The mid-point of diagonal BD is x1 + x2, y1 + y2 i.e. – 2 + 3, 3 + 2 i.e. 12, 5
2 2 2 2 2
As mid-point of both diagonals are same, they intersect each other at their mid-points. i.e. diagonals bisect
each other. Hence, ABCD is a parallelogram.
Example 8:
If the points A(2, 7), B(4, 2) and C(7, 0) are three vertices of a parallelogram. Find the fourth vertex
of the parallelogram which is opposite to B(4, 2).
Let D(x, y) be the fourth vertex of the parallelogram having three A(2, 7)
given vertices A(2, 7), B(4, 2) and C(7, 0). D(x, y)
As diagonals of parallelogram bisect each other, O is the mid-point of
diagonals AC and BD. O
The co-ordinates of mid-point O of AC is 2 + 7, 7 + 0 i.e. 92, 7 B(4, 2)
2 2 2
O is the mid-point of diagonal BD also, so
C(7, 0)
92 4 + x and 72 = 2 + y
= 2 2 Co-ordinate
or, 9 = 4 + x or, 7 = 2 + y
∴ x = 5 ∴ y = 5
Hence, the co-ordinates of the fourth vertex D is (5, 5).
Example 9:
Find the co-ordinates of the points of trisection of the line joining the points M(7, – 3) and N(1, 3).
Points of trisection of a line segment joining two points are those points
N(1, 3)
which divides the given line segment into three equal parts.
Let P(x', y') and Q(x", y") are the points of trisection of the line segment 2
joining the points M(7, – 3) and N(1, 3).
Q(x", y")
So, P(x', y') divides MN in the ratio 1:2 and hence,
(x', y') = m1x2 + m2x1 , m1y2 + m2y1 1 P(x', y')
m1 + m2 m1 + m2
Here, m1 = 1, m2 = 2, x1 = 7, y1 = – 3, x2 = 1 and y2 = 3
(x', y') = 1 × 1 + 2 × 7, 1 × 3 + 2 × (– 3) = 15, – 3 = (5, – 1) M(7, – 3)
1 + 2 1 + 2 33
And Q(x", y") is the mid-point of PN and hence
(x", y") = x1 + x2, y1 + y2
22
Here, x1 = 5, y1 = – 1, x2 = 1 and y2 = 3
(x", y") = 5 + 1, – 1 + 3 = (3, 1)
2 2
Hence, the required points of trisection are P(5, – 1) and Q(3, 1).
Section Formula ~ 97
Theorem
If the points A(x1 , y1), B(x2 , y2), C(x3 , y3) are the vertices of a triangle ABC. Find the co-ordinates of
the centroid of the triangle. (Centroid of a triangle is the intersection of the medians of the triangle.
From geometry, we know that the centroid divides each median in the ratio 2:1 from its vertex).
Proof: A(x1, y1)
Let G(x, y) be the centroid of ∆ABC having vertices A(x1, y1),
B(x2, y2) and C(x3, y3). The co-ordinates of mid-point D of BC is F 2
B(x2, y2)
x2 + x3, y2 + y3 E
2 2 G
1
The centroid G divides median AD in the ratio 2:1. x2 + x3 , y2 + y3 C(x3, y3)
2 2
Hence, (x, y) = m1x2 + mm22x1 , m1y2 + m2y1 D
m1 + m1 + m2
Here, x1 = x1, y1 = y1, x2 = x2 + x3 and y2 = y2 + y3
2 2
2 x2 + x3 + 1×x1 2 y2 + y3 + 1×y1
2 1 , 2 1
i.e. (x, y) = 2+ 2+
(x, y) = x1 + x2 + x3 , y1 + y2 + y3
3 3
∴ is the required centroid.
Example 10:
If the point (2, 0) is the co-ordinates of the centroid of a triangle whose two vertices are at points
(1, 4), (2, – 1), find its third vertex. A(1, 4)
Suppose C(a, b) be the third vertex of a triangle whose two
vertices are A(1, 4), B(2, – 1) and centroid is G(2, 0).
By using centroid formula. G(2, 0)
x = x1 + x2 + x3 and y = y1 + y2 + y3 B(2, – 1) C(x, y)
3 3
Here, (x, y) = (2, 0), (x1, y1) = (1, 4), (x2, y2) = (2, 0) and (x3, y3) = (a, b).
So, 2 = 1 + 2 + a and 0 = 4 – 1 + b
3 3
or, a + 3 = 6 or, 0 = 3 + b
∴ a = 3 ∴ b = – 3
Hence, the co-ordinates of the third vertex is (3, – 3).
Example 11:
The co-ordinates of the mid-points of the sides AB, BC and CA of triangle ABC are (0, 1), (1, – 1)
and (2, 2) respectively. Find the co-ordinates of A, B and C.
98 ~ Perfect Optional Mathematics Class 9
Suppose the vertices of triangle ABC are A(x1, y1), B(x2, y2) and C(x3, y3). The mid-points of sides AB,
BC and CA are P(0, 1), Q(1, – 1) and R(2, 2) respectively.
Hence, x1 + x2 = 0 and y1 + y2 = 1
2 2
or, x1 + x2 = 0 ... ... ... (1) and y1 + y2 = 2 ... ... ... (2)
Similarly, x2 + x3 = 2 ... ... … (3) and y2 + y3 = – 2 ... ... ... (4)
and x3 + x1 = 4 ... ... … (5) and y3 + y1 = 4 ... ... ... (6)
Adding equations (1), (3) and (5), we have,
2(x1 + x2 + x3) = 6 A(x1, y1)
or, x1 + x2 + x3 = 3
or, 0 + x3 = 3 P(0, 1) R(2, 2)
∴ x3 = 3
Substituting value of x3 in equations (3) and (5), B(x2, y2) Q(1, – 1) C(x3, y3)
x2 = – 1 and x1 = 1
Similarly, solving equations (2), (4) and (6), we have Co-ordinate
y1 + y2 + y3 = 2
or, 2 + y3 = 2
or, y3 = 0
and y2 = – 2, y1 = 4
Hence, the vertices of triangle ABC are A(1, 4), B(– 1, – 2) and C(3, 0).
Example 12:
Show that the three points (– 2, 2), (1, 4) and (7, 8) are collinear.
Let P(– 2, 2), Q(1, 4) and R(7, 8) are the given three points. If the points are collinear or lying on the
same straight line, slope of PQ and slope of QR are same.
Suppose P(– 2, 2) = (x1, y1), Q(1, 4) = (x2, y2) and R(7, 8) = (x3, y3)
Slope of PQ = Slope of QR
Now by using slope formula,
i.e. xy22 –– y1 = y3 – y2
x1 x3 – x2
i.e. 1 4– –2 = 8 – 4
(– 2) 7 – 1
or, 1 +2 2 = 4
6
or, 32 = 32, which is true.
Hence, P, Q and R are collinear.
Example 13:
Prove that the line 2x = y – 6 bisects the line segment joining the points (3, 4) and (– 7, 0).
Section Formula ~ 99
Let the end points of given line segment are A(3, 4) = (x1, y1) and B(– 7, 0) = (x2, y2). If C is the mid-
point of line segment AB, then
C(x, y) = x1 + x2, y1 + y2
22
= 3 + 2(– 7), 4 + 0 = 3 – 7, 4 = –2 4, 4 = (– 2, 2)
2 2 2 2
If the line 2x = y – 6 bisects the line segment joining the points A(3, 4) and B(– 7, 0), the mid-point C of
AB lies on the line.
i.e. 2x = y – 6
or, 2 × (– 2) = 2 – 6
or, – 4 = – 4, which is true.
Hence, the given line bisects the line segment AB.
Example 14:
In what ratio does the line of the equation 4x + 5y = 21 divide the line segment joining the points
(– 2, 3) and (4, 5)?
Let the line 4x + 5y = 21 divides the line segment joining the points A( – 2, 3) = (x1, y1) and
B(4, 5) = (x2, y2) in the ratio m1:m2 at point C(x, y).
Now using the section formula, the co-ordinates of C is
C(x, y) = m1x2 + mm22x1 , m1y2 + m2y1
m1 + m1 + m2
= m1 × 4 + +mm2 ×2 (– 2) , m1 × 5 + m2 × 3
m1 m1 + m2
= 4m1 – 2mm2 2 , 5m1 + 3m2
m1 + m1 + m2
Here, the point C lies on the line 4x + 5y = 21, so the co-ordinates of point C satisfy the equation of the
line, i.e. 4 4m1 – 2m2 +5 5m1 + 3m2 = 21
m1 + m2 m1 + m2
or, 4(4m1 – 2m2) + 5(5m1 + 3m2) = 21
m1 + m2
or, 16m1 – 8m2 + 25m1 + 15m2 = 21(m1 + m2)
or, 41m1 + 7m2 = 21m1 + 21m2
or, 41m1 – 21m1 = 21m2 – 7m2
or, 20m1 = 14m2
or, m1 = 14 = 7
m2 20 10
Hence, the line 4x + 5y = 21 divides the line segment joining the points (– 2, 3) and (4, 5) in the ratio 7:10.
100 ~ Perfect Optional Mathematics Class 9
Example 15:
If a line segment AB is divided at the point P(2, – 1) in the ratio 2:3 and at Q(3, 1) in the ratio 1:4.,
find the co-ordinates of points A and B.
Let the line segment joining the points A(x1, y1) and B(x2, y2) is divided at point P(2, – 1) = (x, y) in
the ratio 2:3 = m1:m2, then
m1 = x – x1 Again, m1 = y – y1
m2 x2 – x m2 y2 – y
or, 32 = 2 – x1 or, 23 = – 1 – y1
x2 – 2 y2 – (– 1)
or, 2(x2 – 2) = 3(2 – x1) or, 2(y2 + 1) = 3(– 1 – y1)
or, 2x2 – 4 = 6 – 3x1 or, 2y2 + 2 = – 3 – 3y1
or, 3x1 + 2x2 = 6 + 4 or, 3y1 + 2y2 = – 3 – 2
or, 3x1 + 2x2 = 10 ... ... (1) or, 3y1 + 2y2 = – 5 ... ... (2) Co-ordinate
Again let the line segment joining A(x1, y1) and B(x2, y2) is divided at point Q(3, 1) = (x, y) in the ratio
1:4 = m1:m2, then
m1 = x – x1 Again, m1 = y – y1
m2 x2 – x m2 y2 – y
or, 14 = 3 – x1 or, 14 = 1 – y1
x2 – 3 y2 – 1
or, x2 – 3 = 4(3 – x1) or, y2 – 1 = 4(1 – y1)
or, x2 – 3 = 12 – 4x1 or, y2 – 1 = 4 – 4y1
or, 4x1 + x2 = 12 + 3 or, 4y1 + y2 = 4 + 1
or, 4x1 + x2 = 15 ... ... (3) or, 4y1 + y2 = 5 ... ... (4)
Solving equations (1) and (3), we have Solving equations (2) and (4), we have
3x1 + 2x2 = 10 3y1 + 2y2 = – 5
–8x1 +– 2x2 =–30 –8y1 +– 2y2 =–10
– 5x1 = – 20 – 5y1 = – 15
\ x1 = 4 \ y1 = 3
Now, from equation (3), Now, from equation (4),
4 × 4 + x2 = 15 4 × 3 + y2 = 5
or, 16 + x2 = 15 or, 12 + y2 = 5
or, x2 = 15 – 16 or, y2 = 5 – 12
\ x2 = – 1 \ y2 = – 7
Hence the co-ordinates of points A and B are (4, 3) and (– 1, – 7) respectively.
Section Formula ~ 101
EXERCISE : 4.1A
1. Find the co-ordinates of point which divides the line segment joining the following pair of lines in the
ratio 2:1 internally.
a. (4, 6) and (– 2, 3) b. (– 4, – 3) and (5, 2)
2. Find the co-ordinates of a point which divides the line segment joining the following pair of points in
the ratio 2:3 externally.
a. (– 1, 2) and (1, 1) b. (3, 2) and (0, 1)
3. a. Prove that the point (2, – 1) bisects the line segment joining the points (1, 3) and (3, – 5).
b. If one end of a linesegment is (2, – 5) and mid-point is (– 2, 3), find the co-ordinates of other end.
c. If one end of a linesegment is A(4, 4) and mid-point is B(– 1, 3), then find the co-ordinates of the
other end.
4. a. In what ratio is the line segment joining the points (2, 4) and (– 3, – 2) divided by X– axis?
b. In what ratio does Y-axis divide the line joining (5, 5) and (– 2, – 2)?
c. Find the ratio in which P(– 2, 5) divides the join of (4, 8) and (– 4, 4)?
5. a. Find the co-ordinates of the centroid of a triangle having vertices A(2, 7), B(6, 2) and C(4, 5).
b. If the centroid of a triangle with vertices (2, 5), (a, – 2) and (– 1, – 3) is (2, b), find the values of a
and b.
c. If the centroid of a triangle is (0, 3) and its two vertices are (3, 5) and (– 1, 3), find the third vertex
of the triangle.
6. a. In the given figure, line AB cuts X-axis at N Y Y
such that 2AN = 3NB. If the co-ordinates of A(– 2,6)
A and B are (– 2, 6) and (6, p), find the value A(– 2,6)
Q
of p. M B(a, 2)
b. In the given figure, line AB cut Y-axis at Q X' O X
such that 2AQ = QB. If the co-ordinates of X' O NX Y' Q.N.6(b)
A and B are (– 2, 6) and (a, 2), find the value B(6,p)
Y' Q.N.6(a)
of a.
EXERCISE : 4.1B
1. a. If A(2, 5) and B(– 1, 2) are two points, find the co-ordinates of points which divides AB in the
ratio 2:3 internally and externally.
b. If P(– 2, 3) and Q(4, – 9) are two points, find the co-ordinates of points which divide PQ in the
ratio 1:2 internally and externally.
2. a. If the three points M(2, 4), N(3, 1) and S(5, a) lie on the same straight line, find the value of a.
b. If the three points (– 3, 3), (– 1, 2) and (a, – 1) lie on the same straight line, find the value of a.
c. A line MN with points M(6, – 2) and N(1, 2) is produced to point P(k, 6). Find the value of k.
d. Where does the line joining the points (– 1, 5) and 2, – 25 meet the X– axis ?
102 ~ Perfect Optional Mathematics Class 9
e. Find the co-ordinates of a point on the Y- axis at which the linesegment joining the points (4, – 2)
and (2, 1) meets.
f. A(2, 5), B(x, y) and C(– 1, 2) are collinear points such that AB = 2BC. Find the co-ordinates of B.
3. a. A triangle ABC has its vertices at the points A(5, 5), B(– 5, – 1) and C(3, – 5). Find the length of
the median drawn from vertex A.
b. The co-ordinates of the vertices P and Q of triangle PQR are (4, 3) and (– 4, 1) respectively. If
the co-ordinates of the mid-point of side QR is (0, – 1), find the length of the median drawn from
vertex Q.
4. a. Prove that the quadrilateral having vertices at the points (– 2, 1), (– 3, 3), (1, 1) and (2, – 1) is a
parallelogram.
b. If the three points (– 5, – 2), (– 4, 1) and (1, 3) are the three vertices of a parallelogram, find the
co-ordinates of the fourth vertex which is opposite to the vertex (– 4, 1).
c. The point of intersection of the diagonals of a parallelogram whose two vertices are (7, 16) and
(10, 2) is (2, 6). Find the co-ordinates of the remaining vertices.
d. The point (2, 4) is the point of intersection of the diagonals of a parallelogram of which two Co-ordinate
corners are (8, 6) and (6, 0). Find the co-ordinates of the remaining vertices.
e. If the points A(2, – 2), B(a, 4), C(5, b) and D(– 1, 1) are the vertices of a parallelogram, find the
value of a and b.
5. a Find the co-ordinates of the points of trisection of the line joining the points (25, 10) and (– 5, – 5).
b.
c. Find the co-ordinates of two points which trisect the line segment joining points (5, 3) and (8, 6).
d. Find the co-ordinates of the points which divides the line segment joining the points (– 4, 8) and
(4, – 4) into four equal parts.
6. a.
P, Q, R are collinear points, P and Q are (3, 4) and (7, 7) respectively and PR is equal to 10 units.
Find the length of PQ and find the co-ordinates of R.
If M and N are the mid-points of AB and AC of a triangle ABC whose vertices are A(– 2, 2),
B(2, 4) and C(6, – 4), then show that BC = 2MN.
b. The vertices of ∆PQR are P(2, 5), Q(6, 1) and R(– 2, – 1). Find the length of the longest median.
c. If the mid-points of the sides of a triangle are (2, 4), 12, 7 and 25, 9 , find the co-ordinates of the
2 2
vertices of the triangle.
7. Prove that the following three points line on a straight line.
a. (– 3, 2), (0, 0) and (3, – 2) b. (0, 3), (2, 1) and (4, – 1)
8. a. In what ratio does the line of equation 2y – 3x = 0 divide the line segment joining the points (4, 1)
and (1, 4)?
b. Prove that the line x – y = 0 bisects the line segment joining the points (1, 6) and (4, – 1).
c. A line segment PQ is divided at A(3, 2) in the ratio 2:3 and at B(4, 3) in the ratio 3:2. Find the
co-ordinates of P and Q.
d. Find the centroid of the triangle formed by the equation of its sides x – 2y = 0, x + y = 6 and y = 0.
Section Formula ~ 103
4.2 Locus of a Point
Introduction A P2
O
In the figure (a), A and B are two fixed points, P1, P2, P3 etc. are P1
the points which are equidistant from the fixed points A and B.
There are many other points which are equidistant from A and B
B. Joining the points P1, P2, P3 etc. we get a straight line which P3
is the perpendicular bisector of line segment AB. P4
P5 Fig. (a)
In figure (b), O is a fixed point M1, M2, M3 etc. are the points at
distance 5 units from O. Joining the points M1, M2, M3 etc. we M1
get the path traced out by a moving point which moves so that
its distance from O is always 5 units. M2
M3
It is a circumference of a circle having centre O and radius 5 M4
units.
The path traced out by a moving point which moves under
certain condition is known as the locus of a point.
Fig. (b)
Equation of a Locus
Let O be the fixed point and P the moving point which moves Y
so that OP = 5 units. The locus of a point P is a circle with
centre at O(0, 0) and radius 5 units. If (0, 5)
(3, 4)
P(x, y) be any point on the locus, then (– 3, 4) (4, 3)
(– 4, 3)
(x – 0)2 + (y – 0)2 = 5 X' (5, 0) X
or, x2 + y2 = 25 (– 5, 0) O
Some of the points on the locus are (5, 0), (3, 4), (4, 3), (– 4, – 3) (4, – 3)
(0, 5), etc. (– 3, – 4) (3, – 4)
(0, – 5)
When, x = 5, y = 0, x2 + y2 = 52 + 02 = 25
x = 3, y = 4, x2 + y2 = 32 + 42 = 25 Y'
x = 4, y = 3, x2 + y2 = 42 + 32 = 25
x = 0, y = 5, x2 + y2 = 02 + 52 = 25 etc.
All the points on the locus satisfy the equation x2 + y2 = 25, so x2 + y2 = 25 is the equation of the locus.
"An equation which is satisfied by all the points on the locus is known as equation of locus."
104 ~ Perfect Optional Mathematics Class 9
Method of Finding the Equation of Locus Co-ordinate
To find the equation of a locus of a moving point, the following steps are to be followed:
a. Draw a figure according to given condition and take any point (x, y) on the locus.
b. Write the geometrical condition with which the point moves.
c. Express the given geometrical condition in the form of algebraic expressions.
d. Simplify to get the equation.
Example 1:
Show that the point (5, 0) lies on the locus 2x + 3y = 10.
Here, if given point (5, 0) lies on the locus 2x + 3y = 10, then the point satisfies the given locus.
i.e. 2 × 5 + 3 + 0 = 10
or, 10 + 0 = 10
or 10 = 10, which is true.
Hence, the point (5, 0) lies on the locus 2x + 3y = 10.
Example 2:
Which of the points (5, – 1) and (7, 5) lie on the locus having equation x2 + y2 – 6x – 4y = 12 ?
Substituting the point (5, – 1) in x2 + y2 – 6x – 4y = 12,
i.e. x = 5 and y = – 1, 52 + (– 1)2 – 6 × 5 – 4(– 1) = 12
i.e. 25 + 1 – 30 + 4 = 12
i.e. 0 = 12 which is not true.
Hence, the point (5, – 1) does not lie on the given locus.
Substituting the point (7, 5) in x2 + y2 – 6x – 4y = 12
i.e. x = 7, y = 5, 72 + 52 – 6 × 7 – 4 × 5 = 12
i.e. 49 + 25 – 42 – 20 = 12
i.e. 12 = 12 which is true.
Hence, the point (7, 5) lies on the given locus.
Example 3:
If the point (0, a) lies on the locus 3x + 2y = 25, then find the value of a.
Here, point (0, a) lies on the locus 3x + 2y = 24.
Hence, the co-ordinates (0, a) satisfies the given locus.
i.e. 3 × 0 + 2 × a = 24
or, 0 + 2a = 24
or, 2a = 24
\ a = 12
Locus of a Point ~ 105
Example 4:
If the point (4, 4) lies on the locus having equation y2 = kx, then show that (16, 8) is another point
on the same locus.
Here, the point (4, 4) lies on the locus y2 = kx.
i.e. 42 = k × 4
or, k = 16
4
\ k = 4
Now, the given locus is y = 4x and the another given point is (16, 8).
Again testing the point (16, 8) with given locus,
or, 82 = 4 × 16
or, 64 = 64
Hence, the point (16, 8) lies on the given locus.
Example 5:
Find the equation of the locus of a point which moves so that it is equidistant from the two points
A(– 6, 0) and B(5, 2).
Let P(x, y) be any point on the locus and A(– 6, 0) and B(5, 2) are P(x, y)
the given two fixed points.
Here, AP = BP A(– 6, 0) B(5, 2)
or, AP2 = BP2
or, (x + 6)2 + (y – 0)2 = (x – 5)2 + (y – 2)2
or, x2 + 12x + 36 + y2 = x2 – 10x + 25 + y2 – 4y + 4
or, 22x + 4y + 7 = 0, this is the required equation.
Example 6:
Find the equation of the locus of a point which moves so that its distance from the point (– 2, 3) is
always 5 units. P(x, y)
Let P(x, y) be any point on the locus and A(– 2, 3) be the given point, then PA = 5.
or, AP2 = 25 5 units
or, (x + 2)2 + (y – 3)2 = 25
or, x2 + 4x + 4 + y2 – 6y + 9 = 25 A(– 2, 3)
∴ x2 + y2 + 4x – 6y – 12 = 0. This is the required equation.
Example 7:
Let A(3, – 5) and B(– 3, 3) are two fixed points. Find the locus of a point P such that angle APB is
a right angle.
106 ~ Perfect Optional Mathematics Class 9
Let A(3, – 5) and B(– 3, 3) be two fixed points and P(x, y) be any point on the locus. DPAB is a right
angled triangle with APB = 90°. So, AB2 = AP2 + BP2
i.e. (– 3 – 3)2 + (3 + 5)2 = (x – 3)2 + (y + 5)2 + (x + 3)2 + (y – 3)2 P(x, y)
or, 36 + 64 = x2 – 6x + 9 + y2 + 10y + 25 + x2 + 6x + 9 + y2 – 6y + 9
or, 100 = 2x2 + 2y2 + 4y + 52
or, 2x2 + 2y2 + 4y – 48 = 0 A(3, – 5) B(– 3, 3)
or, x2 + y2 + 2y – 24 = 0, this is the required equation.
Example 8:
A(1, 2) and B(5, 3) are two points. Find the locus of a point 'P' so that PA:PB = 2:3.
Let, P(x, y) be any point on the locus and A(1, 2) and B(5, 3) are given two points.
Here, AP:BP = 2:3
or, AP = 2 A(1, 2) P(x, y)
BP 3
or, 3AP = 2BP Co-ordinate
or, 9AP2 = 4BP2
or, 9{(x – 1)2 + (y – 2)2} = 4{(x – 5)2 + (y – 3)2} B(5, 3)
or, 9(x2 – 2x + 1 + y2 – 4y + 4) = 4(x2 – 10x + 25 + y2 – 6y + 9)
or, 9x2 – 18x + 9y2 – 36y + 45 = 4x2 – 40x + 4y2 – 24y + 136
or, 9x2 – 4x2 + 9y2 – 4y2 – 18x + 40x – 36y + 24y + 45 – 136 = 0
or, 5x2 + 5y2 + 22x – 12y – 91 = 0.
This is the required equation.
Example 9:
Find the equation of the locus of a point whose distance from the point (– 2, 5) is one half of its
distance from Y-axis. A(– 2, 5) Y
Let P(x, y) be any point on the given locus and A(– 2, 5) is the given point. If
PQ is drawn on the Y-axis, PQ = x.
From the given condition AP = 12 PQ P(x, y) Q X
or, 2AP = PQ
or, 4AP2 = PQ2 X' O
or, 4[{(x – (– 2)2} + (y – 5)2] = x2 Y'
or, 4{(x + 2)2 + y2 – 10y + 25} = x2
or, 4(x2 + 4x + 4 + y2 – 10y + 25) = x2
or, 4x2 + 16x + 16 + 4y2 – 40y + 100 = x2
or, 4x2 – x2 + 4y2 + 16x – 40y + 116 = 0
\ 3x2 + 4y2 + 16x – 30y + 116 = 0 is the equation of the required locus.
Locus of a Point ~ 107
Example 10:
Find the equation of the locus of a moving point which divides the line segment joining the points
(– a, 0) and (0, – b) in the ratio 4:5 and making a + b = – 18.
Let P(x, y) be a moving point and A(– a, 0) and B(0, – b) be the given fixed points.
From question, AP = 4 and by using section formula, we have
BP 5
(x, y) = m1x2 + mm22x1 , m1y2 + m2y1
or, m1 + m1 + m2
(x, y) = 4 × 0+5× (– a) , 4 × (– b) + 5 × 0
4+5 4+5
or, (x, y) = – 95a, – 4b
9
Hence, x = – 5a and y = – 4b
9 9
i.e. 9x = – 5a and 9y = – 4b
or, a = – 9x and b = – 9y
5 5
Again, we have a + b = – 18
or, – 59x + – 9y = – 18
5
or, 4 × (– 9x)2+05 × (–9y) = – 18
or, – 36x – 45y = – 18 × 20
or, – 9(4x + 5y ) = – 360
or, 4x + 5y = 40 is the required equation.
EXERCISE : 4.2
1. Which of the following points lie on the locus having equation 2x – 3y + 8 = 0?
a. (0, 1) b. (– 4, 0) c. (1, 2)
d. (1, – 2) e. (4, 0) f. (2, 4)
2. Which of the following points lie on the locus having equation x2 + y2 – 2x – 6y = 15?
a. (1, 8) b. (7, 2) c. (– 4, 3)
d. (5, 7) e. (1, 0) f. (4, – 1)
3. a. If the point (3, 4) lies on the locus with equation 4x – ky + 8 = 0, find the value of k.
b. The point (4, 2) lies in the locus with equation x2 = 2ay. Show that (8, 8) is another point on the
locus.
c. The point (4, 2) lies on the locus x2 = by. Show that the points 2, 1 and 6, 9 also lie on the
2 2
locus.
108 ~ Perfect Optional Mathematics Class 9
4. a. Find the equation of the locus of a moving point which moves so that it is equidistant from two
fixed points (2, 4) and (– 3, – 2).
b. Find the equation of a moving point which moves so that it is equidistant from the two points
(a + b, b – a) and (a – b, a + b).
c. A point moves so that its distance from the point (– 4, – 3) is twice its distance from Y– axis. Find
the equation of the locus.
d. Find the equation of locus of a moving point whose distance from point (– 1, 1) is equal to twice
its distance from X-axis.
e. A(a, 0) and B(– a, 0) are two fixed points, find the locus of a moving point P if 3PA = 4PB.
5. Find the equation of the locus of a moving point P which moves such that:
a. Its distance from Y– axis is always 3.
b. Its distance from X– axis is always thrice its distance from Y– axis.
c. Its distance from origin is always 6. Co-ordinate
d. Its distance from the point (2, – 4) is always 5.
e. It is equidistant from the points (3, – 2) and (5, 1).
6. A(3, 0) and B(2, – 1) are two fixed points. Find the equations of the locus of moving point P such that:
a. AP2 + PB2 = AB2 b. PA2 + 2PB2 = 3AB2 c. PA = 3PB
7. a. A(2, 0) and B(– 2, 0) are two fixed points. Find the locus of a moving point P, such that PA is
perpendicular to PB. [Hints : PA2 + PB2 = AB2]
b. If A(a, 0) and B(– a, 0) are two fixed points, find the equation of the locus of point P such that.
i. PA2 + PB2 = AB2 ii. mPA = nPB (m and n are constants)
8. a. Prove that the line segment joining (– 2, 3) and (– 7, 7) is bisected by the line having equation
2x + 3y = 6.
b. In what ratio does the line 2x + y + 4 = 0 divide the line segment joining the points (– 6, 5) and
(0, 2) ?
9. a. If m + n = 8, find the equation of locus of a moving point which is always the mid-point of (m, 0)
and (0, n).
b. Find the equation of the locus of a moving point which divides the line segment joining the
points (a, 0) and (0, b) in the ratio 1:2 and making a = 10 – b.
Locus of a Point ~ 109
4.3 Equation of Straight Lines
Equation of a Straight Line
Let the equation of the locus of a moving point be 2x – 3y = 6. We can see its nature by taking a set of points
satisfying this equation and plotting them on the graph paper.
Here, 2x – 3y = 6 i.e. x= 6 + 3y Y
2
x 3 0 6 – 3 – 6
y 0 – 2 2 – 4 – 6
(6, 2)
By plotting these points on the graph paper and joining X' O (3, 0) X
them, the nature of the equation represented by the locus of (0, – 2)
a moving point shows a straight line. 2x – 3y = 6 is a linear
equation in x and y and its nature is a straight line. (– 3, – 4)
(– 6, – 6)
In the same way, the graph of any linear equation in x and Y'
y gives a straight line. Conversely, a straight line has its Y
equation linear in x and y is also true.
Equation of X– axis
In the given figure, (– 7, 0), (– 3, 0) (0, 0) (3, 0) etc. are (- 7, 0) (- 3, 0) O (0, 0) (3, 0) X
some points on the X-axis. Here, we get each and every X'
point on X-axis has ordinate or y co-ordinate zero. Hence, Y'
the equation of X-axis is y = 0.
Y
Equation of a Straight Line Parallel to X– axis
AM P(x, y) B
Let AB be a straight line parallel to the X– axis and cuts the
Y– axis at M such that OM = b. b
Consider P(x, y) be any point on the line AB and draw PN^OX X' O N(x, 0) X
then the co-ordinates of point N is (x, 0). As AB//OX, PN = OM
Y'
i.e. y = b
Every point on the line AB satisfies the equation y = b, hence it
is the equation of AB.
Cor.1. If b = 0, then y = 0 which is the equation of X– axis.
110 ~ Perfect Optional Mathematics Class 9
Equation of Y– axis Y X
X
In the given figure, (0, 0), (0, 2) (0, 4) (0, – 3) etc. are some (0, 4)
points on the Y-axis. Here, we get each and every point on (0, 2)
the Y-axis has abscissa or x co-ordinate zero. Hence, the X'
equation of Y-axis is x = 0. O (0, 0) (3, 0)
Equation of a Straight Line Parallel to Y– axis (0, - 3)
Let CD be a straight line parallel to Y– axis and cut X– axis at A Y'
such that OA = a.
Consider P(x, y) be any point on the line CD and draw PB^OY, YC
then the co-ordinates of point B is (0, y)
As CD//OY, PB = OA B(0, y) P(x, y)
i.e. x = a
Every point on the line CD satisfies the equation x = a, hence it X' O a A Co-ordinate
is the equation of CD. Y' D
Example 1:
Find the equation of a straight line parallel to X– axis which is 4 units above the X-axis.
We know that the equation of straight line parallel to X-axis is y = b.
Here, b = 4
So, y = b
i.e y = 4 is the required equation of straight line.
Example 2:
Find the equation of a straight line parallel to Y– axis which is 3 units left to the Y-axis.
We know that the equation of straight line parallel to Y-axis is x = a.
Here, a = – 3
So, x = a
i.e x = – 3 is the required equation of straight line.
Example 3:
Find the equation of a straight line parallel to Y– axis and passing through the point (4, – 3).
The equation of any straight line parallel to Y– axis is
x = a ... ... ... (1)
But this line passes through the point (4, – 3), so it must be satisfied by the point (4, – 3). i.e. 4 = a
Hence, the equation of the required straight line is x = 4.
Equation of Straight Lines ~ 111
Slope or Gradient of a Straight Line
YY Y
B B B
X' O q X Nq N
M X' O M X X' q O X
A
B N AM
Y'
Y' Y'
In above figures, the line AB makes an angle q with the positive X– axis in anticlockwise direction. Thus,
q may be acute or obtuse angle. Here q is called the Angle of Inclination of the line AB.
The tangent of the angle of inclination of a line is called the slope or gradient of the line.
i.e. The slope or gradient of a line is tangent of the angle made by the straight line with positive X– axis
in anticlockwise direction. It is denoted by m.
Hence, m = tan θ
Theorem
Prove that the slope of the line passing through two points (x1 , y1) and (x2 , y2) is m = tan q = y2 – y1
Proof: x2 – x1
If a line passes through two points M(x1, y1) and N(x2, y2) as shown in figure.
Draw MA⊥OX, NB⊥OX and MC⊥NB. If q be the angle of inclination Y (x2, y2) Q
then QRX = NMC = θ. N
MC = AB = OB – OA = x2 – x1 (x1, y1) C
NC = NB – BC = NB – MA = y2 – y1 Mq
The slope of the straight line PQ is q
m = tan θ = NC = y2 – y1 X' RO A BX
MC x2 – x1 P
Hence, the slope of a line passing through points (x1, y1) and (x2 , y2) is Y'
m = y2 – y1
x2 – x1
Note: a. The slope of the X– axis and the lines parallel to X– axis are zero.
b. The slope of the Y– axis and the lines parallel to Y– axis are not defined.
Example 4:
What is the slope of a line which makes an angle of 30° with X-axis ?
Here, angle made by the line with X-axis = q = 30°.
The slope of the line (m) = tan q = tan 30° = 1
3
1
Hence, the slope of the line is 3 .
112 ~ Perfect Optional Mathematics Class 9
Example 5:
Find the angle of inclination of the line whose slope is 3.
Here, slope of the line = 3
or, tan q = 3
or, tan q = tan 60°
\ q = 60°
Hence, the angle of inclination of the line is 60°.
Example 6:
If the slope of a line passing through the points P(2, – 2) and Q(4, a) is – 1, find the value of a.
The slope of the line passing through the points P(2, – 2) and Q(4, a) is given by
m = y2 – y1 Co-ordinate
x2 – x1
Here, x1 = 2, y1 = – 2
x2 = 4, y2 = a
or, – 1 = a + 2
4 – 2
or, a + 2 = – 2
∴ a = – 4
Example 7:
If the points (k, 4), (– 3, 2) and (3, 5) lie on the same straight line, find the value of k. C(3, 5)
Let the given three points are A(k, 4), B(– 3, 2) and C(3, 5). As A, B and
C lie on the same straight line,
The slope of AB = slope of BC B(– 3, 2)
A(k, 4)
or, –23––4k = 5 – 2
3 + 3
or, – 3– 2 k = 3
– 6
or, – 3 – k = – 2 × 2
or, – k = – 4 + 3
∴ k = 1
Equation of Straight Lines ~ 113
EXERCISE : 4.3
1. a. Find the equation of straight line parallel to the X-axis at a distance of 6 units above it.
b. Find the equation of straight line parallel to the X-axis at a distance of 3 units below it.
2. a. Find the equation of straight line parallel to the Y-axis at a distance of 2 units to its right side.
b. Find the equation of straight line parallel to the Y-axis at a distance of 6 units to its left side.
3. a. Find the equation of straight line parallel to the X-axis and passing through the point (4, 0).
b. Find the equation of straight line parallel to the X-axis and passing through the point (– 3, – 2).
4. a. Find the equation of straight line parallel to the Y-axis and passing through the point (– 2, 5).
b. Find the equation of straight line parallel to the Y-axis and passing through the point (3, – 2).
5. Find the slope of the line having following angle of inclination.
a. 30° b. 45° c. 60° d. 120°
6. Find the angle of inclination of the line having following slopes.
a. 1 b. 3 c. ∞ d. – 1
3
7. a. Find the slope of a straight line parallel to X-axis and at a distance of 3 units above it.
b. Find the slope of a straight line parallel to Y-axis and at a distance of 5 units to its right side.
c. Find the slope of a straight line parallel to X-axis and passing through the point (5, 0).
d. Find the slope of a straight line parallel to Y-axis and passing through the point (0, 4).
8. a. Find the slope of the line passing through the points (4, 5) and (2, 3).
b. Find the slope of the line passing through the points (– 1, – 3) and (– 6, 7).
c. Find the slope of the line passing through the points 23, 4 and –2 1, – 1 .
3 3
9. a. If the slope of the straight line passing through the points A(2, 4) and B(7, b) is 52, find the value
of b.
b. If the slope of the straight line joining the points (x, 5) and (– 6, 1) is 12, find the value of x.
10. Prove that the following sets of points are collinear.
a. (3, – 2), (7, 6) and (9, 10) b. (– 2, – 3), (0, 1) and (4, 9)
11. a. If (1, – 4), (5, 0) and (a, 3) lie on a same straight line, find the value of a.
b. If the three points (3, – 2), (5, 2) and (7, b) are collinear, find the value of b.
12. a. If the points (a, 0), (0, b) and (1, 1) are collinear, prove the 1 + 1 = 1.
a b
b. Prove that the three points A(p, 0), B(0, q) and R(3p, – 2q) are collinear.
114 ~ Perfect Optional Mathematics Class 9
4.4 Slope Intercept Form of Equation of Straight Line
Slope Intercept Form
To find the equation of straight line whose slope is 'm' and which Y P(x, y) B
cuts off a given intercept 'c' on the Y– axis.
N
Let line AB intersects X– axis at A and Y– axis at C such that Cq MX
BAX = q and Y intercept (OC) = c .
Slope of line AB (m) = tan q q
Let P(x, y) be any point on the line AB, draw PM⊥OX and X' AO
CN⊥PM, then CN = OM = x P
PN = PM – MN = PM – OC = y – c Y'
From right angled triangle PNC,
tan θ = PN or, m = y – c Co-ordinate
CN x
∴ y = mx + c
This is true for every point on the line AB, so it is the required equation of AB.
Alternately,
Prove that the equation of straight line having slope m and Y-intercept c is y = mx + c.
Proof:
Let a straight line cuts the Y-axis at point Q making Y-intercept of Y
OQ = c units. P(x, y)
Q(0, c)
Now, the co-rdrinates of point Q be (0, c) and making angle of inclination
QRO = q as shwon in the figure.
Suppose, Q(0, c) = (x, y) and any moving point on the line P(x, y) = (x2, y2),
then slope of QP
y2 – y1 X' q X
Slope (m) = x2 – x1 P O
R Y'
–c
or, m = y –0
x
or, y – c = mx
or, y = mx + c
When a line passes through the origin, its Y-intercept (c) = 0 and the equation of the line is y = mx.
Example 1:
Find the equation of straight line having slope (m) = 1 and Y-intercept (c) = 0.
Equation of straight line in slope intercept form is y = mx + c
Here, m = 1 and c = 0, then
Equation is y = 1 × x + 0
or, y = x is the required equation.
Slope Intercept Form of Equation of Straight Line ~ 115
Example 2:
Find the equation of a straight line which is inclined to the X– axis at an angle 30° and cuts off an
intercepts 3 from Y– axis.
Here, angle of the inclination of the line (q) = 30°, slope of the line (m) = tan q = tan 30° = 1 and
3
Y-intercept of the line (c) = 3
The equation of straight line is
y = mx + c
or, y= 1
3 x+3
or, 3y = x + 3 3
or, x – 3y + 3 3 = 0
∴ The required equation of the line is x – 3 y + 3 3 = 0.
Example 3:
Find the equation of straight line AB from the given figure. Y
A
Let ORQ = q in DORQ
Q
q + 30° + 90° = 180°
or, q = 180° – 120° = 60° 30° 3 units
The slope of the line AB is (m) = tan q = tan 60° = 3 and Y-intercept OX
(c) = OC = 3 units q Y'
The equation of straight line AB is y = mx + c X' R
B
or, y = 3x + 3 is the required equation of straight line.
Example 4:
Find the equation straight line cuting an intercept – 3 from Y-axis and inclined at an angle of
1
tan–1 3 with X-axis.
1
Here, the angle of inclination = tan–1 3
1
or, q = tan–1 3
or, tan q = 1
3
1
or, m = 3 and Y-intercept is (c) = – 3.
Now the equation of straight is y = mx + c
1
or, y= 3 x + (– 3)
or, y = x – 3 3
3
\ 3y = x – 3 3 is the required equation.
116 ~ Perfect Optional Mathematics Class 9
Example 5:
Find the equation of straight line passing through the points (0, – 2) and (2, 1).
The straight line passes through the points (0, – 2) and (2, 1).
As (0, – 2) is a point on the Y– axis, its y-intercept (c) = – 2 and the
slope of line (m) = y2 – y1 = 1 + 2 = 3
x2 – x1 2 – 0 2
Hence, the equation of the straight line y = mx + c
or, y = 32 x + (– 2)
or, y = 3x – 4
2
or, 2y = 3x – 4
or, 3x – 2y – 4 = 0
This is the required equation of the straight line.
Example 6: Co-ordinate
Find the equation of straight line passing through (4, 3) and having slope 2.
Here, the slope of the given straight line (m) = 2.
Let the equation of straight line is y = mx + c
or, y = 2x + c
As the given straight line is passing through (4, 3), we get
3 = 2 × 4 + c
or, 3 = 8 + c
or, 3 – 8 = c
or, – 5 = c
Hence, the equation of straight line is y = 2x – 5.
Example 7:
Find the slope and Y-intercept of the line having equation 3x + 4y = 12.
Here, the equation of given line is 3x + 4y = 12.
or, 4y = – 3x + 12
or, y = – 3x + 12
4
– 3x 12
or, y = 4 + 4
or, y = –4 3 x + 3
Comparing this equation with y = mx + c, we have
rhe slope of the line is (m) = – 3 and Y-ntercept (c) = 3.
4
Slope Intercept Form of Equation of Straight Line ~ 117
EXERCISE : 4.4
1. Find the equation of the following straight lines with :
a. slope 1 and Y-intercept 4 b. slope – 1 and Y-intercept 6.
3 2
2. a. Find the equation of a straight line which is inclined to the X-axis at an angle of 45°and cuts off
an intercept – 2 from Y-axis.
b. Find the equation of a straight line which is inclined to the X-axis at an angle of 120° and cuts
1
off an intercept 2 from Y-axis.
3. Find the equation of given straight line AB.
a. Y b. Y c. Y
R 120° BQ
B O
C
X' Q X
2 units 3 units 60°
2 units
X' 30° X PB X' P AX
Y' Y'
DO
P Y'
4. a. Find the equation of a straight line passing through origin and having slope – 3.
b. Find the equation of a straight line passing through origin and having angle of inclination 150°.
5. a. Find the equation of a straight line making angle of inclination tan–1( 3) and cutting off intercept
– 2 on Y-axis.
b. Find the equation of a straight line cutting off an intercept of 5 from the axis of Y and making
1
angle of inclination tan–1 2 with X-axis.
6. a. Find the equation of straight line passing through the points (0, 5) and (2, 3).
b. Find the equation of straight line passing through the points (0, – 3) and (1, 4).
7. a. Find the equation of straight line passing through the point (– 7, – 4) and having slope 3.
b. Find the equation of straight line passing through the point (5, – 4) and having slope 6.
c. Find the equation of straight line passing through the point (0, 4) and making an angle of 120°
with X-axis in the anticlockwise direction and prove that the line passes through the point ( 3, 1).
8. a. Find the equation of straight line making an angle of 60° with X-axis in positive direction and
intersecting the Y-axis at a distance of 5 units.
b. Find the equation of straight line cutting an intercept of – 5 units on Y-axis and equally inclined to the axes.
9. a. Find the equation of the sides of a right angled isosceles triangle whose vertex is at (4, 5) and
equation of the base side is y = 0.
b. Find the equation of the sides of a right angled isosceles triangle whose vertex is at (2, 3) and
equation of the base is x = 0.
10. a. Find the equation of straight line making an angle of 30° with X-axis in anticlockwise direction
and passing through the mid-point the line segment joining the points (– 2, 3) and (8, 5).
b. Find the equation of straight line making slope 1 and passing through the mid-point of the line
segment joining the points (– 2, 3) and (4, 1).
11. a. Find the slope and Y-intercept of the line having equation x – 3y = 3 3.
b. Find the slope of Y-intercept of the line having equation ax + by + c = 0.
118 ~ Perfect Optional Mathematics Class 9
4.5 Double Intercept Form of Equation of Straight Line
Introduction
To find the equation of a straight line which cuts off given Y
intercepts 'a' and 'b' on the X– axis and Y– axis respectively. B(0, b)
Let AB be a straight line which intersects line X– axis at A P(x, y)
and Y– axis at B such that X– intercept (OA) = a, Y– intercept A(a, 0) X
(OB) = b.
Hence, the co-ordinates of A and B are (a, 0) and (0, b) X' O
respectively. Y'
Let P(x, y) be any point on the line AB, then
slope of AP = slope of AB
i.e. y – 0 = b – 0 Co-ordinate
x – a 0 – a
or, bx – ab = – ay
or, bx + ay = ab
or, bxa+b ay = ab
ab
∴ x + y = 1
a b
This is the required equation of straight in double intercepts form.
Example 1:
Find X-intercept and Y-intercept and the equation of line AB from the following figures.
YY
A D
D
6 units 60° 12 units
X' O 8 units CX X' O AX
B
a. Y' b. Y'
a. From figure, X-intercept (a) = OC = 8 units Y
Y-intercept (b) = OD = 6 units A
D
The equation of straight line AB is given by
6 units
xa + y = 1 CX
b X' O B
or, 8x + y 8 units
6 = 1
Y'
Double Intercept Form of Equation of Straight Line ~ 119
or, 3x4+84y = 1
or, 3x + 4y = 24
This is the required equation of straight line AB.
b. In right angled triangle OAB, OBA = 60° Y
Hence, cos 60° = OB
AB
or, 21 = OB B
12
12 60° 12 units
2
\ Y-intercept (b) = OB = = 6 units X' O AX
Using Pythagoras theorem in DOAB,
AB2 = OA2 + OB2 Y'
or, 122 = OA2 + 62
or, 144 – 36 = OA2
or, OA = 108 = 36 × 3 = 6 3
X-intercept (a) = OA = 6 3 units
Equation of straight line AB is given by
xa + y = 1
b
or, 6x3 + y = 1
6
or, x +6 3y = 1
3
or, x + 3y = 6 3 is the required equation.
Example 2:
Find the equation of line which passes through the point (4, – 1) and makes intercepts on the axes
equal in magnitude but opposite in sign.
Let the X-intercept (a) = k and Y-intercept (b) = – k
The equation of straight line in double intercepts form is x + y = 1
a b
or, kx + y = 1
– k
or, x – y = k … … … (1)
The straight line having equation (1) passes through the point (4, – 1), so this point must satisfy the equation (1),
i.e. 4 – (– 1) = k
or, 4 + 1 = k
or, k = 5
Hence, the required equation of straight line is x – y = 5.
120 ~ Perfect Optional Mathematics Class 9
Example 3:
Find the equation of the straight line which passes through the point (2, – 1) and the portion
intercepted by the axes is divided by this point in the ratio 1:2.
Let the line passes through the point (2, – 1) meets the X– axis at A(a, 0) and Y– axis at B(0, b).
As the point M divides AB in the ratio 1:2, we have,
(x, y) = m1x2 + m2x1 , m1y2 + m2y1 Y
m1 + m2 m1 + m2 O
A(a, 0)
Where, (x, y) = (2, – 1), (x1, y1) = (a, 0), (x2, y2) = (0, b), m1 = 1 and m2 = 2 X' X
or, (2, – 1) = 1 × 0 + 2 × a, 1 × b + 2 × 0 M(2, – 1)
1 + 2 1 + 2
or, (2, – 1) = 23a, b B(0, b)
3 Y'
Hence, 2a = 2 and b = – 1
3 3
∴ a = 3 ∴ b = – 3
So, the equation of straight line in double intercepts form is x + y = 1 Co-ordinate
a b
or, x3 + y
– 3 = 1
or, x –3 y = 1
∴ x – y = 3. This is the required equation.
Example 4:
Find the equation of the straight line which passes through the point (1, 2) and makes intercepts on
the axes, the sum of whose length is 6.
As the sum of the intercepts made by the straight line is 6, a + b = 6 where a and b are the intercepts
of the line, so b = 6 – a.
The equation of the straight line in the double intercepts is x + y = 1
or, xa + 6 =1 a b
y
–a
As this line passes through the point (1, 2), it must satisfy the equation of straight line,
or, 1a + 6 2 a = 1
–
or, 6a–(6a–+a2)a = 1
or, 6 + a = a(6 – a)
or, 6 + a = 6a – a2
or, a2 – 5a + 6 = 0
or, a2 – 3a – 2a + 6 = 0
or, (a – 3) (a – 2) = 0
Either a – 3 = 0 or, a – 2 = 0
∴ a = 3 ∴ a = 2
Double Intercept Form of Equation of Straight Line ~ 121
Case I, when a = 3, b = 6 – 3 = 3
Then the equation of the straight line is
3x + y = 1
3
or, x + y = 3
Case II, when, a = 2, b = 6 – 2 = 4
Then the equation of the straight line is
x2 + y = 1
4
or, 2x4+ y = 1
or, 2x + y = 4
The required equations of straight lines are x + y = 3 and 2x + y = 4.
Example 5:
Find the equation of the straight line which makes intercepts twice the intercepts made by the line
2x – 3y + 5 = 0 on the axes.
Here, the given equation is 2x – 3y + 5 = 0
or, 2x – 3y = – 5
or, 2x–– 53y = 1
or, – 25 x + 35 y = 1
or, –x 52 + y = 1
5 x y
3 a b
Comparing this equation with + = 1, we get
X-intercept (a) = – 25 and Y-intercept (b) = 5
3
Now, X-intercept of the required line (a1) = 2 × – 52 = – 5 and
Y-intercept of the required line (b1) = 2 × 5 = 10
3 3
Hence, the equation of line having the intercepts twice the intertercepts of the line 2x – 3y + 5 = 0 is
xa + y = 1
b
or, –x 5 y
+ 10 = 1
3
or, –x 5 + 3y = 1
10
or, – 2x10+ 3y = 1
or, – 2x + 3y = 10 is the required equation.
122 ~ Perfect Optional Mathematics Class 9
Example 6:
Find the length of portion intercepted between the axes of the line passing through points (2, 3)
and (6, – 3). x y
Here, a b
let the equation of given straight line in double intercept form is + = 1 ... .. (1)
As this line passes through point (2, 3), we get
2a + 3 = 1 ... ... (2)
b
Again, line (1) passes through point (6, – 3), then
6a – 3 = 1 ... ... (3)
b
Adding equation (2) and (3) , we get Now, from equation (2),
2a + 3 + 6 – 3 = 1 + 1 2a + 3 = 1
b a b b
or, 24 + 3
or, 2a + 6 = 2 b = 1
a
or, 21 + 3
or, 2 +a 6 = 2 b = 1
or, 8a = 2 or, 3b = 1 – 1 Co-ordinate
2
or, 2a = 8 or, 3b = 2 – 1 = 1
2 2
or, a = 4. or, b = 6.
So, X-intercept of the given line (a) = 4 and its Y-intercept (b) = 6.
The line cuts X-axis at point A(4, 0) and Y-intercept at point B(0, 6).
The length of the portion intercepted between the axes is
AB = (x2 – x1)2 + (y2 – y1)2
or, AB = (0 – 4)2 + (6 – 0)2
or, AB = 16 + 36
\ AB = 52 = 2 13 units
EXERCISE : 4.5
1. a. Find the equation of straight line having X-intercept (a) = 3 and Y-intercept (b) = 2.
b. If a = X-intercept = 12 and b = Y-intercept = 10 for a line, find the equation of the line.
2. Find X-intercept and Y-intercept and the equation of line AB in the following.
a. Y b. Y c. Y
A B A
D D
30° 12 units
2 units 10 units
C
C
X' 3 units X X' A X X' 8 units X
B Y' B
Y' Y'
Double Intercept Form of Equation of Straight Line ~ 123
3. a. Find the equation of the line cutting X-axis at point (3, 0) and Y-axis at point (0, – 3).
b. Find the equation of the line cutting X-axis and Y-axis at points 32, 0 and 0, 3 recpectively.
4
4. Find the equation of straight lines with following intercepts.
a. X– intercept = 2 and Y– intercept = 3
b. X– intercept = – 3 and Y– intercept = 4
5. a. Find the equation of a straight line passing through the point (– 6, 8) and making equal intercepts
on the axes.
b. Find the equation of straight line which passes through the point (4, 3) and has intercepts on the
axes.
i. equal in magnitude and sign
ii. equal in magnitude but opposite in sign.
c. Find the equation of the straight line which passes through the point (3, 2) and makes an intercept
on the X– axis is twice as long as that on the Y– axis.
6. a. Find the equation of the line through (2, – 3) so that the segment of the line intercepted between
the axes is bisected at this point.
b. Find the equation of the line which passes through the point (2, – 2) and the portion of the line
intercepted between the axes is divided by the point in the ratio 1:2.
c. Find the equation of the straight line which passes through point (3, 1) and the portion of the line
intercepted between the axes is divided by the point in the ratio 1:3.
d. Find the equation of the straight line which passes through the point (– 4, 3) such that the portion
of it between the axes is divided at (– 4, 3) in the ratio 5:3.
7. a. Find the equation of the straight line which passes through the point (2, 2) and makes intercepts
on the axes the sum of whose length is 9.
b. Find the equation of straight line passes through the point (6, 2) and cuts off Y– intercept greater
than its X– intercept by 7.
c. a and b are the intercepts made by a line on the co-ordinate axes. If 2a + b = 6 and the line passes
through the point (1, 1), find the equation of the straight line.
d. Find the equation of the straight line which passes through the point (2, 1) and makes intercepts
on the axes whose product is 8.
8. a. Find the equation of straight line which makes intercepts twice the intercepts made by the line
4x + 5y – 20 = 0 on the axes.
b. Find the equation of straight line which makes intercepts half of the intercepts made by the line
2x + 3y = 12 on the axes.
9. a. Find the length of the portion intercepted between the axes by the line passing through the points
(4, 0) and (2, 3).
b. Find the length of the portion intercepted between the axes by the line passing through the points
(1, 3) and (5, – 1).
124 ~ Perfect Optional Mathematics Class 9
4.6 Perpendicular Form of Equation of Straight Line
Normal or Perpendicular form of Equation of Straight Line
To find the equation of the straight line in terms of the length Y
of the perpendicular 'p' from origin upon it and the angle 'α' the
perpendicular makes with the positive direction of the X– axis. B
M
Let AB be the line which meets X-axis at A and Y-axis at B. P(x, y)
Draw OM^AB so that OM = p and MOA = α. Suppose P(x, y)
be any point on AB and draw PN⊥OX, then ON = x, PN = y. a a A
In right angled triangles OMA and PNA, X' O N(x, 0) X
MOA = NPA = α (why ?) Y'
From right angled DPNA,
tan a = NA ⇒ NA = y tan a ... ... ... (1) Co-ordinate
PN
From right angled ∆MOA,
cos a = OM
OA
OM
or, cos a = ON + NA
or, cos a = x + p α
y tan
or, x cos a + y tan a cos a = p
∴ x cos α + y sin α = p
This is the required equation of straight line in the normal form.
Alternately, Y
Let AB be the line which meets X-axis at A and Y-axis (0, p cosec a)
at B. Draw OM^AB so that OM = p, MOA = α, then
BOM = 90° – α and BM
p
OBM = 90° – (90° – α) X' O a (p sec a, 0)
AX
= 90° – 90° + α = α Y'
From ∆BOM, From ∆OAM
sin α = OOMB cos α = OM
OA
or, sin α = OpB p
or, OA= cos a = p sec α
or, OB = p a = p cosec a The co-ordinates of A is (p sec a, 0).
sin
The co-ordinates of point B is (0, p cosec a).
Perpendicular Form of Equation of Straight Line ~ 125
Equation of line AB is
OxA + y = 1
OB
or, p sexc a + p y a = 1
cosec
or, x cos α + y sin α = 1
p p
or, x cos α + y sin α = 1
p
or, x cos a + y sin α = p, this is the required equation.
Example 1:
Find the equation of the straight line having perpendicular distance from origin 4 units and the
perpendicular makes an angle of 30° with X-axis. Y
B
Here, the perpendicular distance of the line from origin (p) = 4 unts.
Angle made by the perpendicular with X-axis in positive direction is
a = 30°.
Now, the equation of straight line in perpendicular form is M
x cos a + y sin a = p
or, x cos 30° + y sin 30° = 4 4
X' O 30° A X
or, x× 3 + y × 1 = 4 Y'
2 2
or, 3x2+ y = 4
or, 3x + y = 8 is the required equation.
Example 2:
Find the equation of the straight line having p = 3 units and a = 45°.
Here, the perpendicular distance of the line from origin (p) = 3 units.
Angle made by the perpendicular with X-axis in positive direction is a = 45°.
Now, the equation of straight line in perpendicular form is
x cos a + y sin a = p
or, x cos 45° + y sin 45° = 3
or, x× 1 + y × 1 = 3
2 2
or, x +2y = 3
or, x + y = 3 2 is the required equation.
126 ~ Perfect Optional Mathematics Class 9
Example 3:
Find the equation of line AB from the given figure. Y
A
Let a be the angle made by OM with X-axis in positive direction i.e. MOX.
Q
From triangle OPM, MOX = MPO + PMO M
or, a = 30° + 90° = 120° OX
Y'
And perpendicular distance (p) = OM = 3 units X' 30° 3 units
The equation of straight line AB is
x cos a + y sin a = p B P
or, x cos 120° + y sin 120° = 4
or, x× – 12 +y× 3 = 3
2
or, – x + 3y = 3
2
or, – x + 3y = 6 is the required equation of line AB.
Example 4: Co-ordinate
Find the equation of the straight line whose perpendicular distance from origin is 1 and passes
through the point (2, 0). Y
B(0, b)
Let the line AB cuts the X– axis at A(2, 0) and the Y– axis at
B(0, b), OM is drawn perpendicular to AB where OM = 1. From
right angled triangle OMA,
cos α = OM M
OA
1
or, cos α = 1 X' a A(2, 0) X
2
or, cos α = cos 60° O
∴ α = 60° Y'
Here, OM = p = 1, hence the equation of straight line AB is
x cos α + y sin α = p
or, x cos 60° + y sin 60° = 1
or, 12 x + 23 y = 1
∴ x + 3 y = 2, this is the required equation.
Example 5:
Find the equation of the straight line such that the length of the perpendicular from origin upon
it is 4 2 units and the angle made by the perpendicular with X-axis in positive direction is tan–1(1).
The perpendicular distance of the line from origin (p) = 4 2 units.
The angle made by the perpendicular with X-axis in positive direction (a) = tan–1(1).
As we know tan 45° = tan 255° = 1, tan–1(1) = 45° or 225°
Perpendicular Form of Equation of Straight Line ~ 127
Case I, Case II,
Taking tan a = 1 = tan 45° Taking tan a = 1 = tan 225°
or, a = 45° or, a = 225°
The equation of straight line in perpendicular form is The equation of straight line in perpendicular form is
x cos a + y sin a = p x cos a + y sin a = p
or, x cos 45° + y sin 45° = 4 2 or, x cos 225° + y sin 225° = 4 2
or, x× 1 + y × 1 = 4 2 or, x× – 1 +y× – 1 =4 2
2 2 2 2
or, x +2y = 4 2 or, – x – y = 4 2
or, x + y = 4 2 × 2 = 4 × 2 2
or, – x – y = 4 2 × 2 = 4 × 2
\ x + y = 8 is the required equation. or, – x – y = 8
\ x + y = – 8 is the required equation.
Example 6:
Find the equation of the straight lines which are at a distance of 3 units from the origin and pass
through the point (2, 0).
The perpendicular distance of the line from origin (p) = 3 units.
The equation of the line in perpedicular form is
x cos a + y sin a = p
As the line passes through the point (2, 0), we get
or, 2 cos a + 0 × sin a = 3
or, 2 cos a = 3
or, cos a = 3
2
or, cos a = cos 30° or cos 330°
Case I, Case II,
Taking a = 30° Taking a = 330°
The equation of straight line in perpendicular form is The equation of straight line in perpendicular form is
x cos a + y sin a = p x cos a + y sin a = p
or, x cos 30° + y sin 30° = 3 or, x cos 330° + y sin 330° = 3
or, x× 3 + y × 1 = 3 or, x× 3 + y × – 21 = 3
2 2 2
or, 3 x2+ y = 3 or, 3 x2– y = 3
\ 3x + y = 2 3 is the required equation. \ 3x – y = 2 3 is the required equation.
128 ~ Perfect Optional Mathematics Class 9
EXERCISE : 4.6
1. Find the equation of the straight lines in the following conditions where p denotes the length of
perpendicular drawn from the origin and a denotes the angle of inclination of the perpendicular with
the positive X– axis.
a. p = 2, α = 30° b. p = 1, α = 60° c. p = 32, α = 150° d. p = 3, α = 135°
2. a. The length of perpendicular drawn from the origin upon the line is 5 and the inclination of the
perpendicular to the X– axis is 30°. Find the equation of the line.
b. Find the equation of straight line whose perpendicular distance from origin is 3 and the inclination
of the perpendicular to the X-axis is 240°.
3. a. The straight line AB cuts the X-axis at P and Y-axis Y
b. at Q. OM is drawn perpendicular to AB from
the origin O. If OM = 3 units and OPQ = 30°, B Y
determine the equation of the line AB. MQ
B
From the given figure, find the equation of straight X' P 30° 3 O X 4C
line AB. Given OC = 4 units and COA = 60°, 3(a)
A Q.N. Y' 60°
then prove that it passes through the point (5, 3 ).
X' O A X Co-ordinate
Y' Q.N. 3(b)
c. From the given figure, find the equation of straight Y
d. line AB, where OP = 4 units and AEX = 30°. B
The perpendicular distance from the origin to the Y 4 C
straight line is 4 units and the line makes an angle B
150° with the positive X– axis as shown in figure, D 150°
P X' O X
i. Find the equation of the straight line. A
4 Y'
ii. Does the line pass through ( 3 , – 1)? Q.N. 3(d)
X' O E 30° X
Y' Q.N. 3(c) A
4. a. The length of perpendicular drawn from the origin upon the line is 2 units and the perpendicular
makes 45° with the X– axis. Does the line pass through the point ( 2, 2 )?
b. The length of perpendicular from the origin on the line is 2 3 and the perpendicular makes an
angle 150° with the X– axis. Find the equation of the of the straight line and prove that the line
passes through the point (2, 6 3 ).
5. a. Find the equation of the straight line whose distance from origin is 2 units and passes through the
point (2 2 , 0).
b. The slope of a line is – 1 and its perpendicular distance from the origin is 5 units. Find the
equation of the line.
6. a. Find the equation of straight line which is at a distance of 5 units from origin, so that perpendicular
distance from the origin to the line makes an angle of tan–1 3 with X-axis in positive direction.
4
b. Find the equation of straight line which is at a distance of 3 units from origin, so that perpendicular
origin line an angle tan–1 12
distancefrom the to the makes of 5 with positive direction of X-axis.
Perpendicular Form of Equation of Straight Line ~ 129
4.7 Reduction of the Linear Equation Ax + By + C = 0
Reduction to Slope Intercept Form
The given equation of a line is
Ax + By + C = 0
or, By = – Ax – C
or, y = – A x – C ... ... ... ... (1)
B B ... ... ... ... (2)
Which is of the form y = mx + c
Comparing (1) and (2), we have
Slope (m) = – A = – coeff. of x
B coeff. of y
and Y– intercept (c) = – C = – constant term
B coeff. of y
Reduction to Double Intercepts Form
The given straight line is
Ax + By + C = 0
or, Ax + By = – C
or, –A Cx + By = – C
– C – C
or, – ACx/A + y = 1 ... ... ... (1)
– C/B ... .... ... (2)
Which is of the form x + y = 1
a b
Comparing (1) and (2)
X– intercept (a) = – C
A
Y– intercept (b) = – C
B
Reduction to Normal (Perpendicular) Form
The given equation of a line is
Ax + By + C = 0 ... ... ... ... (1)
The equation of a line in normal form is
x cos α + y sin α = p ... ... ... ... (2)
Straight lines (1) and (2) will represent the same straight line if the coefficients of corresponding terms are
proportional.
i.e. coAs a = sin a = – P = k (say)
B C
130 ~ Perfect Optional Mathematics Class 9
∴ cos α = Ak, sin α = Bk, p = – Ck
We have, cos2α + sin2α = 1
or, k2(A2 + B2) = 1
or, k=± 1
A2 + B2
The sign of the value of k should be chosen such that p is always positive.
The equation of a line in the normal form is
± Ax B2 ± By B2 = ± C
A2 + A2 + A2 + B2
If C > 0, the required equation is
Ax B2 + By B2 = C
A2 + A2 + A2 + B2
If C < 0, the required equation is
– Ax – By B2 = – C B2 ... ... ... (3)
A2 + B2 A2 + A2 +
Note: To reduce the equation of a straight line to the normal form, each term must be divided by A2 + B2 Co-ordinate
making right hand side positive.
Example 1:
Find the slope and the Y– intercept of the equation 3 x – 4y + 6 = 0.
3 x – 4y + 6 = 0
or, 4y = 3 x + 6
or, y= 43 x + 3
2
Comparing this equation with, y = mx + c
Slope (m) = 3 and Y– intercept (c) = 3
4 2
Example 2:
Find the intercepts on the axes of the equation 5x + 3y + 12 = 0.
The given equation of a straight line is
5x + 3y + 12 = 0
or, 5x + 3y = – 12
or, –5 1x2 + 3y = – 12
– 12 – 12
or, – x12 y
+ – 4 = 1
Comparing 5 equation with x + y = 1, we have
this a b
X– intercept (a) = – 12
5
Y– intercept (b) = – 4.
Reduction of the Linear Equation Ax + By + C = 0 ~ 131
Example 3:
Find the length of perpendicular from the origin on the line x – 3 y + 7 = 0 and the angle made by
the perpendicular with positive X– axis.
The given equation of straight line is
x – 3y + 7 = 0
or, x – 3y = – 7
or, – x + 3y = 7
Dividing both sides by (– 1)2 + ( 3)2 = 2
–2 1 x + 23 y = 7
2
Comparing this equation with x cos a + y sin a = p , we have
cos α = – 1 = cos 120° or cos 300°
2
and sin α = 3 = sin 60° or sin 120° and p = 7
2 2
Since, cos α is negative and sin α is positive, α lies in the second quadrant and α = 120°, p = 7
2
Example 4:
If the straight line 2x – 3y + 10 = 0 intersect the X– axis at A and Y– axis at B, find the area of DAOB.
As this line intersect X– axis at A and Y– axis at B. We have to reduce it into double intercept form to
have OA and OB for the calculation of area of ∆AOB.
The given equation of straight line is Y
2x – 3y + 10 = 0
or, 2x – 3y = – 10 B(0, b)
or, – 2x + 3y = 10
or, –1 20x + 3y = 10 X' A(a, 0) OX
10 10 Y'
or, –x 5 + y =1
10
3 x y
a b
Comparing this equation with + = 1, we have
X– intercept (a) = OA = – 5
Y– intercept (b) = OB = 10
3
Area of ∆AOB = 1 × OA × OB
2
1 10
= 2 × (– 5) × 3
132 ~ Perfect Optional Mathematics Class 9
= – 25 sq. units.
3
25
Area of triangle can not be negative, hence ∆AOB = 3 sq. units.
Example 5:
P and Q are two points on the line x + y – 1 = 0 and are at a distance of 5 units from the origin. Find
the area of triangle OPQ.
Here, the given line is x + y – 1 = 0
or, x + y = 1 Y
Dividing both sides of the equation by (1)2 + (1)2 = 1 + 1 = 2, we have P1
R2
x +y = 1
2 2 5Q
or, x2 + y = 1 X' O 5 X
2 2
or, 12 x + 1 y = 1 Y' Co-ordinate
2 2
Comparing this equation with x cos a + y sin a = p, we get
p= 1 = OR and OP = 5 = OQ
2
From right angled triangle ORQ, OR2 + RQ2 = OQ2
or, 12 2 + RQ2 = (5)2
or, 21 + RQ2 = 25
or, RQ2 = 25 – 1
2
50 – 1 49
= 2 = 2
\ QR = 49 = 7
2 2
From isosceles triangle OPQ, PQ = 2QR = 2 × 7 = 7 2 units
2
1
Area of DOPQ = 2 × OR × PQ
= 1 × 1 × 7 2
2 2
= 7 sq. units
2
Reduction of the Linear Equation Ax + By + C = 0 ~ 133
EXERCISE : 4.7
1. Reduce the following equation to y = mx + c and find slope (m) and Y-intercept (c).
a. x + 4 = 3y b. 3x + 4y = 12
c. y = 0 d. y = 3
e. x + 3y – 4 = 0 f. 2 3x – 6y = 9
2. Reduce the following equation to x + y = 1 and find X-intercept (a) and Y-intercept (b).
a b
a. 2x + 3y = 9 b. 3x + y = 2 3
c. x – y + 2 = 0 d. x – 3y = 12
3. Reduct the following equation to x cos a + y sin a = p and find a and p.
a. 2x + 23 y = 8 b. –x – y = 5 2
c. 3 x – y + 2 = 0 d. x + 3 y = – 4
4. a. Reduce the equation of straight line ax + by + c = 0 in the form of y = mx + c and also find m and c.
b. Find the slope and Y-intercept of line x + y = 1.
a b
c. Reduce the equation of the line x cos a + y sin a = p in the form of y = mx + c and also find m and c.
5. a. Reduce the line with equation y = mx + c in the double intercepts form and find a and b.
b. Reduce the equation of line x cos a + y sin a = p in the form of x + y = 1 and find a and b.
a b
6. a. Find the perpendicular form of the line with equation y = mx + c.
b. Find the perpendicular form of the line with equation x + y = 1.
a b
7. a. Find the intercepts of the line having equation x cos q + y sin q = 2 sin q cos q.
b. If p be the perpendicular dropped from the origin on the line x + y = 1, prove that 1 + 1 = p12.
a b a2 b2
8. a. Reduce the following equations to slope intercept form, double intercepts form and perpendicular form:
i. 3x – 4y = 5 ii. 2 3 x + 2y = 11
b. Find the length of the perpendicular from the origin on the line 3 x + y + 6 = 0 and the angle
9. a. made by the perpendicular with the X– axis.
b.
If the line 2x + y = 6 cuts X-axis at A and Y-axis at B, find the area of DAOB.
10. a.
b. If the straight line 2x – 5y = 12 cuts the X– axis and Y– axis at the points A and B respectively,
find the area of DAOB.
A and B are two points on the line 2x + y = 2 and are at distance of 10 units from the origin.
Find the area of ∆OAB.
If a straight line passes through the points A(– 3, 4) and B(5, – 2), find the area of triangle OAB
where O is the origin.
134 ~ Perfect Optional Mathematics Class 9
4.8 Special Forms of Lines
Introduction
We can obtain the equation of a straight line if
a. Its slope and a point through which it passes are known.
b. Two points through which it passes are known.
Slope Point Form Y
C
To find the equation of a straight line passing through a given
point (x1, y1) a given slope (m). P(x, y)
Let q be the angle of inclination of the straight line AC which B D(x1, y1)
passes through the point D(x1, y1).
Hence, the slope of the line AC is m = tan q. If P(x, y) be any point X' q X Co-ordinate
on the line, then we have
P AO
Slope AC = Slope of DP
Y'
or, m= y – y1
x – x1
∴ y – y1 = m(x – x1)
This is the required equation of straight line.
Two Points Form
To find the equation of a straight line passing through two given points (x1, y1) and (x2, y2).
Let AB be the straight line passing through the points M(x1, y1) and Y
N(x2, y2). If P(x, y) be any point on the straight line AB. N(x2, y2)
As the point M, N and P lie on the same straight line.
Slope of MP = slope of MN P(x, y)
y– y1 = y2 – y1 X' O M(x1, y1) X
or, x – x1 x2 – x1 B
∴ y – y1 = y2 – y1 (x – x1), this is the required equation. Y'
x2 – x1
Example 1:
Find the equation of straight line passing through (– 2, 3) and having angle of inclination 60°.
The angle of inclination of the line (q) = 60°
Slope of the line (m) = tan θ = tan 60° = 3
As the line passes through the point (– 2, 3), the equation of the straight line is
y – y1 = m(x – x1)
Special Forms of Lines ~ 135
or y – 3 = 3 (x + 2)
or y – 3 = 3 x + 2 3
∴ 3 x – y + 3 + 2 3 = 0
This is the required equation of the straight line.
Example 2:
If (– 2, 3), (1, 0) and (3, 8) are the vertices of a triangle, find the equation of the median drawn
through the first vertex.
Let A(– 2, 3), B(1, 0) and C(3, 8) be the vertices of the given triangle, the co-ordinates of the mid-point
of side BC is D 1 + 3, 0 + 8 i.e. (2, 4).
2 2
A(– 2, 3)
For the equation of median AD, A(– 2, 3) = (x1, y1), D(2, 4) = (x2, y2),
the equation of AD is
y – y1 = y2 – y1 (x – x1) B(1, 0) D(2, 4) C(3, 8)
x2 – x1
or, y – 3 = 4 – 32 (x + 2)
2 +
or, 4(y – 3) = x + 2
∴ x – 4y + 14 = 0, this is the required equation.
Example 3:
Find the ratio in which the line joining the points (2, 0) and (– 3, 5) divide the line segment joining
the points (– 1, – 1) and (5, 5).
Let A(2, 0) and B(– 3, 5) be two points then equation of AB is
y – y1 = y2 – xy11 (x – x1)
or, x2 –
y – 0 = –5 3––02 (x – 2)
or, y = – 1(x – 2)
or, x + y = 2 ... ... ... (1)
If the line AB interects the line segment joining C(– 1, – 1) and D(5, 5) at M such that (CM:MD = m1:m2).
The co-ordinates of point M is
M m1mx12 + mm22x1, m1y2 + m2y1
+ m1 + m2
i.e. M m1 × 5 + m2 × – 1 m1 × 5 + m2 × – 1
m1 + m2 , m1 + m2
136 ~ Perfect Optional Mathematics Class 9
i.e. M 5m1 – mm22, 5m1 – m2
m1 + m1 + m2
As M lies in line (1), so
x + y = 2
or, 5mm11+– m2 + 5m1 – m2 = 2
m2 m1 + m2
or, 5m1 – m2 + 5m1 – m2 = 2
m1 + m2
or, 10m1 – 2m2 = 2m1 + 2m2
or, 10m1 – 2m1 = 2m2 + 2m2
or, 8m1 = 4m2
∴ m1 = 4 = 1
m2 8 2
Co-ordinate
Hence, the required ratio is 1:2.
Example 4:
Find the equation of straight line passing through (2, 7) and point of intersection of two lines
x + y = 2 and 2y + 3x = 1.
Let the given two straight lines intersect each other at B.
The co-ordinates of B can be obtained by solving given two equations
2[x + y = 2] x+y=2
–3x +– 2y –= 1
–x =3 A(2, 7)
\ x = – 3
B
Hence, by substituting the value of x in the first equation, 2y + 3x = 1
we have y = 5.
Now equation of straight line passing through A(2, 7) and B(– 3, 5) is
y – y1 = y2 – xy11 (x – x1)
x2 –
or,
y – 7 = –5 3––72 (x – 2)
or,
or, y – 7 = – 2 (x – 2)
– 5
5y – 35 = 2x – 4
or, 2x – 5y – 4 + 35 = 0
∴ 2x – 5y + 31 = 0, this is the required equation.
Special Forms of Lines ~ 137
Example 5:
The point P(a, b) lies on the line x – 2y = 2 and the point Q(b, a) lies on the line 2x + y = 6. Find
the equation of PQ.
Since, P(a, b) lies on the line x – 2y = 2,
a – 2b = 2 ... ... ... (1)
Again, since Q(b, a) lies on the line 2x + y = 6,
2b + a = 6 ... ... ... (2)
Solving the equations (1) and (2), we have
a – 2b = 2
2b + a = 6
2a = 8
\ a =4
Substituting the value of a in equation (1), we get
4 – 2b = 2
∴ b = 1
The co-ordinates of the points P and Q are (4, 1) and (1, 4) respectively, the equation of the line passing
through P and Q is
y – y1 = y2 – y1 (x – x1)
or, x2 – x1
y – 1 = 4 – 14 (x – 4)
1 –
or, y – 1 = –3 3 (x – 4)
or, y – 1 = – x + 4
∴ x + y – 5 = 0, this is the required equation.
EXERCISE : 4.8
1. Find the equation of straight line passing through the following points and having following slopes.
a. Point (2, – 3) and slope (m) = – 31 b. Point 3, 1 and slope (m) = 1
3
1 d. point 13, 2 and slope (m) = 3
c. Point (– 4, 1) and slope (m) = 3 5
3
2. Find the equation of straight lines passing through the following pair of points.
a. (2, 1) and (3, – 4) b. 31, 4 and –4 3, 1
c. (5, 2) and (– 3, 7) d. (– 4, 1) and (– 3, 0)
138 ~ Perfect Optional Mathematics Class 9
3. a. Find the equation of the straight line AB from Y Y
the given figure. B B
(3, 2) (3, 4)
b. In the figure, the co-ordinates of A is (6, 0), find X' O 60° X X' O AX
the equation of line AB. Y' Q.N. 3(b)
A
Y' Q.N. 3(a)
4. a. Find the equation of a straight line having slope 1 and passing through the mid-point joining the
points (– 4, 7) and (– 2, 3). 2
b. Find the equation of straight line passes through the point (– 2, 1) and having angle of inclination
5. a. 60°. Also prove that the line pass from the point ( 3 – 2, 4). Co-ordinate
b. Find the equation of a straight line passing through the point (1, 4) and the mid-point of the line
segment joining (– 3, 2) and (– 5, 0).
6. a.
Find the equation of straight line passing throught the point (0, 0) and the mid-point of the line
b. segment joining the points (2, 4) and (6, 2).
7. a. Find the equation of straight line passing through the point (– 3, – 3) and equally inclined with the
axes of co-ordinates.
b.
Find the equation of the straight line passing through the point (4, 1) having slope 2 and also find
8. a. the mid-point of the portion of the line intercepted between the axes of co-ordinates.
b. Find the equation of the straight line passing through the points (3a, 0) and (0, 3b). Also show
that the line passes through the point (a, 2b).
9. a.
Find the equation of the straight line passing through the points (– 4, 6) and (8, – 3) and also find
b. the length intercepted between the co-ordinate axes.
10. a.
b. A(– 2, 8), B(1, 2) and C(7, – 1) are the vertices of DABC. Find the length and equation of the
11. a. median passing through the vertex A.
b. If P(x1, y1) lies on the line 6x – y = 1 and Q(y1, x1) lies on the line 2x – 5y = 5, find the length and
the equation of the line segment PQ.
12. a.
Find the ratio in which the line joining (– 1, 0) and (3, 2) divide the line segment joining the
b. points (– 1, 5) and (2, 1).
In which ratio does the line joining (3, 2) and (5, 0) divide the join of (2, 1) and (5, 2) ?
Find the ratio in which the X-axis divides the line joining the points (4, 6) and (– 3, – 4).
Find the ratio in which the Y-axis divides the line joining the points (– 2, 1) and (– 5, 4).
Find the equation of straight line joining the point (2, 3) to the point of intersection of
2x + 3y – 11 = 0 and 3x – 4y = 8.
Find the equation of straight line passes through origin and the point of intersection of the lines
x – y = 3 and 2x – 3y = 1.
Find the equation of straight line passes through the intersection of x – y + 2 = 0 and x + y = 6
and the intersection of x = – 3 and y + x = 0.
Find the equation of straight line passes through the intersection of X-axis and the line
2x + 3y = 6 and the intersection of lines x – 2y = 0 and x + y = 3.
Special Forms of Lines ~ 139
4.9 Length of Perpendicular
To find the length of perpendicular drawn from point (x1, y1) to the line x cos α + y sin α = p
Let AB be a straight line having equation
x cos α + y sin α = p ... ... ... (1)
Let P(x1, y1) be given point, draw PN perpendicular to AB.
Through P, draw a line CD parallel to AB which meets X– axis and Y– axis at C and D respectively.
Here, OQ^AB and OQ = p and QOA = α
Y Y
D B
BR P(x, y) D Q
R N
Q
N a P(x, y)
X' a A X X' OC X
O C A
Y' Fig. (a) Y' Fig. (b)
The line OQ meets CD (produce OQ if necessary) at point R as shown in the figure and OR = p'
The equation of CD is x cos α + y sin α = p'
Again, the line CD passes through the point P(x1, y1), so
x1 cos α + y1 sin α = p' ... ... ... (2)
PN = OR – OQ or OQ – OR [according as OR > OQ or OR < OQ]
or, PN = ± (OR – OQ) ( ‘+’ or ‘– ’ sign is chosen to make PN positive)
or, PN = ± (p' – p)
∴ PN = ± (x1 cos α + y1 sin α – p)
Cor.1. Perpendicular distance between a line Ax + By + C = 0 and a point P(x1, y1) is ± Ax1 + By1+ C
Given equation of the line is Ax + By + C = 0 ... ... ... (1) A2 + B2
Perpendicular form of the equation (1) is
± Ax B2 ± By B2 = ± – C
A2 + A2 + A2 + B2
Comparing it with x cos a + y sin a = p,
cos α = ± A B2 , sin α = ± B B2 , p = ± – C
A2 + A2 + A2 + B2
Hence, the required perpendicular distance from point P(x1, y1) to the line Ax + By + C = 0 is given by
d = ± (x1 cos α + y1 sin α – p) = ± x1 A + y1 B+ C
A2 + B2 A2 + B2 A2 + B2
d = ± Ax1 + By1 + C
A2 + B2
140 ~ Perfect Optional Mathematics Class 9
Example 1:
Find the perpendicular distance from a point (2, – 3) to the line 3x – 4y + 7 = 0.
The given equation of straight line is 3x – 4y + 7 = 0 ... ... ... (1)
Comparing (1) with Ax + By + C = 0, we have A = 3, B = – 4 and C = 7
The perpendicular distance from the point (2, – 3) to the line 3x – 4y + 7 = 0 is
d = ± Ax1 + By1 + C P(2, – 2)
A2 + B2
=± 3 × 2 + (– 4) × (– 3) + 7 Q 3x – 4y + 7 = 0
32 + (– 4)2
=± 6 + 12 + 7
5
= 25 = 5 units.
5
Example 2: Co-ordinate
Find the distance between the pair of parallel lines 2x + 5y = 10 and 2x + 5y + 6 = 0.
The given two parallel lines are,
2x + 5y = 10
2x + 5y = 10 ... ... ... (1)
2x + 5y = – 6 ... ... ... (2) A(0, 2)
For the distance between two parallel lines, we find a point on the 2x + 5y + 6 = 0
first line. For this put x = 0, then y = 2.
B
Hence, (0, 2) is a point on the first line. As given two lines are
parallel to each other, we can calculate the distance between them
anywhere.
Now, perpendicular distance from a point (0, 2) to the line 2x + 5y = – 6 or 2x + 5y + 6 = 0 (Ax + By + C = 0)
d = ± Ax1 + By1 + C =± 2×0+5×2+6 =± 16
A2 + B2 22 + 52 29
Hence, the distance between the given two parallel lines is 16 units.
29
Example 3:
Find the value of m if the distance between the line mx – 2y + 14 = 0 and the point (3, 2) is 3 units.
Here, the equation of straight line mx – 4y + 14 = 0 (i.e. Ax + By + C = 0) and the given point is (3, 2).
The distance of the line from the given point is = 3 units. A(3, 2)
i.e. ± Ax1 + By1 + C = 3
A2 + B2
or, ± m × 3 + (– 4) × 2 + 14 =3 B mx – 4y + 14 = 0
m2 + (– 4)2
Length of Perpendicular ~ 141
or, ± 3m – 8 + 14 =3 Squaring both sides, we get
m2 + 16 2
or, ± 3m + 6 =3 or, {± (m + 2)}2 = m2 + 16
m2 + 16 or, (m + 2)2 = m2 + 16
or, m2 + 4m + 4 = m2 + 16
or, ± 3(m + 2) = 3 m2 + 16 or, 4m = 16 – 4
or, 4m = 12
or, ± (m + 2) = m2 + 16 \ m = 3
Hence, the value of m is 3.
EXERCISE : 4.9
1. Find the length of the perpendicular drawn from the point to the respective straight line.
a. (5, 0), x – y + 4 = 0 b. (– 2, – 7), 8x + 6y = 15
c. (7, 4), 3x – 5y = – 28 d. (2, 3), 3x – 4y – 9 = 0
2. a. Find the perpendicular distance of the line 4x + 3y – 10 = 0 from origin.
b. Find the perpendicular distance of the line 12x – 5y = 0 from the point (2, – 3).
c. Find the perpendicular distance of the line 3x – y + 10 = 0 from the point (7, 1).
d. Find the pependicular distance of the line 3x + 4y + 3 = 0 from the point of intersection of the
lines 3x – y – 1 = 0 and 2x + y – 4 = 0.
3. Find the distance between the following pair of parallel straight lines.
a. 3x + 4y + 4 = 0 and 3x + 4y – 16 = 0 b. 6x – 8y – 3 = 0 and 6x – 8y + 12 = 0
c. 9x + 40y = 20 and 9x + 40y + 21 = 0 d. 4x – 3y = 6 and 8x – 6y = 15
4. a. If the length of the perpendicular drawn from the point (a, 3) on the line 3x + 4y + 7 = 0 is 8 units.
Find the value of a.
b. If the perpendicular distance from a point (2, 3) on the line 3x + 4y + k = 0 is 6 units, find the
possible values of k.
c. If the perpendicular distance from the point (2, – 4) to the line 4x – 3y + n = 0 is 5 units find the
value of n.
5. Find the perpendicular distance from origin to the line with equation.
a. y = mx + c b. x + y = 1 c. x cos a + y sin a = p
a b
142 ~ Perfect Optional Mathematics Class 9
4.10 Area of Triangles and Quadrilaterals
Area of Triangle Y A(x1, y1)
Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of a B(x2, y2) C(x3, y3)
triangle ABC. Draw perpendiculars AM, BN and CP from A, PX
B, and C to OX meeting at the points M, N and P respectively X' O N M
then, Y'
OM = x1, AM = y1, ON = x2, BN = y2, OP = x3, CP = y3.
Now, ∆ABC = Trap.BNMA + Trap.AMPC – Trap.BNPC
= 12 MN (BN + AM) + 21 MP (AM + CP) – 21 NP (BN + CP) Co-ordinate
= 12 (OM – ON) (BN + AM) + 12 (OP – OM) (AM + CP) – 21 (OP – ON) (BN + CP)
= 12 {(x1 – x2) (y2 + y1) + (x3 – x1) (y1 + y3) – (x3 – x2) (y2 + y3)}
= 12 [x1y2 + x1y1 – x2y2 – x2y1 + x3y1 + x3y3 – x1y1 – x1y3 – x3y2 – x3y3 + x2y2 + x2y3]
∴ ABC = 1 [(x1 y2 – y1 x2) + (x2y3 – y2 x3) + (x3 y1 – y3 x1)]
2
The area of triangle can be remembered as follows:
D = 1 x1 x2 x3 x1
2 y1 y2 y3 y1
Note: The above formula gives a positive result if the vertices of a triangle are arranged in anticlockwise
direction. The area will be negative if the points are arranged in the clockwise direction. But the area
is always positive, so we have to take magnitude only.
If the area of triangle formed by three points is zero then the three points lie in one straight line i.e. the
points are collinear.
Area of Quadrilateral
Let A(x1, y1), B(x2, y2), C(x3, y3) and D(x4, y4) are the Y
vertices of a quadrilateral ABCD. Draw AQ, BP, CR and A(x1, y1)
DS perpendiculars to OX then,
D(x4, y4)
OQ = x1, AQ = y1, B(x2, y2)
OP = x2, BP = y2, X' OPQ C(x3, y3) X
OR = x3, CR = y3 and Y' RS
OS = x4, DS = y4.
Area of quadrilateral ABCD
= Trap.ABPQ + Trap.AQSD – Trap.BPRC – Trap.CRSD.
Area of Triangles and Quadrilaterals ~ 143
= 21 PQ (BP + AQ) + 21 QS (AQ + DS) – 21 PR (BP + CR) – 21 RS (CR + DS)
= 12 [(x1 – x2) (y1 + y2) + (x4 – x1) (y1 + y4) – (x3 – x2) (y2 + y3) – (x4 – x3) (y3 + y4)]
= 21 [x1 y1 + x1 y2 – x2y1 – x2y2 + x4y1 + x4y4 – x1y1 – x1y4
– x3y2 – x3y3 + x2y2 + x2y3 – x4y3 – x4y4 + x3y3 + x3y4]
= 12 [(x1y2 – y1x2) + (x2y3 – y2x3) + (x3y4 – y3x4) + (x4y1 – y4x1)]
Area of quad. = 12 x1 x2 x3 x4 x1
Notes:
y1 y2 y3 y4 y1
a. Area of quadrilateral can be obtained by adding the areas of triangles formed by any diagonal.
b. Before finding the area of quadrilateral, we must see that whether the vertices of the quadrilateral are
taken in order or not. For this, we must plot the points in order roughly and we arrange the vertices
in order, then apply the formula otherwise the result may be wrong.
Example 1:
Find the area of triangle whose vertices are A(– 2, 1), B(3, 3) and C(1, 5).
Here x1 = – 2, y1 = 1, x2 = 3, y2 = 3, x3 = 1 and y3 = 5.
A B C A
– 2
3 1 – 2
1 3 5 1
Area of triangle ABC
= 21 [(– 2 × 3 – 1 × 3) + (3 × 5 – 3 × 1) + {1 × 1 – 5 × (– 2)}]
= 21 [(– 6 – 3) + (15 – 3) + (1 + 10]
= 12 [– 9 + 12 + 11]
= 12 × 14 = 7 sq. units.
Example 2:
If the points (7, a), (– 5, 2) and (3, 6) are collinear, find a.
Let the given three collinear points are A(7, a), B(– 5, 2) and C(3, 6), then the area of DABC is zero.
A B C A
7 – 5 3 7
a 2 6 a
∆ABC = 12 [{7 × 2 – a × (– 5)} + (– 5 × 6 – 2 × 3) + (3 × a – 6 × 7)]
i.e. 0 = 12 [14 + 5a – 30 – 6 + 3a – 42]
i.e. 0 = 8a – 64
∴ a = 8.
144 ~ Perfect Optional Mathematics Class 9
Example 3:
If the co-ordinates of the points A, B and C are (3, 7), (2, 1) and (5, 4) respectively and P is the point
(x, y), show that ∆PBC = y – x + 1.
∆ABC 5
As the co-ordinates of points A, B, C and P are (3, 7), (2, 1), (5, 4) and (x, y) respectively.
A B C A A(3, 7)
3 2 5 3
7 1 4 7
∆ABC = 21 [(3 – 14) + (8 – 5) + (35 – 12)] = 21 [46 – 31]
P(x, y)
= 125 sq. units. B(2, 1)
Again, P B C P C(5, 4)
x 2 5 x
y 1 4 y
∆PBC = 12 [(x – 2y) + (8 – 5) + (5y – 4x)] Co-ordinate
= 21 (3y – 3x + 3)
= 3(y –x + 1) sq. units.
2
23 (y – x + 1)
Now, ∆PBC = 125
∆ABC
= 3(y –x + 1) × 2 = y – x + 1
2 15 5
Example 4:
Find the area of the quadrilateral whose vertices are (1, 1), (5, – 2), (3, 4) and (4, – 7).
Plotting the points on the co-ordinates plane roughly, we Y (3, 4)
have known the given four vertices of a quadrilateral are not
given in order.
Hence let A(1, 1), B(4, – 7), C(5, – 2) and D(3, 4) be the four X' (1, 1) X
O
vertices of the given quadrilateral, then (5, – 2)
A B C D A
1 4 5 3 1
1 – 7 – 2 4 1 (4, – 7)
Area of ABCD Y'
= 21 [(– 7 – 4) + (– 8 + 35) + (20 + 6) + (3 – 4)]
Area of Triangles and Quadrilaterals ~ 145
= 21 [– 11 + 27 + 26 – 1]
= 21 × 41 = 41 sq. units. EXERCISE : 4.10A
2
1. Find the area of the polygons shaded in the following figures.
a. YA b. Y c. Q Y P
A
R
B C D
X' O X
X' O B C X X' O S X
Y' Y' Y'
2. Find the area of the triangle with vertices.
a. (2, 3) (4, 5) and (6, 2) b. (2, – 3), (3, 2) and (– 2, 5)
c. (3p, 0), (0, 3q) and (p, q) d. (a, a + b), (b, b + c) and (c, c + a)
3. Prove that the following three points are collinear.
a. (1, 1), (– 2, – 5) and (– 1, – 3) b. 2, – 21 , 0, 23 and – 21, 2
c. (3x, 0), (0, 3y) and (x, 2y) d. (a, b + c), (b, c + a) and (c, a + b)
4. a. The area of triangle formed by the points P(5, 1), Q(m, 3) and R(2, 7) is 18 square units. What
will be the value of m?
b. For what value of k will the triangle with the vertices (– 1, 4), (– 3, 8) and (– k + 1, 3k) have the
area 10 sq. units?
c. Find the value of a if the area of the triangle formed by the points (a, 1), (2, – 3) and (4, 4) is 15
sq. units.
5. a. For what value of a will the points (2, 6), (3, 8) and (– 1, a) be on the same straight line?
b. For what value of m are the points (m, 2 – 2m), (1 – m, 2m) and (– m – 4, 3m + 9) collinear?
6. a. If the points (a, 0), (0, b) and (3, 3) lie on the same straight line show that 1 + 1 = 13.
a b
b. If the points (a, 1), (1, 2) and (0, b + 1) are collinear, show that 1 + 1 = 1.
a b
c. If the point (h, k), (h', k') and (h – h', k – k') are collinear, show that hk' = h'k.
d. If the points P(a, 0), Q(2, 2) and R(0, b) are collinear prove that 2a + 2b = ab.
7. Find the area of the quadrilateral formed by the points:
a. (– 3, 0), (0, 4), (5, 0) and (0, – 6) b. (3, – 2), (– 2, 5), (3, 3) and (7, 1)
146 ~ Perfect Optional Mathematics Class 9
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