by the Y
line on y-axis is called y-intercept. It is usually denoted
by b or c. The y-intercept b is considered positive if it is B(0, b)
measured above the origin and negative, if it is measured
below the origin. In the adjoining figure, OB = b and X' b A(a, 0)
OC = – b. So, the coordinates of B and C are (0, b) and D(–a, 0) aX
(0, – b) respectively.
–a O
Note:
(1) A horizontal line has no x-intercept. In the figure, the line –b
C(0, –b)
l1 does not have x-intercept.
Y Y' l4 COORDINATE
(2) A vertical line has no y-intercept. In the figure, the line l2 l3 l2 GEOMETRY
does not make y-intercept but its has only x-intercept.
l1
(3) An oblique line which does not pass through the origin has
both x-intercept and y-intercept. In the figure, the line l3 OX
makes both intercepts.
(4) An oblique line, which passes through the origin, has
no intercept. In the figure, the line l4 does not make any
intercept.
Example 1 : Find the slopes of the lines AB and CD from the adjoining figure.
Solution : For the line AB: Y B
X
The running to the right of A belonging to B (a) = 4.C A
O
The rising up of A belonging to B (b) = 3.
Y'
Now, we have X' Q
The slope of AB (m) = Rising up (b) = 3
Running (a) 4
For the line CD: D
The running to the right of C belonging to D (a) = 6
The rising down of C belonging to D (b) = –7
Now, we have
Now, the slope of CD (m) = Rising up (b) = –7
Running (a) 6
Equation of Straight Line 125
Example 2 : Find the slope of straight line which makes an angle of 120o with Y
the x-axis in positive direction.
Solution: Here, 120o
The angle made by the straight line with x-axis (θ) = 120o. OX
Now, the slope of the line (m) = tan θ = tan 120o = – 3
Example 3 : Find the slope of the line passing through the origin and the point (2, 6).
Solution : Here, the line passes through the point (2, 6).
Then x–coordinate (x) = 2 and y coordinate (y) = 6
Now, the slope of the line (m) = y = 6 = 3.
x 2
Example 4 : Find the slope of a line joining the points A(3, –1) and B(–2, 4). Also find the angle of
inclination of the line.
Solution : Here, the line contains the points A(3, – 1) and B(– 2, 4). Y
Suppose A(3, – 1) = (x1, y1) and B(– 2, 4) = (x2, y2)
Now, we have y2 – y1 4+1 5 (–1, 4)
x2 – x1 –2–3 –5 O
The slope of the line (m) = = = = –1
COORDINATE Again, if θ is the angle of inclination of the line, then
GEOMETRY
m = tan θ X' θ
X
or, – 1 = tan θ
(3, –1)
or, tan 135o = tan θ Y'
∴ θ = 135o.
Hence, the angle of inclination of the line is 135o.
Example 5 : If the slope of the line passing through the points (2, k) and (3, – 4) is 3 then find the
value of k. 2
Solution: Here, (2, k) = (x1, y1) and (3, –4) = ( x2, y2)
Now, we know that
The slope of the line (m) = y2 – y1
or, 23 = –34––2k x2 – x1
or, 3(3 – 2) = 2(– 4 – k)
or, 3 = – 8 – 2k
or, 2k = – 8 – 3 Y A
∴ k = – 211. O B
Example 6 : Prove that the points A(4, 5), B(5, 1) and C(7, – 7) are collinear. Y'
C
Solution: Here, the given three points are A(4, 5) , B(5, 1) and C(7, –7). X'
Now, The slope of AB (m1) = 1 – 5 = –4 = – 4.
5 – 4 1
The slope of BC (m2) = –7–1 = –8 = – 4.
7–5 2
Here, m1 = m2. So, the given three points are collinear.
126 Illustrated Optional Mathematics-9
Example 7 : Find the slope of the line making the x-intercept 5 and y-intercept 3.
Solution : Here, the line makes the x-intercept 5 and the y-intercept 3. So, Y
the coordinates of the points intersected by the line on x-axis and (0, 3)
y-axis are (5, 0) and (0, 3) respectively.
Now, we have 3
The slope of the line (m) = y2 – y1 O θ (5, 0) X
x2 – x1 5
= 3 – 0 = – 3 .
“Alternatively” 0 – 5 5
Here, the line having the x-intercept 5 and y-intercept 3 makes an angle of θ with x-axis in
clockwise direction.
So, the slope of the line (m) = tan (– θ) = – tan θ = y-intercept = – 3 .
x-intercept 5
EXERCISE - 4.4 (A)
1. (a) Write the formula to find the slope of a straight line if y-intercept and x-intercept are given.
(b) Write the formula to find out the slope of straight line PQ, where P and Q are (x1, y1) and (x2, y2) COORDINATE
respectively. GEOMETRY
(c) What are x-intercept and y-intercept?
(d) What is the slope of the straight line parallel to x-axis?
(e) Write down the angle of inclination of the line parallel to y-axis.
2. Find the slope of the lines in the following figures.
(a) Y (b) Y (c) Y
B P M
X' O X X' O X
X' O X A
X
N
A Q Y'
Y' Y'
3. Find the slope of the straight line which makes the following angles with x-axis:
(a) 60o in positive direction, (b) 150o in negative direction.
4. Find the slope of the line AB in the following figures:
(a) Y (b) Y (c) Y
A
A 150o
30o X
O O X O
B 45o B
B
Equation of Straight Line 127
(d) Y (e) (f) Y A
Y
45o A B
A
O XO X
O X
B
B
5. Find the slope of the line segment joining the origin and the following points:
(a) (2, 2) (b) (– 2, 5) (c) (0, – 2) (d) (– 3, – 5)
6. Find the slope of the line joining the pair of the following points.
(a) (2, – 3) and (4, 1) (b) (– 1, 0) and (0, – 2)
7. Find the angle of inclination of a line joining the following points.
(a) (0, 2) and (3, – 1) (b) (2, 3) and (– 3, – 4)
COORDINATE 8. (a) If the slope of the line passing through the points (2, 3k) and (– 3, – 1) is 2, find the value of k.
GEOMETRY (b) If the slope of the line passing through the points (–1, 3) and (2a, a) is 1, find the value of a.
9. Prove that the following points are collinear:
(a) A(– 3, 0) B(– 1, – 2) and C(0, – 3) (b) P(1, 4) Q(5, 0) and R(7, 2)
10. Determine whether the pair of the following lines are parallel or not.
(a) (4, 2) and (0, 2); (0, – 3) and (2, – 1) (b) (1, 2) and (– 3, 6); (5, 0) and (2, 3)
11. Find the slope of the lines which makes the following intercepts on axes:
(a) x-intercept 3 and y-intercept – 4 (c) both intercepts are equal but in opposite signs.
(b) both intercepts are equal (d) x-intercept is twice as y-intercept
ANSWERS
2. (a) 6 (b) – 3 (c) 5 3. (a) 3 (b) – 1
5 2 7 3
(c) 1 (d) – 1
4. (a) 1 (b) – 1 3 (e) 0 (f) no slope
3 (b) – 2
(b) – 5 (c) no slope (d) 5 6. (a) 2
5. (a) 1 2 3
7. (a) 315o (b) tan–1 7 8. (a) 3 (b) – 4
5
10. (a) Not parallel (b) Parallel 11. (a) – 43 (b) 1 (c) – 1 (d) 1
2
128 Illustrated Optional Mathematics-9
4.4 (B) Equations of Straight Lines Parallel to Axes
Learning Objectives
At the end of this topic, the students will be able to:
• find the equations of the straight lines parallel to the axes.
a. Equation of Straight Line Parallel to x-axis Y
Let AB be a straight line parallel to the x-axis and intersects the y-axis at C P(x, y) B
C. Then OC = b units. Take P(x, y) on the line AC and draw PQ ⊥ OX. So, A b
PQ = y since OA = PQ. Therefore, y = b, which is the required equation OQX
of the straight line parallel to the x-axis (or perpendicular to the y-axis).
YB
b. Equation of Straight Line Parallel to y-axis
Q P(x, y)
Let AB be a straight line parallel to the y-axis which intersects the x-axis
at the point C. Then OC = a units. Take a point B(x, y) on the line AB and a C X COORDINATE
draw PQ ⊥ OY. So, PQ = x since OC = PQ. Therefore, x = a, which is A GEOMETRY
required equation of the straight line parallel to y-axis (or perpendicular to O
x-axis).
Example 1 : Find the equation of the line parallel to the x-axis at Y
the distance of 2 units above it. y=2
Solution : Here, the y-intercept (b) = 2.
The equation of the line parallel to the x-axis is y = b OX
Y
or, y = 2.
Example 2 : Find the equation of the straight line parallel to the
y-axis and passing through the point (– 2, – 3).
Solution : Let the equation of the straight line parallel to the y-axis be X' O X
x = a ….......... (i)
Since, the line passes through the point (– 2, – 3). (–2, –3)
Y
So, the x-coordinate (a) = – 2.
The required equation of the line is x = – 2. Y'
B
or, x + 2 = 0.
P(2, 2)
Example 3 : What is the equation of the straight line passing through
(2, – 1), (2, 2) and (2, – 5) ?
Solution : The straight line passes through (2, – 1), (2, 2) and (2, – 5). OX
P(2, –1)
Here, the x-coordinate of each point is 2.
So, the equation of the line having x-coordinate 2 is x = 2. P(2, –5)
Equation of Straight Line 129
Example 4 : In the adjoining figure, ΔABC is an equilateral triangle, Y A
where BC is parallel to x-axis. Find the slope of its sides. 60o
Solution : Here all the angles of an equilateral triangle are 60o.
i.e., ∠ABC = ∠BCA = ∠CAB = 60o. B 60o 60o C
60o 60o
For BC: BC is parallel to x-axis. X
O
Then the angle of inclination of BC (θ) = 0O
∴ The slope of BC (cm) = tan 0o = 0
For AB: The angle of inclination of AB(θ) = 60o ∴ The slope of AB(m) = tan 60o = 3
For AC: The angle of inclination of AC(θ) = 120o ∴ The slope of AC (m) = tan 120o = – 3
EXERCISE - 4.4 (B)
1. Write the equation of straight line parallel to: (a) x-axis (b) y-axis
2. Write down the equation of the line for the following conditions:
(a) Parallel to x-axis at the distance of 3 units below it.
(b) Parallel to x-axis at the distance of 5 units up it.
COORDINATE (c) Parallel to y-axis at the distance of 2 units to the right.
GEOMETRY
(d) Parallel to y-axis at the distance of 4 units to the left.
3. What is the equation of the line for the following conditions?
(a) Parallel to x-axis and passing through the point (–1, –3).
(b) Parallel to x-axis and passing through the point (2, 3).
(c) Parallel to y-axis and passing through the point (2, –4).
(d) Parallel to y-axis and passing through the point (–2, 5).
4. Find the equation of the straight line,
(a) which makes an intercept 4 along positive x-axis and parallel to y-axis.
(b) which makes an intercept 3 along positive y-axis and parallel to x-axis.
(c) which makes an intercept 2 along negative x-axis and parallel to y-axis.
(d) which makes an intercept 5 along negative y-axis and parallel to x-axis.
5. Find the equation of the straight line passing through the following points:
(a) (3, –1), (3, 0) and (3, 5) (b) (3, –1), (2, –1) and (– 4, –1)
6. In the given figure, AB is parallel to the y-axis , find the slope AC and BC.
(a) Y C (b) Y C
45o 45o 30o
A B A B
O X C
30o
O X
(c) Y A (d) Y
B
60o X A B
O X
C
O
130 Illustrated Optional Mathematics-9
7. Find the slopes of the sides of the following quadrilateral from the following figures.
(a) Y (b) Y D
DC AC
A B B
O X OX
8. (a) If the points (a, 0), (0, b) and (1, 1) are collinear, show that 1 + 1 = 1.
a b
(b) If the points (p, 0), (0, q) and (4, 4) are collinear, prove that 1 + 1 = 41.
p q
x x
(c) If (x, y) is any point on a straight line which passes through (a, 0) and (0, b), show that a + b = 1.
ANSWERS
1. (a) y = 0 (b) x = 0 2. (a) y + 3 = 0 (b) y – 5 = 0
3. (a) y + 3 = 0 (b) y – 3 = 0
(c) x – 2 = 0 (d) x + 4 = 0 (b) y – 3 = 0 COORDINATE
(b) y + 1 = 0 GEOMETRY
(c) x – 2 = 0 (d) x + 2 = 0 4. (a) x – 4 = 0
(d) 2 + 3, –2 – 3
(c) x + 2 = 0 (d) y + 5 = 0 5. (a) x – 3 = 0
6. (a) 1, – 1 (b) no slope, – 3 (c) no slope, 1
3
7. (a) 0, no slope, 0, no slope (b) –1, 1, –1, 1
4.4 (C) Equation of Straight Line in Slope-intercept Form
Learning Objectives
At the end of this topic, the students will be able to:
• find the equation of the straight line in slope–intercept form.
Three Standard Forms of Equation of Straight Line
The general form of the equation of a straight line is ax + by + c = 0 where a, b and c are arbitrary constants.
The straight line ax + by + c = 0 either has an angle of inclination with x-axis and makes y-intercept or
makes the intercepts on both axes or makes an angle with the x-axis by the perpendicular drawn from the
origin to it shown in the figures below:
YY Y
B B B
c b p
α
O θo θo
A O
X O aA X A X
Slope – intercept form Double intercepts form Perpendicular or Normal form
Equation of Straight Line 131
Equation of a Straight Line in Slope-intercept Form (y = mx + c)
To find the equation of a straight line having the slope (m) and cutting the y-intercept (c).
Let AB be a straight line which makes a positive angle θ with x-axis and cuts the y-axis at Q.
Take a point P(x, y) on the line AB and draw PM ⊥ OX and QR ⊥ PM. Then ∠ABX = ∠AQR = θ and
OQ = MR = c, Y A
OM = QR = x and PM = y. Qθ P(x, y)
Therefore, PR = PM – MR = y – c R
Now,
In right-angled triangle ΔPQR, we have X' θ c X
The slope of the line AB (m) = tanθ B O
M
or, m = QPRR or, m = y – c or, mx = y – c
x
or, y = mx + c, which is the required equation of the straight line AB in slope–intercept form.
'' Alternative Method ''
COORDINATE Suppose a straight line AB makes an angle θ with x-axis and cuts the y-axis at θ such that OQ = C. So, the
GEOMETRY
the coordinate of the point Q is (0, c) and ∠ QBO = θ. Take a point P(x, y) on the line AB. Then the points
B, Q, P and A lie on the same line AB. So, they are collinear, Y
Now, we have A
P(x, y)
The slope of AB = The slope of QP
or, m= y – c [∴ m = y2 – y1 ] Q (0, c)
x – 0 x2 – x1
c
or, mx = y – c or, y = mx + c. X' θ X
B
This is the required equation of the line in the slope– intercept form. O
Special Cases
Case I : If c = 0 that means the line will pass through the origin then the equation of the line will be
y = mx + 0 or, y = mx.
Case II : If m = 0 that means the line parallel to x = axis, then the equation of the line will be
y = 0 × x + c or, y = c.
Example 1 : Find the equation of the line that makes an angle of 60o with x-axis and cuts of
y-intercept equal to 3 units. Y
Solution : Here, y-intercept (C) = 3 units slope of the line (m) = tan 60o = 3
Now, we have B
The equation of the line is y = mx + c. 3 unit
or, y = 3x + 3 X' 60o X
or, 3x – y + 3 = 0 AO
Y'
132 Illustrated Optional Mathematics-9
Example 2 : Find the equation of the straight line PQ which cuts the y-axis at the point (0, 4) and
makes an angle of 45o with x-axis in clockwise direction. Y
Solution: Here,
The y-intercept cut by the straight line AB B(0, 4)
at the point (0, 4), (c) = 4 units.
4 units
The slope of the line (m) = tan (180o – 45o) = tan 135o = –1. 45o
OA
Now, we have X
The equation of the line AB is y = mx + c.
or, y = –1x + 4
or, x + y = 4.
Example 3 : Find the equation of the line cut with equal intercepts on positive axes and passing
through the point (–2, 5).
Solution : The line which cuts with equal intercepts, makes the angle of 45o with x-axis and the
y-intercept c.
The slope of the line AB (m) = tan(180o – 45o) = tan135o = –1 Y
∴ The equation of the line AB is y = mx + c. (–2, 5) B A X COORDINATE
or, y = –x + c ………(i) a GEOMETRY
The equation contains the point (–2, 5). Oa
So, 5 = + 2 + c,
or, c = 3
Putting the value of c in equation (i), we get
y=–x+3
or, x + y = 3, which is required equation of the line.
EXERCISE - 4.4 (C)
1. (a) What is the equation of straight line in slope-intercept form?
(b) Write the equation of the line passing through the origin in slope-intercept form.
(c) What is the equation of the line parallel to x-axis?
2. Find the equation of the straight line under the following conditions:
(a) The slope (m) = 1 and the y-intercept (c) = 2
3
(b) The slope (m) = 3 and the y-intercept (c) = –3
2
3. Find the equation of the straight line under the following conditions:
(a) The slope (m) = tan 135o and the y-intercept (c) = –3
(b) The slope (m) = tan 150o and the y-intercept (c) = 2
Equation of Straight Line 133
4. Find the equation of the line AB from the given figures. AY
(a) Y (b) Y (c)
A
X' (–3, –2) O X
A
X' O (0, 4)
2 units X OX
150o X' 45o
30oB Y'B B
Y' Y'
COORDINATE
GEOMETRY
5. Find the equation of the straight line under the following conditions:
(a) The angle with x-axis (θ) = –60o and the y-intercept (c) = 5
(b) The angle of inclination (θ) = 150o and the y-intercept (c) = – 4
(c) The angle with positive y-axis (θ) = 60o and the y-intercept (c) = 7
(d) The angle with negative y-axis (θ) = 45o and the y-intercept (c) = – 3
6. Find the equation of the straight lines passing through the origin and making the following angles
with x-axis:
(a) 60o (b) 120o (c) 150o (d) 0o
7. Find the equation of the straight lines for the following conditions:
(a) Parallel to x-axis and y-intercept (c) = 4 (b) Parallel to x-axis and y-intercept = – 2
8. Find the equation of the straight lines for the following conditions:
(a) Having slope 1 and passing through (2, 5).
3
(b) Making an angle of 30o with x-axis and passing through (4, – 3).
9. (a) Find the equation of a straight line passing through the point (2, – 1) and having the y-intercept 3 units.
(b) Find the equation of the straight line having the y-intercept –2 units passing through the point (4, 4).
10. (a) Find the equation of the straight line that cuts equal intercepts on the positive axes and passes
through the point (2, 3).
(b) Find the equation of the straight line passing through the point (1, 2) and cutting equal intercepts
on the axes but in opposite sign.
11. (a) A straight line makes an angle of 30o with x-axis and passes through (0, 5). Prove that the line
also passes through the point (–3 3, 2).
(b) A straight line makes 60o with positive y-axis in anticlockwise direction and passes through the
point ( 3, 0). Prove that the line also passes through the point (– 3, 2).
ANSWERS
2. (a) x – 3y + 2 3 = 0 (b) 3x – 2y – 6 = 0 3. (a) x + y + 3 = 0 (b) x + 3y – 2 3 = 0
4. (a) x – 3y – 2 3 = 0 (b) x – y + 4 = 0 (c) 3x + y + 2 + 3 3 = 0
5. (a) 3x + y – 5 = 0 (b) x + 3y + 4 3 = 0
(c) x – 3y + 7 3 = 0 or, x + 3y – 7 3 = 0 (d) x + y + 3 = 0 or, x – y – 3 = 0
6. (a) 3x – y = 0 (b) 3x + y = 0 (c) x + 3y = 0 d) y = 0
7. (a) y – 4 = 0 (b) y + 2 = 0 8. (a) x – 3y + 13 = 0 (b) x – 3y – 3 3 – 4 = 0
9. (a) 2x + y – 3 = 0 b) 3x – 2y – 4 = 0 10. (a) x + y – 5 = 0 (b) x – y + 1 = 0
134 Illustrated Optional Mathematics-9
4.4 (D) Equation of Straight Line in Double-intercept Form
Learning Objectives
At the end of this topic, the students will be able to:
• find the equation of a straight line in double-intercept form.
Equation of a Straight Line in Double-intercept Form ( x + y = 1)
a b
To find the equation of a straight line which makes the intercepts on the axes.
Suppose a straight line AB cuts the x-axis at A and the y-axis at B such that OA = a and OB = b. Take a
point P(x, y) on the line AB and draw PQ ⊥ OX. Then OQ = x, PQ = y Y
∴ QA = OA – OQ = a – x. B
Now, in the similar ∠OAB and ∠QAP, OA = OB P(x, y)
QA PQ b
a b
or, a – x = y AX
or, ay = ab – bx O Qa COORDINATE
GEOMETRY
or, ay + bx = ab
or, bx + ay = 1
ab ab
or, x + y = 1, which is required equation of the line in the double-intercept form.
a b
“ Alternative Method”
Suppose a straight line AB cuts the x-axis at A and y-axis at B such that OA = a and OB = b then the co–
ordinate of A and B are (a, 0) and (0, b) respectively. Take a point P(x, y) on the line AB. So, the points A,
P and B are collinear. Then slope of AP = Slope of PB Y
or, y – 0 = b – y B (0, b)
x – a 0 – x
or, – xy = bx – xy – ab + ay P(x, y)
b
or, bx + ay = ab
A(a, 0)
bx ay X
or, ab + ab = 1 Oa
or, x + y = 1
a b
This is the required equation of the straight line in double-intercept form.
Special Cases
Case I : If the line parallel to x-axis, then the x-intercept (a) is an undefined i.e. a = ∞. So the equation
of the line becomes x + y = 1, or, 0 + y = 1 or, y = b.
∞ b b
Case II : If the line parallel to y-axis then the intercept (b) is undefined i.e., b = ∞. The equation of the
line becomes x + y = 1, or, x + 0 = 1 or, x = a.
a ∞ a
Equation of Straight Line 135
Example 1 : If a line cuts the x-axis and the y-axis at the points A(– 2, 0) and B(0, 3), find the equation of
line AB.
Solution: Here, Y
The line AB cuts the x-axis at A(–2, 0) and the y-axis at B(0, 3).
So the x-intercept = –2 the y-intercept (b) = 3. B(0, 3)
Now, we know that
The equation of the line AB is x + y = 1
a b
or, –x2 + 3y = 1 θ
X' X
A(–2, 0) O
or, – 3x6+ 2y = 1 Y'
or, – 3x + 2y = 6
COORDINATE or, 3x – 2y + 6 = 0
GEOMETRY
“Alternatively”
The slope of the line AB passing through the points A(–2, 0) and B(0, 3) is
(m) = y2 – y1 = 0 3 –0 = 3
x2 – x1 – (– 2) 2
The y-intercept cut by the line at (0, 3) (c) = 3 units.
Now, the equation of the line AB is y = mx + c.
or, y = 3 x + 3
2
or, 2y = 3x + 6
or, 3x – 2y + 6 = 0.
Example 2 : Find the equation of the line cut with equal intercepts on positive axes and passing
through the point (– 2, 5).
Solution : Suppose the line AB passing through the point (–2, 5) that cuts off equal intercepts 'a' on
both axes. Y
So, its equation is x + y = 1 x + y = 1
a a a b
(–2, 5)
x + y = a ..... (i) B
The equation (i) contains the point (–2, 5). a
So, (– 2) + 5 = a
O A X
a
or, a = 3
Putting the value of a in the equation (i), we get
3x + y3 = 1
or, x + y = 3, which is required equation of the line.
136 Illustrated Optional Mathematics-9
Example 3 : The point (2, 4) bisects the line segment intercepted between the positive axes. Find the
equation of the line.
Solution : Let AB be a line segment intercepted between the positive axes such that OA = a and
OB = b then the coordinates of A and B are (a, 0) and (0, b) respectively. Here the point
(2, 4) is the mid-point of AB, then by using mid-point formula, we get
(2, 4) = a + 0, 0 + b Y
2 2
or, (2, 4) = 2a, b B(0, b)
2
∴ a = 2 and b = 4 b (2, 4)
2 2
a = 4 and b = 8 O a A(a, 0) X
Now, the equation of the line AB is x + y =1
or, 4x + 8y = 1 a b or,
2x + y = 8.
Example 4 : Find the equation of a straight line which passes through the point (8, 2) and cuts an
intercept on the x-axis twice as long as the intercept on the y-axis in magnitude but in
the opposite sign. Y COORDINATE
GEOMETRY
Solution: Let, y-intercept (b) = – k then x-intercept (a) = 2k (8, 2)
X
Now, we have X' O 2k
–k
The equation of the line is x + y = 1 A
or, 2xk + –yk = 1 a b
B
or, x – 2y = 2k ……..(i) Y'
The equation contains the point (8, 2). Then,
8 – 2 × 2 = 2k or, k = 2
Putting the value of k in equation (i), we get
x – 2y = 2 × 2
or, x – 2y – 4 = 0, which is the required equation of the line.
Example 5 : Find the equation of a straight line passing through the point (1, 8) such that the sum
of intercepts on the axes is 15.
Solution: Suppose the x-intercept = a and y-inetercept = b. Then a + b = 15 or, b = 15 – a
Now, we have
The equation of the straight line in double intercept form is x + y = 1. Y
a b
or, xa + 15y– a = 1 ...............
(i)
Since the line (i) passes through the line (1, 8) so, (1, 8)
b
1a + 158– a = 1
or, 15 – a + 8a = 15a – a2 or, a(a – 5) – 3(a – 5) = 0
or, a2 – 8a + 15 = 0 or, (a – 5) (a – 3) = 0 Oa X
or, a2 – 5a – 3a + 15 = 0 ∴ a = 3 or 5.
Equation of Straight Line 137
When a = 3, b = 15 – 3 = 12. Therefore, the eqn of the line is x + y = 1.
3 12
or, 4x + y = 12.
When a = 5, b = 15 – 5 = 10. Therefore, the eqn of the line is x + y = 1.
5 10
or, 2x + y = 10.
EXERCISE - 4.4 (D)
1. (a) Write down the equation of a straight line in double-intercept form.
(b) What is the equation of the straight line parallel to y-axis?
2. Find the equation of a straight line for the following intercepts on the axes:
(a) x-intercept = 3 and y-intercept = 5
(b) x-intercept = – 4 and y-intercept = 7
3. (a) If a straight line cuts the x-axis and y-axis at the points (2, 0) and (0, – 4), find the equation of the line.
(b) Find the equation of the line passing through the points (– 3, 0) and (0, – 2)
COORDINATE 4. (a) Find the equation of the straight line which makes intercepts 3 and – 4 on the axes. Does the line
GEOMETRY pass through the point (1, 2)?
(b) Find the equation of the straight line which cuts intercepts 5 and 7 on the axes from the origin.
Does it pass through the point (–5, 14)?
(c) Find the equation of the straight line passing through the points (3, 0) and (0, 4) and prove that
the line also passes the point (6, – 4).
5. (a) Find the equation of a straight line passing through the point (0, 4) and having the x-intercept 3.
(b) Find the equation of a straight line passing through the point (– 2, 0) and having the y-intercept – 5.
6. (a) Find the equation of the straight line that cuts equal intercepts on the positive axes and passes
through the point (3, 4).
(b) Find the equation of the straight line passing through the point (– 2, 3) and cutting equal
intercepts on the axes but in opposite sign.
7. (a) The point (2, 5) bisects the line segment intercepted between the axes. Find the equation of the line.
(b) The point (– 2, 4) bisects the line segment intercepted between the axes. Find the equation of the line.
8. (a) The point P(2, 3) divides the line segment AB intercepted between the axes at A and B
respectively in the ratio 1:2. Find the equation of the line AB.
(b) The point P(– 2, 2) divides the line segment AB intercepted between the axes at A and B
respectively such that AP:PB = 3:4. Find the equation of the line AB.
(c) Find the equation of the line AB from the given figures:
(i) Y (ii) Y
A
(–3, 2) 3 a b2 (p, q)
3
X' 2 X X' Oa X
bO
B
Y' Y'
138 Illustrated Optional Mathematics-9
9. (a) Find the equation of the line which passes through the point (– 3, 1) and cuts an intercept on COORDINATE
the x-axis twice as long as the intercepted on the y-axis in the magnitude. GEOMETRY
(b) A straight line cuts an intercept on the y-axis twice as long as the intercept on the x-axis in magnitude
but in opposite sign which passes through the point (10, – 2). Find the equation of the line.
10. (a) Find the equation of the straight line that passes through the point (2, 1) and the sum of whose
intercept on the axes is 6.
(b) Find the equation of the straight line passing through the point (4, – 1) such that intercept on
the x-axis exceeds the intercept on the y-axis by 3.
11. (a) Find the equation of the straight line passing through the point of intersection of the lines 2x –
y + 4 = 0 and x + y + 2 = 0 and cut with the equal intercept on the axes.
(b) A straight line passing through the point of intersection of the lines 4x + 2y – 1 = 0 and x + y = 4
cuts the intercept on the x-axis thrice as long as the intercept on the y-axis. Find the equation of
the line.
(c) Find the equation of a straight line in which the reciprocal of whose intercepts on the axes are
m and n respectively.
ANSWERS
2. (a) 5x + 3y – 15 = 0 (b) 7x – 4y + 28 = 0 3. (a) 2x – y – 4 = 0 (b) 2x + 3y + 6 = 0
4. (a) 4x – 3y – 12 = 0, No (b) 7x + 5y – 35 = 0, Yes c) 4x + 3y = 12
5. (a) 4x + 3y – 12 = 0 (b) 5x + 2y + 10 = 0 6. (a) x + y – 7 = 0 (b) x – y + 5 = 0
7. (a) 5x + 2y – 20 = 0 (b) 2x + y – 8 = 0 8. (a) 3x + 4y – 18 = 0 (b) 2x + 3y – 7 = 0
(c) (i) x + y – 5 = 0 (ii) 2qx + 3py – 5pq = 0
9. (a) x + 2y + 1 = 0 (b) 2x – y – 22 = 0
10. (a) x + y – 3 = 0, x + 2y – 4 = 0 (b) 2x + 5y – 10 = 0, x + 2y – 6 = 0
11. (a) x + y + 2 = 0 (b) x + 3y – 19 = 0 (c) mx + ny = 1
Project Work
Take a square-shaped graph and draw axes having equal quadrants. Draw the seven line-segments in
each quadrant such that the y-intercept is double the x-axis and colour with different colours.
Equation of Straight Line 139
4.4 (E) Equation of Straight Line in Normal (Perpendicular) Form
Learning Objectives
At the end of this topic, the students will be able to:
• find the equation of a straight line in normal (perpendicular) form.
Equation of the Straight Line in Perpendicular (or normal) Form
To find the equation of a straight line in which the perpendiculars (p) drawn from the origin to the line
makes an angle (α):
Let AB be a straight line which cuts the x-axis at A and B respectively. Draw the perpendicular line OQ
drawn from the origin to the line AB such that OQ = P makes an angle α with the x-axis, i.e., ∠AOQ = α.
Take a point P(x,y) on the line AB and draw PM ⊥ OX. Then OM = x and PM = y.
In similar right-angled ΔAOQ and ΔAPM, ∠AOQ = ∠APM = α. Y
B
Now, in right-angled ΔPAM,
tan α = AM
PM
COORDINATE or, MA = y tan α [ PM = y] Q
GEOMETRY P(x, y)
Again, in right-angled ΔAOQ p α
cos α = OQ or, cos α x + y sin α = p α MA X
OA cos α O
or, cos α = OM p MA or, cos α x cos α + y sin α =p
+ cos α
or, cos α = x + p α or, x cos α + y sin α = p where, cos α ≠ 0.
y tan
This is the required equation of the straight line in perpendicular (or normal) form.
Note: (1) The perpendicular line drawn from the origin to the line is always positive.
(2) The value of α will be from 0o to 360o.
Case I : If the angle α is equal to 0o, the equation of the line becomes x cos 0o + y sin 0o = p
Special x = p, which is parallel to the y-axis.
Cases Case II : If the angles equal to 90o, the equation of the line becomes x cos 90o + y sin 90o = p
or, y = p, which is parallel to the x-axis.
Example 1 : The perpendicular distance from the straight line is 5 units and the angle made by the
perpendicular with the x-axis is 120oF, find the equation of the line. Y
Solution: Here the length of perpendicular (p) = 5 units.
The angle made by the perpendicular with x-axis (α) = 120o
Now, the equation of the line is, 120o
5
x cos 120o + y sin 120o = 5 [ x cos α + y sin α = P]
X' OX
or, x – 1 +y 3
2 2 = 5 or, –x + 3 y = 10
or, x – 3 y + 10 = 0, which is the required equation of the line.
140 Illustrated Optional Mathematics-9
Example 2 : Find the equation of a straight line at a distance of 5 units from Y
the origin so that the slope of the perpendicular is 43.
Solution: Here, the perpendicular distance from the origin to
the line (p) = 5 units. 180o + α 3 4
α
3 p X' X
tan α = 4 = b 3O 5
α5
∴ h = p2 + b2 = 16 + 9 = 25 = 5 units.
4
Since the value of tan α is positive.
So, α lies other first quadrate or third quadrate. Y'
For the first quadrant,
sin α = p = 4 and cos α = b = 3
h 5 h 5
∴ The eqn of the line in normal form is x cos α + y sin α = p
or, x . 3 + y . 4 = 5 or, 3x + 4y = 25
5 5
For the third quadrant,
We have, sin α = – 4 and cos α = – 3 COORDINATE
5 5 GEOMETRY
3 4
∴ The eqn of the line in normal form is x – 5 + y – 5 =5
or, – 3x – 4y = 25 or, 3x + 4y = – 25
Hence, the required equation of the line is 3x + 4y = ± 25
Example 3 : The perpendicular distance from the origin to a straight line is 2 units. If the line
passes through the point (3, 0), find its equation. Y
Solution: Here,
The perpendicular distance from the origin to the line (p) = 2 units.
Let α be the angle made by the perpendicular to the x-axis, 2
then the equation of the line is x cos α + y sin α = p
or, x cos α + y sin α = p X' α X
O
α (3, 0)
2
or, x cos α + y sin α = 3 ............ (i)
Since, the line (i) passes through the point (3, 0).
So, 3cos α + 0 . sin α = 2 Y'
or, cos α = 2
3
4 5
∴ sin α = ± 1 – cos2 α = ± 1 – 9 = ± 3
Putting the values of sin α and cos α in eqn (i), we get
or, x 2 5
3 +y ± 3 =2
or, 2x ± 5 y = 6.
Hence, the required equation of the line 2x + 5 y = 6
or, 2x – 5y = 6.
Equation of Straight Line 141
EXERCISE - 4.4 (E)
1. Find the equation of the straight line for the given values of p and α:
(a) p = 4 units, α = 30o (b) p = 7 2 units, α = 135o (c) p = 5 units, α = 90o
2. (a) The perpendicular distance drawn from the origin to a straight line is 6 units. The perpendicular
makes 45o with the x-axis. Find the equation of the straight line.
(b) The length of the perpendicular from the origin on a straight line is 3 units and the perpendicular
line is inclined to the x-axis at 60o. Find the equation of the line and prove that the line passes
through the point (3, 3).
3. Find the equation of the line AB from the given figures.
(a) Y (b) Y
A
X' 4O
60o X
135o 2 X
X' O
COORDINATE Y' Y'
GEOMETRY
4. (a) If the perpendicular distance of a straight line from the origin is 6 units and the slope of the
perpendicular is 3, find the equation of the line.
(b) The length of perpendicular drawn from the origin to a straight line is 5 units and the slope of
the perpendicular is –1. Find the equation of the line.
5. (a) The length of perpendicular distance from the origin to a straight line is 4 units and the slope
of the perpendicular line is 3 with x-axis, find the equation of the line.
(b) The perpendicular distance from the origin to a straight line having the slope –1 is 7 units. Find
the equation of the line.
6. (a) Find the equation of a straight line such that the perpendicular distance from the origin to the
straight line is 5 2 units and the angle made by the perpendicular with x-axis is tan–1 ( 3).
(b) Find the equation of a straight line at a distance of 3 units from the origin such that the
7. (a) perpendicular from the origin to the line makes an angle of c52otu–n1 it1s85frowmitthhepoosriitgivine x-axis.
Find the equation of a straight line which is at a distance of and passes
through the point (0, 4).
(b) The perpendicular distance from the origin to a line is 2 3 units and line passes through the
point (–3, 0). Find the equation of the line.
ANSWERS
1. (a) 3 x + y – 8 = 0 (b) x – y + 14 = 0 (c) y – 5 = 0
2. (a) x + y – 6 2 = 0 (b) x + 3 y = 6 3. (a) x + y – 2 2 = 0 (b) 3 x + y – 8 = 0
(b) x + y – 7 2 = 0
4. (a) x + 3 y – 12 = 0 (b) x – y + 5 2 = 0 5. (a) x + 3 y – 8 = 0 (b) 2x – y + 10 = 0
6. (a) x + 3 y – 10 2 = 0 (b) 15x + 8y – 51 = 0 7. (a) 39x + 5y – 20 = 0
142 Illustrated Optional Mathematics-9
4.4 (F) Reduction of Given Equation into Standard Forms
Learning Objectives
At the end of this topic, the students will be able to:
• reduce the given equation ax + by + c = 0 in standard forms.
Using the different conditions, we obtained different forms of equations of a straight line. The equation
ax + by + c = 0 is the first degree in x and y where a, b and c are constants, represents a straight line. It is
called general equation of the first degree in x and y. So, it is called linear equation. We can reduce this
general equation into three different forms as given below:
a. Reduction to Slope-intercept Form
We have,
The equation of a straight line is ax + by + c = 0
or, by = – ax – c
or, y = – a x – c .............. (i)
b b
Comparing the equation (i) with y = mx + c, we get COORDINATE
GEOMETRY
m = – a and c = – bc, b ≠ 0
b
i.e., slope of the line (m) = – a = – coefficient of x and y-intercept (c) = – c = – constant y
b coefficient of y b coefficient of
Thus, ax + by + c = 0 can be reduced to y = – a x – c , b ≠ 0.
b b
b. Reduction to Double-intercept Form
We have, the equation of a straight line is ax + by + c = 0
or, ax + by = –c
or, a x + b y = – c
–c –c – c
or, x + y = 1 .................. (i)
– c – c
a b
x y
Comparing the equation (i) with a + b = 1, we get
x-intercept (A) = – c and y-intercept (B) = – c
a b
Thus, ax + by + c = 0 can be reduced to x + y = 1.
– c – c
a b
c. Reduction to Perpendicular (or Normal) Form
We have,
The equation of straight line is ax + by + c = 0 ……………. (i)
Multiplying the equation (i) by k, we get
k (ax + by + c) = 0 ……………. (ii)
Equation of Straight Line 143
The equation of straight line in perpendicular form is xcos α + ysin α = P
or, xcos α + ysin α – P = 0 …………….. (iii)
The equation (ii) and (iii) are identical. Then,
x cos α + ysin α – P = k(ax + by + c)
or, x cos α + ysin α – P = kax + kby + kc
∴ cos α = ka ……………. (iv)
sin α = kb ……………. (v)
– p = kc …………… (vi)
Squaring and adding equation (iv) and (v), we get
or, cos2 α + sin2 α = k2 a2 + k2 b2
or, 1 = k2 (a2 + b2)
or, k2 = a2 1 b2 ∴ k=± 1
+ a2 + b2
COORDINATE Substituting the value of k in equation (iv), (v) and (vi), we get
GEOMETRY
∴ cos α = ± a b2 , sin α = ± b b2 , – p = ± c b2 i.e., p = ± c
a2 + a2 + a2 + a2 + b2
Thus, the equation ax + by + c = 0 is reduced to ± a x + ± b + b2 y + ± c =0
a2 + b2 a2 a2 + b2
Note: (1) The perpendicular distance should be positive. So, take the + sign that makes the perpendicular distance
into positive from the expression mentioned above.
(2) (i) If the value of sin α, cos α and tan α are all positive, then α lies in the first quadrant.
(ii) If the value for sin α is positive and cos α and tan α are negative, then α lies in the second quadrant.
(iii) If the value for tan α is positive and sin α and cos α are negative, then α lies in the third quadrant.
(iv) If the value for cos α is positive and sin α and tan α are negative, then α lies in the fourth quadrant.
Example 1 : Reduce the equation 3x – 2y + 1 = 0 to three standard forms of equation of a straight
line.
Solution : Given,
The equation of the straight line is 3x – 2y + 1 = 0 ………(i)
Now,
(i) Reduction to the slope intercept form:
3x – 2y + 1 = 0
or, – 2y = – 3x – 1
or, y = – 3x – 1
–2
3 21,
or, y = 2 x + which is the required form of the equation in y = mx + c,
where slope (m) = 3 and y-intercept (c) = 12.
2
144 Illustrated Optional Mathematics-9
(ii) Reduction to the double intercept form:
3x – 2y + 1 = 0
or, 3x – 2y = –1
or, –31 x – –21y = 1 [ Dividing by –1 on both the sides] x y
worh, ere x–-xi31nt+ery21ce=pt1(,aw) h=ic–h31isatnhdeyr-eiqnuteirrecdepetq(uba)ti=on12.in the form of a b
+ = 1,
(iii) Reduction to the perpendicular (or normal) form:
3x – 2y + 1 = 0
or, 3x – 2y = –1
∴ a2 + b2
= 32 + (– 2)2
= 9+4 COORDINATE
GEOMETRY
= 13
Dividing by 13 on both the sides, we get
3 x – 2 y = – 1
13 13 13
or, – 3 x + 2 y = 1
13 13 13
x cos 146.31o + y sin 146.31o = 1 , which is the required equation of the line in the
13
perpendicular form where the perpendicular distance from the origin to the line (p) = 1
units and the angle made by perpendicular line to the x-axis (α) = 146.31o. 13
Example 2 : Find the area of the triangle formed by the line 3x + 4y – 12 = 0 with the positive axes
of the coordinates. Y
Solution : Given, the equation of the line is 3x + 4y – 12 = 0
or, 3x + 4y = 12 3x + 4y – 12 = 0
or, 31x2 + 142y = 1 b=3
or, x4 + y3 = 1 ................... (i) X' a=4 X
O
Comparing the equation (i) with x + y = 1, we get Y'
a b
x-intercept (a) = 4 and y-intercept (b) = 3.
Now, the area of the triangle formed by the given line with the axes of coordinates is
1 ab = 1 × 4 × 3 = 6 square units.
2 2
Equation of Straight Line 145
xy 11 1
Example 3 : Reduce a + b = 1 into perpendicular form and prove that a2 + b2 = p2 .
x y
Solution : Here, the given equation is a + b = 1 ................ (i)
or, bx + ay = 1
ab
or, bx + ay = ab
Now,
Dividing equation (ii) by b2 + a2 on both sides, we get
b2b+ a2 x + a a2 y = ab .................. (ii)
b2 + b2 + a2
Comparing the equation (iii) with x cos α + y sin α = p, we get
p= ab
b2 + a2
Squaring on the both sides, we get
COORDINATE p2 = a2 b2
GEOMETRY b2 + a2
or, p12 = ba22+ba22
or, p12 = ab2b22 + a2
a2b2
or, p12 = a12 + 1
b2
i.e., 1 + 1 = 1 . Proved.
a2 b2 p2
EXERCISE - 4.4 (F)
1. (a) What is the slope-intercept form of the straight line ax + by + c = 0?
(b) What is the double-intercept form of the straight line ax + by + c = 0?
(c) What is the slope of the line ax + by + c = 0?
(d) Write the y-intercept of the line ax + by + c = 0.
(e) What is the perpendicular distance from the origin to the line ax + by + c = 0?
2. Write the following equations into the slope intercept form:
(a) 3x – 2y + 1 = 0 (b) 1 x+ 2 y + 2 = 0 (c) 3y – 2 = 0
2 3
3. Write the following equations into the double intercept form:
(a) 2x + 3y – 6 = 0 (b) 4x – y + 5 = 0 (c) 4x + 1 = 0
4. Write the following equations into the perpendicular (or normal) form:
(a) 3 x + y = 10 (b) x – y + 5 3 = 0 (c) 2y + 5 = 0
5. Find the slope and y-intercept of the following straight lines:
(a) 4x + 2y – 3 = 0 (b) 3x – 3y + 3 = 0 (c) 2y + 3 = 0
146 Illustrated Optional Mathematics-9
6. Find the x-intercept, y-intercept and the angle with x-axis made by the following equations:
(a) 3x – 3y + 2 = 0 (b) y = 3 (c) x cos α + y sin α = sin2 α
7. Calculate the length of the perpendicular distance from the origin to the following straight lines and
also find the angle made by the perpendicular with x-axis:
(a) x – y + 2 = 0 (b) 2x + 3 = 0 (c) 3y – 1 = 0
8. (a) Find the length of the line segment of a straight line 2x + 3y = 6 between the axes.
(b) Calculate the length of the line segment of a straight line 3x – 4y + 12 = 0 between the axes.
9. Find the area of the triangle formed by the following lines with the axes:
(a) 2x + 3y – 6 = 0 (b) 4x – 3y + 1 = 0
10. (a) Find the equation of a straight line whose intercepts on axes are twice of the intercepts of the
straight line 2x + 3y – 6 = 0.
(b) Find the equation of a straight line whose intercepts on x-axis and y-axis are respectively
double and triple of those by the straight line 4x – 5y – 20 = 0.
11. (a) Reduce ax + by = 1 into the normal form and prove that a2 + b2 = 1 .
p2
(b) Reduce x + y = 1 into the normal form and prove that a2 + b2 = 1 . COORDINATE
a2 b2 b2 a2 p2 GEOMETRY
(c) Reduce x – y = 1 into the normal form and prove that a– 2 + b– 2 = p– 2.
b a
ANSWERS (b) y = (– 3 )x + (–3) (c) y = 0.x + 2
4 3
2. (a) y = 3 x + 1
22 (b) x + y = 1 (c) x = 1
–5/4 5 – 1/4
3. (a) x + y = 1 x cos 5
32 (b) x cos135o + y sin135o = 5 (c) 270o + y sin 270o =
2
4. (a) x cos30o + y sin30o = 5 (c) – 2, – 3
5. (a) – 2, 3 (b) 3, –1 32
2 (b) no x-intercept, 3, 0o or 180o (c) sin α . tan α, sin α, 90o + α
6. (a) – 2, 2, 45o
(b) 3 units, 180o (c) 1 units, 90o
33 2 3
7. (a) 2 units, 315o
(b) 5 units
8. (a) 13 units (b) 1 sq. units
9. (a) 3 sq. units
24
10. (a) 2x + 3y – 12 = 0
(b) 6x – 5y – 60 = 0
Equation of Straight Line 147
4.4 (G) Distance between Point and Line
Learning Objectives
At the end of this topic, the students will be able to:
• find the perpendicular distance from a point to the given line.
There are two conditions to find the perpendicular distance from a point (x1 y1) to the given line. They are
based on the form of equation i.e., general form and perpendicular form:
a. Distance Between a Point (x1, y1) and a Line x cos α + y sin α = p
The equation of a straight line in perpendicular form is xcos α + ysin α = p
where, OM = p, OM ⊥ AB and ∠ MOA = α Y
COORDINATE Draw another straight line CD such that AB//CD. Also, show D X
GEOMETRY ON⊥CD where O, M and N are collinear and ON = p and ∠NOC
= α and the equation of CD is xcos α + ysin α = p. Take a point BN
P(x1, y1) on the line CD and draw PQ ⊥ AB. P(x1, y1)
Now, the distance from the point P(x1, y1) to the line AB is pM
PQ = MN = ON – OM = p' – p when p' > p and PQ = p – p' when
p > p'. Thus, PQ = ± (p' – p). αQ
O AC
Here, proper sign is taken so as to make PQ positive. Since the point (x1, y1) lies on the CD.
So, x1 cos α + y1sinα = p'.
Hence, ± PQ = (x1cos α + y1sin α – p), which is the required perpendicular distance between the
point (x1, y1) and the line x cos α + y sin α = p.
b. Distance Between a Point (x1 , y1) and the Line ax + by + c = 0
The general equation of the first degree in x and y is ax + by + c = 0 ………(1)
The perpendicular form of the equation (1), we have
or, a b2 x + b b2 y + c b2 = 0 ................... (ii)
a2 + a2 + a2 +
Comparing equation (ii) with x cos α + y sin α – P = 0, we get
cos α = a b2 , sin α = b b2 , p = – c
a2 + a2 + a2 + b2
Now, the perpendicular distance from a point (x1, y1) to the line x cos α + y sin α = p is
d = ± (x1cos α + y1sin α – p).
± x1 a + y1 b + c = ax1 + by1 + c
a2 + b2 a2 + b2 a2 + b2 a2 + b2
Hence, the distance between the point (x1, y1) and the line ax1 + by1 + c = 0 is d = ax1 + by1 + c .
a2 + b2
Note: The symbol | | always gives the numerical value as positive or absolute and which is called modulus.
148 Illustrated Optional Mathematics-9
Example 1 : Find the length of the perpendicular drawn from the point (2, –1) to the line 8x – 6y + 7 = 0.
Solution : Let d be the length of the perpendicular drawn from the point (2, –1) to the line
8x – 6y + 7 = 0. Then,
(2, –1)
d= ax1 + by1 + c = 8 × 2 – 6 × (–1) + 7 d
a2 + b2 82 + 62
= 16 + 6 + 7 = 29 = 29 = 2.9 units. X' 8x – 6y + 7 = 0 X
64 + 36 100 10
Example 2 : Find the distance between the parallel lines 2x + 3y – 6 = 0 and 4x + 6y = 3.
Solution : Here, the equations of the line are 2x + 3y – 6 = 0 ……… (i) P(x1, y1)
and 4x + 6y = 3
or, 4x + 6y – 3 = 0 ……….. (ii) 2x + 3y – 6 = 0
d
Let any point on the line (i) be (x1, y1) X' 4x + 6y = 3 X
Then, 2x1 + 3y1 – 6 = 0 COORDINATE
GEOMETRY
or, 2x1 + 3y1 = 6 ………. (iii)
Now, the perpendicular distance from the point P(x1, y1) of the line (i) to the another line
4x + 6y – 3 = 0 is,
d= 4x1 + 6y1 – 3 = 2(2x1 + 3y1) – 3
42 + 62 16 + 36
= 2×6–3 =2 9 units.
52 13
Alternatively
By inspection, (3, 0) lies on the equation (i) so,
4×3+6×0–3
d = 42 + 32
= 9 = 2 9 units
52 13
EXERCISE - 4.4 (G)
1. (a) What is the distance from the point (x1, y1) to the straight line ax + by + c = 0?
(b) Write the distance from the point (a, b) to the line px + qy = r.
2. Find the perpendicular distance for the following conditions:
(a) From the point (3, 2) to the line 4x + 3y + 2 = 0
(b) From the point (0, –3) to the line 3x + 4 = 0
(c) From the point (–1, 2) to the line 2y – 5 = 0
Equation of Straight Line 149
3. Find the distance between the following parallel lines:
(a) 2x – 3y = 7 and 4x – 6y = 2 (b) 4x + 3y = 21 and 3x + 6y = 2
(c) 3x + 3y = 4 3 and x + y – 3 = 0 (d) 2x + 1 = 0 and 4x – 3 = 0
4. (a) The length of perpendicular drawn from the point (2a, 3) to the line 3x – 4y + 13 = 0 is 5 units.
Find the value of a.
(b) The length of perpendicular drawn from the point (2k, –k) to the line 2x – 3y = 1 is 13 units.
Find the value of k.
(c) Find the possible value of p if the perpendicular distance from the point (2, – 3) to the line
4x – 3y + p = 0 is 3 units.
(d) If the perpendicular distance from the point (3, –2c) to the line 3x + 4y – 3c = 0 is 4 units, find
the value of c.
5. (a) Which of the straight lines y = 2x + 3 or 4y = x – 7 is far from the origin? Are they the same
sides of the origin?
(b) Which of the lines 2x – 3y + 17 = 0 or 3x + y – 6 = 0 is far from the point (2, 3)? Are they the
same sides of the point (2, 3)?
6. (a) Prove that the straight lines 3x – y + 6 = 0 and 6x + 13y – 21 = 0 are equidistant from the origin.
COORDINATE (b) Prove that the point (1, 1) is equidistant from the lines x + y – 6 = 0 and x + y + 2 = 0
GEOMETRY
7. (a) Find the points on the line 3x + y = 3 that lies at a distance of 1 unit from the line 3x + 4y = 7.
(b) Find the coordinates of the points on the line x + 5y – 13 = 0 at a distance of 2 units from the
line 12x – 5y + 26 = 0.
8. (a) If the distance between the parallel lines 3x + 2y – k = 0 and 3x + 2y + 3 = 0 is 13units, find
the value of k.
(b) If the distance between the parallel lines 4x – 3y – 21 = 0 and 8x – 6y + 3p = 1 is 5 units, find the value
of p.
9. (a) If the points A and B lying on the straight line 4y = 3x + 15 are at a distance of 5 units from the
origin, find the area of ∆AOB.
(b) The distance from the origin to two points lying on the line x + 2y = 15 5 are equal to 17
units. Find the area of triangle form by these two points and origin.
10. If the length of perpendicular line drawn from (1, 1) to the line ax + by – c = 0 is 1, prove that:
1 + 1 – 1 = 21ab.
a b c
ANSWERS
2. (a) 4 units (b) 4 units (c) 4 units
3
3. (a) 6 units (b) 4 units (c) 1 units (d) 1.25 units 4. (a) 4 (b) 2 (c) 2 (d) –1
13 6
5. (a) Second line, No (b) First line, Yes 7. (a) 10 , – 1 (b) 1, 12 , – 3, 16
(0, 3), 9 3 5 5
8. (a) 10 (b) 3 9. (a) 12 sq. units (b) 120 sq. units
150 Illustrated Optional Mathematics-9
The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.
Final-for-print-Illustrated-OPT-Math-9-2076-2-1
Discover the best professional documents and content resources in AnyFlip Document Base.
Search