Infinity Optional mathematics book 9 Final for CTP 2077

(b) Obtain the equation of the locus of a point which moves such that its
distance from (–2, 3) is 3 units.

9. (a) Find the equation of the locus of a point which is equidistant from (0, 2)
and x -axis.

(b) Find the equation of the locus of a point which is equidistant from (3, 0)
and y-axis.

(c) A point moves such that the algebraic sum of its distance from the
coordinate axes is a constant quantity 'z'. Find the equation of the locus.

(d) Find the equation of the locus of a point which moves in such a way that
its distance from the fixed point (h, k) is always 'r' units.

Section 'C'

10. (a) A(5, 0) and B(–5, 0) are two fixed points. Find the equation of the locus
of a point 'P' under the condition.

(i) PA2 + PB2 – AB2 = 0 (ii) 2PA = PB

(b) A (0, 5) and B(0, –5) are two fixed points. Find the equations of the locus
of a point P which moves so that AB always subtends right angle at P.

11. (a) Find the equation of the locus of a point which moves so that its distance
from the point (2, 3) is two times its distance from the point (4, 1).

(b) Obtain the equation of the locus of a point which moves so that its
distance from the point (3, 0) is three times its distance from the point
(0, 2).

(c) Find the equation of the locus of a point which moves so that its distance
from the point (–2, 5) is half of its distance from y-axis.

(d) Find the equation of the locus of a point which moves so that its distance
from the point (–2, 3) is two third of its distance from the point (4, –5).

12. (a) M(2, 3) and N(3, 1) are two fixed points. Find the equation of the locus of
point P such that PM: PN = 2:3.

(b) C(–2, 0) and D(2, 0) are two fixed points. Find the equation of the locus
of a point P such that PC: PD = 1:3.

(c) Find the equation of the locus of a point which moves so that it is
equidistance form the points (–2, 1) and (4, 1).

13. In which area locus can be used? Make a report and present in the class.

Infinity Optional Mathematics Book - 9 101

4.2 Section Formula

What do you mean by ratio? Discuss on it.
Let us study the following figures and discuss on the following questions.

AC B
fig. (i)

M PN
fig. (ii)
(i) How the point C divides the line segment AB in figure (i)? Measure the length

of AC and BC then find AC:BC.
(ii) Similarly, in figure (ii) measure the length of MP and NP. Which one is

greater? Find MP: NP.
(iii) Is their any difference between their ratio?

Internal division

Let, AB be a straight line. The point P lies within AB then P divides AB internally
in the ratio of AP:PB. Therefore, AP and PB are two parts of AB.

To find the coordinates of the point which divide
internally, the line joining two points (x1, y1) and A P B
(x2, y2) in the given ratio m1: m2.

Let, A(x1, y1) and B(x2, y2) be two points and (x, y) be the coordinates of the point P
which divides AB in the ratio of m1:m2. Then AP:PB = m1:m2.

Draw AL, PN and BM perpendiculars on OX. Similarly, AQ and PR are also

perpendiculars on PN and BM respectively, Y B(x2,y2)

Now, m2
In x-axis,
P(x,y)
OL = x1, ON = x, OM = x2 m1 R
∴ LN = AQ = ON – OL
)

= x – x1 ......... (i) ,y
and NM = PR = OM – ON
1

A(x Q

1

= x2 – x ......... (ii) X' O L N MX
Similarly, in y -axis, we have

A L = y1, PN = y BM = y2 Y'

102 Infinity Optional Mathematics Book - 9

Now,

PQ = PN – QN

= y – AL [ QN = AL]

PQ = y – y1 .......... (iii)
and

BR = BM – RM

= y2 – PN [ RM = PN]

BR = y2 – y .............. (iv)

Since, PQ//BR, ∆APQ and ∆BPR are similar so their corresponding sides ares

proportional.

∴ AP = AQ = PQ
BP PR BR

or, mm12 = x – x1 = y – y1 [ From equation (i), (ii), (iii), (iv) and given)
x2 – x y2 – y

Taking, 1st and 2nd ratios, we get

m1 = x – x1
m2 x2 – x

or, m1(x2 – x) = m2 (x – x1)
or, m1x2 – m1x = m2x – m2x1
or, m1x2 + m2x1 = m1x + m2x
or, m1x2 + m2x1 = x(m1 + m2)


or, x = m1x2 + m2x1
m1 + m2

Similarly, taking 1st and 3rd ratio, we get

m1 = y – y1
m2 y2 – y

or, m1(y2 – y) = m2 (y – y1)
or, m1y2 – m1y = m2y – m2y1
or, m1y2 + m2y1 = m1y + m2y
or, m1y2 + m2y1 = y(m1 + m2)

or, y= m1y2 + m2y1
m1 + m2

Infinity Optional Mathematics Book - 9 103

Hence, the coordinates of division point P are

P(x, y) = P m1x2 + mm22x1, m1y2 + m2y1
m1 + m1 + m2

External division

Let AB be a straight line and C be a point.

If the point C lies outside the line AB and A BC

if on producing AB, it meets C, then C is said to divide AB externally in the ratio of

AC:BC.

∴ AC and BC are two parts, when C Y m 1 m 2 R(x, y)
divides AB. P(x 1,y 1)
Q(x 2,y 2) K
To find the coordinates of the point which
divides the line joining the two points L
(x1, y1) and (x2, y2) externally in the given
ratio m1:m2.

Let, P(x1, y1) and Q(x2, y2) be two given X' OM N SX
points. Let, R(x, y) divides the line PQ Y'

externally in the ratio of m1:m2.

Then, PR : QR = m1: m2.
or, QPRR = m1
m2

From P, Q and R draw PM, QN and RS perpendicular to OX. Similarly, from P. draw
PL perpendicular to RS and QK is also perpendicular to RS. ( QN||RS)

In x-axis, ........(i)
OM = x1, ON = x2, OS = x .......... (ii)
Now, QK = NS = OS – ON
or, QK = NS = x – x2 [ KN = PM]
and PL = MS = OS – OM ........... (iii)
or, PL = MS = x – x1
Similarly, in y-axis
PM = y1, QN = y2, RS = y
Now, RK = RS – KS
or, RK = y – QN
or, RK = y – y2

104 Infinity Optional Mathematics Book - 9

and RL = RS – LS

or, RL = y – PM [ LS = PM]

or RL = y – y1 ............... (iv)

∆PQK and ∆PRL are similar so corresponding sides are proportional.

PR = PL = RL
QR QK RK

m1 = x – x1 = y – y1 [ From equation (i), (ii), (iii), (iv) and given]
m2 x – x2 y – y2

Taking above first and second ratios, we get

m1 = x – x1
m2 x – x2

or, m1 (x – x2) = m2(x – x1)

or, m1x –m1x2 = m2x – m2x1

or, m1x – m2x = m1x2 – m2x1

or, x(m1 – m2) = m1x2 – m2x1

or, x = m1x2 – m2x1
m1 – m2

Similarly, from the above first and last ratios, we get

m1 = y – y1
m2 y – y2

or, m1 (y – y2) = m2(y – y1)

or, m1y – m1y2 = m2y – m2y1

or, y(m1 – m2) = m1y2 – m2y1

or, y = m1y2 – m2y1
m1 – m2

Hence, the coordinates of the external point R are

R(x, y) = R m1x2 – m2x1 , m1y2 – m2y1
m1 – m2 m1 – m2

Infinity Optional Mathematics Book - 9 105

Section formula Application

1. Midpoint formula: How do you relate section formula with midpoint?

If P(x, y) is the midpoint of the line joining the points A(x1, y1) and B(x2, y2)

then AP = BP and AP:BP = m1:m2 = 1:1 and the coordinates of P are

+ + mm22y1 m2 B(x2,y2)
x = m1mx21 + m2x1 m1y2 + m1 P(x, y)
m2 and y = m1

= 1 × x2 + 1 × x1 = 1 × y2 + 1 × y1
1 + 1 1 + 1
A(x1,y1)
x = x1 +2 x2 y = y1 + y2
2 x1 + y2, y1 + y2
2 2
Therefore, the coordinates of the midpoint P(x, y) = P

2. To find the coordinates of the point which divide a straight line

joining two points (x1, y1) and (x2, y2) in the ratio of k:1
Let, A(x1, y1) and B(x2, y2) be the two given point and (x, y) be the coordinates

of point P which divides AB in the ratio k:1.

Then, m1:m2 = k:1
or, m1 = km2
By formula,

x = m1x2 + m2x1 and y = m1y2 + mm22y1
km2 + m2 m1 +

= kmm2x12++mm22x1 = km2y2 + mm22y1
km2 +

= mm2(2k(xk2++1x)1) = mm2(2k(yk2++1y)1)

x = k xk2++1x 1 y= ky2 + y1
k+1

Therefore, the coordinates of P(x, y) are

=P kxk2++1x1, ky2 + y1
k+1

106 Infinity Optional Mathematics Book - 9

3. Centroid formula

What is centroid? How do we get the A(x1, y1)
G(x,y) E
centroid? Discuss.
D
Let, A(x1, y1), B(x2, y2) and C(x3, y3) be
the vertices of a triangle ABC. D, E and F

be the midpoints of sides BC. AC and AB F
B(x2, y2)
respectively, so that AD, BE and CF are

called medians. G(x, y) is the intersection

point of all three medians is called centroid

of ∆ABC. C(x3, y3)

Since, D is the mid point of BC, so by

midpoint formula, the coordinates of D are
x2 + x3 , y2 + y3
D 2 2

From the plane geometry, we know that the centroid of a triangle divides the

median in the ratio of 2:1 from the vertex to the midpoint.

∴ AG : GD = m1:m2 = 2:1
For AD :

Let, A(x1, y1) = (x1, y1)

D x2 + x3 , y2 + y3 = (x2, y2)
2 2

m1:m2 = 2 : 1
By using internal section formula

m1x2 + m2x1 m1y2 + mm22y1
m1 + m2 m1 +
x = and y =

2 × x2 + x3 + 1 × x1 2 × y2 + y3 + 1 × y1
2 2
= =
2+1 2+1

= x2 + x33 + x1 = y2 + y3 + y1
3

x= x1 + x2 + x3 y = y1 + y2 + y3
3 3

Therefore, the coordinates of the centroid of ∆ABC
x1 + x2 + x3, y1 + y2 + y3
G(x, y) = G 3 3 .

Infinity Optional Mathematics Book - 9 107

Identification of some geometrical figures by using section
formula



S.N. Figure Conditions

1. Parallelogram Diagonals bisect to each other.

2. Rectangle Diagonals are equal and bisect to each other.
All sides are equal and diagonals bisect to each other.
3. Rhombus


Points of Trisection:

Let, AB be a straight line. Two points P and Q which divides line segment AB

into three equal parts such that 1 1 1
AP = PQ = QB A P QB

then these points P and Q are called the trisection points of AB. Here, P divides
AB in the ratio of 1:2 and Q divides AB in the ratio of 2:1.

WORKED OUT EXAMPLES

1. Find the coordinates of the point which divides the line segment
joining the points (5, -2) and (-2, -2) in the ratio of 4:3 (i) internally (ii)
externally.

Solution: Here,

Case I:

Let, P(x, y) divides the line joining the points A(5, -2) and B(-2, -2) internally

in the ratio of 4:3. 4 P(x, y) 3
Then, A(5, –2) B(–2, –2)

Let, A(5, –2) = (x1, y1)

B(–2, –2) = (x2, y2)

m1 : m2 = 4 : 3

By using internal section formula, we get

x = m1x2 + mm22x1 and y = m1y2 + m2y1
m1 + m1 + m2

= 4 × (–2) + 3(5) = 4 × (–2) + 3 × (–2)
4+ 3 4 + 3
– 8+ 15 = – 8– 6
= 7 7

108 Infinity Optional Mathematics Book - 9

= 7 = –14
7 7

= 1 = –2
∴ The required point P(x, y) = P(1, –2) 4
Case II: 3
Let, C(a, b) divides the line joining B(–2, 2) C(a, b)

the points A(5, –2) and B(–2, –2)
externally in the ratio of 4:3.
Then, by using external section A(5, –2)
formula, we get

x = m1x2 – mm22x1 and y = m1y2 – m2y1
m1 – m1 – m2

= 4 × (–2) + 3 × 5 = 4 × (–2) – 3 × (–2)
4–3 4 – 3

= – 8– 15 = – 8+ 6
1 1

= –23 = –2

Therefore, the required external point coordinate C(a, b) = C(–23, –2)

2. If one end of a line is A(–3, 4) and the midpoint is P(4, 3) then what are
the coordinates of the other end? Find it.

Solution: Here, A(–3, 4) P(4, 3) B(a, b)

Let, B(a, b) be the coordinates of other end.

Let, A(–3, 4) = (x1, y1)

B(a, b) = (x2, y2)

P(4, 3) = (x, y)

By mid point formula,

x = x1 + x2 and y = y1 + y2
2 or, 2
or,
or, 4 = – 3+ a 3 = 4 + b
2 2

or, 8 + 3 = a 6=4+b

or, a = 11 or, b = 2

Therefore, the coordinates of other end are (11, 2).

Infinity Optional Mathematics Book - 9 109

3. Find the ratio in which the point B(3, b) divides the line joining M(7,

–3) and N(–2, 5). Also, find the value of 'b'. m2 N(–2, 5)
Solution: Here,

Let, the point B(3, b) divides the line joining the
m1 B(3, b)

points M(7, –3) and N(–2, 5) in the ratio of m1:m2.

M(7, –3)

Alternative method

Let, M(7, –3) = (x1, y1) M(7, –3 = (x1, y1)
N(–2, –5) = (x2, y2)
B(3, y) = (x, y) N(–2, –5) = (x2, y2)
B(3, y) = (x, y)

By using section formula

By using section formula x = m1x2 + m2x1
or, 3 = m1 +(–m2)2+ m2
m1 = x – x1 m1 + m2 × 7
m2 x2 – x m1 ×

= 3–7 or, 3m1 + 3m2 = –2m1 + 7m2
–2–3

= –4 or, 3m1 + 2m1 = 7m2 – 3m2
–5

= 4 or, 5m1 = 4m2
5

∴ m1:m2 = 4:5 or, m1 = 4
m2 5

Again, by section formula ∴ m1:m2 = 4:5

y = m1y2 + m2y1 Again, by section formula
m1 + m2

or, b = 4 × (–5) + 5 × (–3) y = m1y2 + m2y1
4 + 5 or, b m1 + m2
or, b
or, b = – 20 – 15 ∴ b = 4 × (–5) + 5(–3)
9 4+5

∴ b = – 35 = – 20 – 15
9 9

= – 35
9

Hence, m1:m2 = 4:5 and b = – 35
9

110 Infinity Optional Mathematics Book - 9

4. Find the coordinates of the point of trisection of the line joining the

points A(1, –2) and B(–3, 4). 1 2
Solution: Here,

Let, P(x. y) and Q(a, b) be the A(1, –2) P(x, y) Q(a, b) B(–3, 4)

points of trisection of the line

joining the points A(1, –2) and B(–3, 4).

The point P(x, y) divides the line AB in the ratio of 1:2

Let, A(1, –2) = (x1, y1)

B(–3, 4) = (x2, y2)

m1 : m2 = 1:2
By using section formula

x = m1x2 + mm22x1 and y = m1y2+ m2y1
m1 + m1 + m2

= 1 × (–3) + 2 (1) = 1 × 4 + 2(–2)
1+2 1 + 2

= – 3+ 2 = 4 – 4
3 3

= – 1 =0
3

∴ P(x, y) = P – 1 , 0
3

Again, Q(a, b) is the midpoint of line segment PB

Let, P – 1 , 0 = (x1, y1)
3

B(–3, 4) = (x2, y2)

By using mid point formula

x = x1 + x2 and y = y1 + y2
2 2

–1 + (–3)
2
or, a= 3 b = 0 + 4
2

or, a = –610 b= 2

∴ a = –35 ∴ Q(a, b) = Q – 5 , 2
3

Hence, P – 1 ,0 and Q – 5 , 2 are the points of trisection of the line AB.
3 3

Infinity Optional Mathematics Book - 9 111

5. The points (3, 7), (5, –7) and (–2, 5) are three vertices of a parallelogram.

Find the coordinates of the remaining vertex which is opposite to

(5, –7). Also, find the midpoint of diagonal.

Solution: Here,

Let, B(3, 7), E(5, –7) and S(–2, 5) be the three vertices of a parallelogram

BEST. Let, the remaining vertex T(a, b)

By mid point formula, B(3, 7) T(a, b)
the coordinates of the mid point of diagonal

BS = 3 + ( –2), 7 + 5  (x, y) = x1 + x2, y1 + y2
2 2 2 2

= 21, 6 E(5, –7) S(–2, 5)

Similarly, the coordinate of the midpoint of diagonal

ET = 5 + a, –7 + b  (x, y) = x1 + x2, y1 + y2
2 2 2 2

Since, the diagonals of a parallelogram bisect to each other i.e. they have

common midpoint. So,

or, 21 = 5 + a and 6= –7 + b
2 2

or, 1 – 5 = a or, 12 = – 7 + b

a = –4 or, 12 + 7 = b

b = 19
Therefore, the mid point of diagonal = 21, 6 and the coordinates of remaining

vertex is (–4, 19).

6. Find the coordinates of the centroid of triangle whose vertices are
(2, 5), (–4, 6) and (2, –2).

Solution: Here, A(2, 5)

Let, G(x, y) be the centroid of triangle ABC whose

vertices are A(2, 5), B(–4, 6) and C(2, –2).

Let, A(2, 5) = (x1, y1) G(x,y)

B(–4, 6) = (x2, y2)

C(2, –2) = (x3, y3) B(–4, 6) C(2, –2)

112 Infinity Optional Mathematics Book - 9

By centroid formula

x = x1 + x2 + x3 and y = y1 + y2+ y3
3 3

= 2 – 4 + 2 = 5+6–2
3 3

= 0 y = 3

Therefore, the coordinates of centroid are (0, 3).

7. The middle points of the sides of a triangle are (1, 1), (2, 3) and (4, 1),

find the coordinates of the vertices of the triangle.

Solution: Here, M(x1, y1) T(4, 1)
Let, M(x1, y1), A(x2, y2) and N(x3, y3) be the P(1, 1)

vertices of ∆MAN. Let, P(1, 1), O(2, 3) and
T(4, 1) be the midpoints of sides AM, AN

and MN respectively.

From the figure,
PT//AN and PO//MN [The line joining the A(x2, y2) O(2, 3) N(x3, y3)

mid point of any two sides of ∆ is parallel

to third side]

i.e. PT//ON and PO//TN

∴ PONT is a parallogram.

By midpoint formula,

The coordinates of the midpoint of diagonal OT

= 2 + 4, 3 + 1 = (3, 2)
2 2
1 + x3 1 + y3
The coordinates of the midpoint of diagonal PN = 2 , 2

Since, the diagonal of a parallelogram bisect to each other so,

1 + x3 = 3 ⇒ 1 + x3 = 6 ∴ x3 = 5
2

and 1 + y3 = 2 ⇒ 1 + y3 = 4 ∴ y3 = 3
2

∴ The coordinates of N(x3, y3) = N(5, 3).

Infinity Optional Mathematics Book - 9 113

Again, in MN line T(4, 1) is the mid point of MN, So by midpoint formula,

x = x1 + x2 and y = y1 + y2
2 3

or, 4 = x1 + 5 or, 1 = y1 + 3
2 2

or, 8 – 5 = x1 or, 2 – 3 = y1

∴ x1 = 3 ∴ y1 = – 1

Therefore, the coordinates of M(x1, y1) = M(3, –1)

In MA line P(1, 1) is the mid point of MA, So by midpoint formula,

x = x1 + x2 and y = y1 + y2
2 3

1 = 3 + x2 1 = –1 + y2
2 2

or, 2 – 3 = x2 or, 2 = –1 + y2

x2 = –1 or, 2 + 1= y2

y2 = 3

∴ The coordinates of A(x2, y2) = A(–1, 3)

Hence, (3, –1), (–1, 3) and (5, 3) are the required coordinates of the vertices of

∆MAN.

Exercise 4.4

Section 'A'

1. (a) What do you mean by internal division? Illustrate it with figure.
(b) Define centroid of triangle. Is it inside the triangle?
(c) What are the application of section formula? Write them.
2. (a) What are the condition to prove rectangle by section formula? Write

them.
(b) Define point of trisection of the line. Illustrate it with figure.
3. (a) If the midpoints of the diagonals of a parallelogram are 32, 5 and

114 Infinity Optional Mathematics Book - 9

32, –5 + b then find the value of 'b'.
2

(b) Find the coordinates of the midpoint of the line joining the points (–3, –6)

and (1, –2).

Section 'B'

4. (a) Find the coordinates of a point dividing the line joining the points.

(i) (–3, 9) and (1, –3) internally in the ratio of 3:1.

(ii) (–1, 7) and (4, –3) internally in the ratio of 2:3.

(iii) (4, 6) and (–2, 3) internally in the ratio of 2:1.

(b) Find the coordinates of a point dividing the line joining the points.

(i) (–1, 2) and (1, 1) externally in the ratio of 2:3.

(ii) (–3, 2) and (4, –4) externally in the ratio of 4:3.

(iii) (–5, –2) and (–3, 2) externally in the ratio of 7:5.

(c) (i) Find the coordinates of the point which divides the line joining the

points (3, 4) and (8, 7) in the ratio 2:5. Is this division internal or

external? Give reason for your answer.

(ii) Find the coordinates of the point which divides the line joining

(–3, –4) and (8, 7) in the ratio 3: –4. Is this division internal or

external? Give reason.

5. (a) Find the coordinates of the middle points of the line joining the points.

(i) (7, –2) and (–3, 6) (ii) ( 3, 2) and (– 3, 3 2).

(b) If the coordinates of the midpoints of the line joining the point P(3, 5)

and Q(x, y) is the point (1, –1), find the values of x and y.

(c) The vertices of the ∆ABC are (1, –2), (–3, 4) and (5, –6). Find the midpoints

of the sides of the triangle.

6. (a) Find the ratio in which the point A(1, 1) divides the line joining the

points (–6, 8) and (8, –6).

(b) Find the ratio in which the point (a, 1) divides the line segment joining

the points (6, –1) and (–4, 4).

7. (a) Find the coordinates of the centroid of triangle whose vertices are

(i) (0, 0), (3, 3) and (9, – 9) (ii) (2, 7), (6, 2) and (4, 5)

(b) The centroid of a triangle is (2, 3). If two of its vertices are (5, 4) and (–3,

3), find the third vertex.

Section 'C'

8. (a) Find the ratio in which the point P(x, –1) divide the line joining the
points M(4, 4) and N(7, 7). Also find the value of x.

Infinity Optional Mathematics Book - 9 115

(b) Find the ratio in which the line segment joining the points (1, –5) and
(–4, 5) is divided by the x-axis. Also find the coordinates of the point of
division.

(c) Find the ratio in which the y-axis divides the line segment joining the
points (5, –6) and (–1, –4). Also find the point of intersection of the line
and y-axis.

9. (a) Find the coordinates of the points of trisection of the line joining the

following points.

(i) (4, –3) and (–2, –3) (ii) (2, –2) and (–7, 4)

(b) The line AB is trisected by the points (2, 3) and (5, 4). Find the coordinates

of the points A and B.

(c) Find the coordinates of the point which divide the line segment joining

A(–2, 2) and B(2, 8) into four equal parts.

10. (a) Prove that the following points represent the vertices of a parallelogram.

(i) (2, 1), (5, 2), (6, 4) and (3, 3) (ii) (3, –2), (4, 0), (6, –3) and (5, –5)

(b) Prove that the following points represent the vertices of a rectangle.

(i) A(–1, 1), B(–2, 2), C(4, 8) and D(7, 5)

(ii) T(2, –2), A(8, 4), S(5, 7) and K(–1, 1)

(c) If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken

in order, find the values of x and y.

(d) Three of the four vertices of a parallelogram are given below. Find the

remaining (fourth) vertex of a parallelogram.

(i) (2, 3), (4, –1) (0, 5) (ii) (2, 6), (6, 2) and (12, 4)

11. (a) P(3, 4), Q(7, 7) and R are three collinear points. The distance between

P and R is 10 units. Find the distance between P and Q. Also, find the

coordinates of R.

(b) A(2, 4), B(3, 3) and C(a, b) are three collinear points and AC is equal to

2 2 units. Find the length of AB and the coordinates of C.

12. If A and B are (–2, –2) and (2, –4) respectively of a line segment AB. Find the
3
coordinates of the point P such that AP = 7 AB and P lies on the line segment

AB.

13. If P(1, 1), Q(7, –3), C(12, 2) and D(7, 21) are the vertices of a quadrilateral

PQRS. Prove that the lines joining the midpoints of the sides of a quadrilateral

PQRS taken in order form a parallelogram.

116 Infinity Optional Mathematics Book - 9

14. (a) The coordinates of the midpoints of the sides CA, AT and TC of triangle
CAT are (0, 1), (1, –1) and (2, 2) respectively. Find the coordinates of the
vertices of ∆CAT.

(b) A(4, –2), B(–2, 3) and C(6, 4) are the vertices of a triangle ABC. Find the
length of the median drawn from 2nd vertex B on AC.

15. What are the characteristics of geometrical figure, which can be prove by using
section formula? Prepare a small report by different group of students and
discuss in your classroom.

4.3 Equation of Straight Lines

Let, us observe the given figures.

Figure (I) Figure (II)
What does figure (I) and figure (II) represent ? Discuss.

Again,

Let us discuss on the following questions.

(i) Plot the pair of points in a graph from the given table.

x0 1 –1 2 –2
y0 1 –1 2 –2

What type of figure formed by joining them?

(ii) From the given table plot the pair of points in a graph.

x0 1 –1 2 –2
y0 1144

What type of figure formed by joining them?

(iii) What is the difference between two graph obtained from (i) and (ii) table?

Infinity Optional Mathematics Book - 9 117

Equation of straight lines parallel to coordinate axes

(i) Equation of straight line parallel to x-axis. Y

Let, AB be a straight line parallel to x-axis

which cuts y -axis at N such that ON = b.

Let, P(x, y) be a point on AB. Draw PM A N P(x,y) B
perpendicular to OX so that PM = y.
by
But, PM = ON
or, y = b X' O xM X

∴ y = b is the required equation of the Y'
straight line AB (parallel to x-axis).

⇒ This relation is true for any point (x, y) on AB.

If the straight line AB lies b units below from line x-axis then what would be

the equation of line AB? Discuss on it.

Cor. 1 If the line AB lies b units below from x-axis then the equation of line is y = –b.

Cor. 2 If b = 0 then the straight line AB coincides with x-axis i.e. y = 0 is the equation

of AB. Also, y = 0 is the equation of x-axis. YA
(ii) Equation of straight line parallel to y-axis.

Let, AB be a straight line parallel to y-axis N x P(x,y)
which cuts x-axis at M such that OM = a. y

Let, P(x, y) be any point on AB.

From P, draw PN perpendicular to OY so that

PN = x X' O a MX

But, PN = OM Y' B
x = a

∴ x = a is the required equation of the

straight line AB (parallel to y-axis).

⇒ This relation is true for any point (x, y) on AB. If the straight line AB lies 'a'

units left from y-axis then, what would be the equation of AB? Discuss on it.

Cor.1 If the straight line AB lies 'a' units on the left side of y-axix, then the equation

of AB is x = –a.

Cor.2 If the straight line AB coincides with y-axis then a = 0 and the equation of

AB is y = 0. Also, y = 0 is called the equation of y-axis.

118 Infinity Optional Mathematics Book - 9

Slope of a line (Gradient) : A
Do you have seen ladder? Generally, how to put (adjust) B
it? Observe and analyze about it.

The real fact of the position of ladder is called the slope θ
of ladder with the ground. In the adjoining figure, AC is a
ladder which makes an angle θ with x-axis in the positive C
direction (i.e. anticlockwise direction) with ground BC.
Then tanθ is called the slope or gradient of the ladder.

Again, Let us observe the following figures. Y
YB B

X' O θ X X' θ
Y' A Y' X

A

The above figures shows that the angle formed by AB in different positions with

OX. Slope of the line is denoted by the letter 'm'. Thus, if a line makes an angle θ

with x-axis in positive direction, the slope (m) = tanθ

∴ Slope(m) = tanθ
The angle (θ) lies between 0° to 180°. If θ is acute, then slope is positive; and if θ is
obtuse, then the slope is negative.

Slope of a straight line joining two given points.

Let, A(x1, y1) and B(x2, y2) be two given points. Y B(x2, x2)
Join AB. Produce BA to meet X-axis at C so

that ∠XCB = θ. A(x1, x1) θ
R
Then, slope of CB = m = tanθ. From A and
B, draw AP and BQ perpendicular to OX. X' O θ X
Similarly, from A, draw AR perpendicular to Y' CP Q
BQ.

∠RAB = ∠XCB = θ [ AR||CQ]

We have, AR = PQ = OQ – OP = x2 – x1 and

BR = BQ – RQ = y2 – AP = y2 – y1 [ RQ = AP]

From right angled ∆ABR

Infinity Optional Mathematics Book - 9 119

tanθ = BR = y2 – y1
∴ AR x2 – x1
∴
Slope (m) = tanθ = y2 – y1
x2 – x1

Slope (m) = y2 – y1
x2 – x1

Cor.1 Two straight lines with equal slopes are said to QY
be parallel. B

Cor.2 If three points A(x1, y1), B(x2, y2) and C(x3, y3)
lie in a same straight line (collinear) then,

Slope of AB = Slope of BC = Slope of AC

Intercepts: O AP X
Let PQ be a straight line which intersects X' Y'

X-axis at A and Y-axis at B. Then OA is said to be

the intercept of the line PQ on x-axis or simply called

x-intercept. Similarly, OB is said to be the intercept of the line PQ on Y-axis or

simply called y-intercept. AB is called the portion of line PQ intercepted between

the axes. X-intercept and Y-intercept are generally denoted by x-intercept (OA) = a

and y-intercept (OB) = b

The equation of straight lines in standard forms :

i) Slope - Intercept form (y = mx + b) : YB
To find the equation of straight line which cuts (0, b) M P(x, y)
the y-axis at a distance of 'b' from the origin and
makes an angle θ with the x-axis in positive direction.

Let, AB be a straight line which cuts y-axis at b X
M such that OM = b and makes an ∠XAB = θ so that X' q
AO
Slope of line AB = m = tanθ and the coordinates
of M are (0, b). If P(x, y) be any point on AB, then Y'

slope of PM = y2 – y1 = y – b = y– b
x2 – x1 x – 0 x

Since, A, M, P and B are collinear points so slope of AB = slope of PM

m = y – b
x

120 Infinity Optional Mathematics Book - 9

or, y – b = mx

or, y = mx + b is the required equation of line AB.

This relation is true for any point (x, y) on AB.

Note : Alternative methods are always accepted.

Cor.1 If the straight line passes through origin, b = 0 and the equation becomes
y = mx + 0
or, y = mx is the required equation of straight line passing through origin.
Cor.2 If θ = 0° then slope of line (m) = tanθ = tan0° = 0 then the equation becomes
y = 0.x + b
or, y = b is the equation of straight line parallel to x-axis.

WORKED OUT EXAMPLES

1. Find the equation of a straight line parallel to x-axis at a distance of
5 units below it.

Solution : Here,

The equation of straight line parallel to x-axis is Y

y = b .............(i) X' OX
Since, the straight line lies 5 units below x-axis.

∴ b = –5 5b
Now, substituting b = –5 in equation (i), we get
y = –5 A B

or, y + 5 = 0 is the required equation straight line. Y'

2. Find the equation of a straight line parallel to

y-axis and passing through the point (4, –2). YA
Solution: Here,

The equation of the straight line parallel to y-axis is

x = a ............. (i) X' a
X
Since, the straight line passes through the point
4 (4, –2)
(4, -2)

∴ a = 4 Y' B


Infinity Optional Mathematics Book - 9 121

Now, substituting a = 4 in equation (i) we get
x = 4
or, x – 4 = 0 is the required equation of straight line.

3. Find the slope of a straight line

(i) making 120° with the x-axis in positive direction.

(ii) Passing through the points ( 4, 5) and (–8, 9) Y
Solution: Here,
(i) Angle of inclination (θ) = 120° 120°
Now, X
slope of line (m) = tanθ = tan120° = – 3
(4, 5)
(ii) Let, X' O X
Y'
(4, 5) = (x1, y1), (–8, 9) = (x2, y2)
To find: Slope of line (m) (–8, 9) Y

By formula, X' O
Y'
slope of line (m) = y2 – y1
x2 – x1

= 9– 5 = 4 = –1
–8 – 4 –12 3

4. The side BC of an equilateral ∆ABC is

parallel to the x-axis. Find the slopes of the sides BC, AB and AC.

Solution: Here, In an equilateral triangle each angle = 60° A
Y

∴ ∠CBA = ∠XPA = 60°

[Corresponding angle BC // PQ] B 60° 60° C

∴ Slope of AB(m) = tanθ = tan60° = 3 X' 60° 60° QX

OP

and ∠XQA = 180° – 60° [ By sum of linear pair] Y'

= 120°

∴ Slope of AC(m) = tanθ = tan120° = – 3

Since, the side BC is parallel to x-axis, so angle between BC and OX is 0°.
∴ θ = 0°
∴ Slope of BC (m) = tanθ = tan0° = 0

122 Infinity Optional Mathematics Book - 9

5. Prove that the points (2, 3), (8, –6) and (4, 0) are collinear.
Solution: Here,

Let, A(2, 3), B(8, –6) and C(4, 0) be three given points

Then, slope of AB = y2 – y1 = –6 – 3 = –9 = –3
x2 – x1 8–2 6 2

Slope of BC = y2 – y1 = 0– –(–86) = 6 = –3
x2 – x1 4 –4 2

Since, slope of AB = slope of BC and B is a point common to them.
∴ A, B and C are collinear.

6. Find the equation of the straight line cutting of an intercept –5 from

y-axis and inclined at 30° with x-axis. Y
Solution: Here, y-intercept (b) = –5

Angle of inclination (θ) = 60° 1 O 30° X
3 5
Now, slope of line (m) = tanθ = tan30° = X'

The equation of straight line is (0, –5)

y = mx + b Y'

or, y= 1 x + (–5)
3
x –5
or, y = 3 3

or, 3 y = x – 5 3
or, x – 3y = 5 3 is the required equation of straight line.

7. Find the equation of the straight lines cutting off an intercept 4 from
Y
the y-axis and equally inclined to the axes.

Solution: Here, B
Let, AB and BC be the two straight lines which are
45° 45°
equally inclined to the axes. So, AB and BC makes

angle 45° with the axes. X' A 45° 45° X
For the line AB: OC
Inclination (θ) = 45°

Slope (m) = tanθ = tan45° = 1 Y'
y-intercept (b) = 4

∴ The equation of line AC is, y = mx + b

or, y = 1 × x + 4

or, x – y + 4 = 0 is the required equation.

For line BC:

Inclination (θ) = 180° – 45° = 135°

Infinity Optional Mathematics Book - 9 123

Slope (m) = tan135° = –1

y-intercept (b) = 4

The equation of line BC is, y = mx + b

y = (–1) x + 4

or, y = –x + 4

or, x + y = 4 is the required equation.

Hence, the required equation of straight lines are x – y + 4 = 0 and x + y = 4.



Exercise 4.3

Section 'A'

1. (a) Define slope of a straight line.

(b) What is y-intercept of a straight line.

(c) Write the equation of the straight line in slope intercept form.

(d) Write the slope of the straight line passing through the points (a, b) and
(c,d).

2. (a) Find the equation of the straight line parallel to x-axis which is at a
distance of 6 units about it.

(b) Find the equation of the straight line parallel to y-axis which is at a
distance of 6 units to the right.

3. (a) Find the slope (gradient) of line whose inclination is

(i) 60° (ii) 90° (iii) 135° (iv) 150°

(b) Find the inclination of the line whose gradient is:

(i) 0 (ii) 1 (iii) –1 (iv) – 3
3

4. (a) Find the slope of the line AB in the given figures.

(a) Y (b) Y (c) Y

X' BB O
A 30°
135° A X
(c) X' Y' B
X
60° X X' A
O Y'

Y'

(b) Find the slope of line passing through the following pair of points.
(i) (4, 5) and (1, 2) (ii) (2, 3) and (6, 7)

124 Infinity Optional Mathematics Book - 9

5. (a) Find the equation of the straight line passing through origin and having
slope 3.

(b) Find the equation of the straight line with slope 1 and y-intercept 4.

Section 'B'

6. (a) Find the equation of the straight line parallel to x-axis and passing
through the point (–2, 4).

(b) Find the equation of the straight line parallel to y-axis and passing

through the point (–7, –8).

7. (a) Find the angle of inclination of the line through each pair of the following

points.

(i) (2, 3) and (4, – 3) (ii) (4, 5) and (5, 6)

(b) Prove that the following set of points are collinear.

(i) (3, –2), (2, 2) and (1, 6) (ii) (2, 3), (–1, –1) and (8, 11)

(c) If the points A(5, 6), B(4, 5) and C(3, b) are collinear, find the value of b.

(d) If the points (x, 7), (3, 5) and (1, 3) are collinear, find the value of x.

8. (a) If the slope of a straight line joining the points (8, k) and (10, 2) is -1.

Find the value of k.

(b) If the slope of the line joining the two points (2, m) and (4, 6) is –3 find
2

the value of m. Also find the distance between them.

9. (a) Find the equation of the straight line when

(i) y- intercept (b) = 5 and inclination = 45°

(ii) y - intercept = –1 and inclination = – 60°

(iii) y- intercept = 7 and inclination = 150°
3

(b) Find the equation of the straight line passing through origin and meeting

x-axis at an angle of

(i) 30° (ii) 60° (iii) 135° (iv) tan-1 1
(c) 3

Find the equation of straight line having y-intercept 3 units and parallel

to x-axis

Infinity Optional Mathematics Book - 9 125

Section 'C'

10. (a) In the given figure, BC is parallel to x-axis. Find the slope of AC, BC and
AB.

(i) Y (ii) Y

AA

B 45° 60° C B C
X' O X
X' O X
Y'
Y'

(b) The side PQ of a square PQRS is parallel to the y-axis and side PS is

parallel to x-axis. Calculate the slopes of PS, PR, PQ, QS, QR and RS.

11. (a) If the points (3, 3), (h, 0) and (0, k) are collinear then show that

1 + 1 = 1
h k 3

(b) If (x, y), (0, b) and – mb , 0 are collinear, then show that y = mx + c.

12. (a) Find the equation of the straight line cutting of an intercept 3 from y-axis

and inclined at 45° with x-axis. Also, show that the line passes through

the point (1, 4).

(b) Find the equation of the straight line passing through the point (0, 4)

and making an angle 120° with x-axis in the anticlockwise direction and

prove that it passes through the point ( 3, 1).

13. (a) Find the equation of the straight lines passing through the point (0, 6)
and equally inclined to the axes.

(b) Find the equation of the straight lines passing through the point (0, –4)
and equally inclined to the axes.

(c) Find the equation of the straight lines bisecting the angle between the
axes.

126 Infinity Optional Mathematics Book - 9

(ii) Double intercept from (Intercepts form) x + y =1 :
To find the equation of a straight line making a b

intercepts 'a' and 'b' on x-axis and y-axis NY
respectively. B(0, b)
Let, MN be a straight line which cuts
P(x, y)

on x-axis at A and y-axis at B such that X' O A(a, 0)
x-intercept (OA) = a and y-intercept (OB) = Y' X
b. Then, the coordinates of A and B are (a, M
0) and (0, b) respectively. Let P(x, y) be any

point on the line AB.

Now, Slope of AB = y2 – y1 = b – 0 = –b
x2 – x1 0 – a a

and slope of BP = y2 – y1 = y – 0 = y – b
x2 – x1 x – 0 x

Since, A, P and B lie on a same straight line, they are collinear. So,

Slope of BP = Slope of AB

y – b = –b = 1
x a

or, ay – ab = –bx
or, bx + ay = ab
Dividing both sides by ab, we get

bx + ay = ab
ab ab ab

or, ax + y =1
b

This relation is satisfied by any point on MN.

So, x + y =1 represents the equation of the straight line MN.
a b

Can you derive x + y = 1 by another method? Teachers or facilitators are
a b

requested to encourage the students to do by searching another method.
.

Infinity Optional Mathematics Book - 9 127

WORKED OUT EXAMPLES

1. Find the equation of a straight line which cuts off intercept –3 and 2
from the axes. Also, prove that the line passes through the point (0, 2).

Solution: Here, x-intercept (a) = –3 Y
and y - intercept (b) = 2 B
Now, the equation of straight line in double

intercept form is x + y = 1 2
a b OX
Y'
or, x + y = 1 X' A 3
–3 2

or, 2x––63y = 1

2x – 3y + 6 = 0 ........... (i) is the required
equation of line.

If it passes through the point (0, 2) it should satisfy equation (i)
Now, put x = 0 and y = 2, in equation (i), we get
or, 2 × 0 – 3 × 2 + 6 = 0
or, 0 – 6 + 6 = 0
0 = 0 which is true.
Hence, it is proved that the line passes through the point (0, 2).

2. Find the equation of a straight line passing through the point (3, –4)

and has intercepts on the axes equal in magnitude but opposite in

sign. Y
Solution: Here,

Let, x-intercept (a) = k then y -intercept X' O k AX
(b) = –k

Here, k (3, –4)
the equation of the straight line is

x + y = 1 B
a b Y'

or, x + y = 1
k –k

or, x – y = 1
k k

or, x – y = k ............ (i)

128 Infinity Optional Mathematics Book - 9

Since, the line passes through the point (3, –4)
So, it should satisfy equation (i)
Now, put x = 3 and y = – 4 then from equation (i) we get
or, 3 – (–4) = k
or, k = 7
Substituting k = 7 in equation (i),we get
x – y = 7 is the required equation of line.
3. Find the equation of a straight line passing through the point (–1, 3)

whose intercept on x-axis is thrice that on y-axis.
Solution: Here,

Let, y - intercept (b) = k then

x - intercept (a) = 3k Y

Now, the equation of line is B

x + y = 1 (–1, 3) k
a b OX
x y Y'
or, 3k + k = 1 X' A 3k

or, x + 3y =1
3k

x + 3y = 3k .......... (i)

Since, the line passes through the point (–1, 3) so it should satisfy equation (i)

Now, put x = –1 and y = 3 then from equation (i), we get

–1 + 3 × 3 = 3k

k= 8
3
8
Substituting k = 3 in equation (i) we get

x + 3y = 3k

or, x + 3y = 3 × 8
3

or, x + 3y = 8 is the required equation of line.

Infinity Optional Mathematics Book - 9 129

4. Find the equation of a straight 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.

Solution: Here,

Y

Let, AB be a straight line intercepted between X' O A (a, 0) X

the axes such that x-intercept (OA) = a and

(2, –2)

y-intercept (OB) = b. Then the coordinates of A B(0, b)
and B are (a, 0) and (0, b) respectively. Y'

The point (2, –2) divides AB in the ratio of 3:4.

Now,

let, A(a, 0) = (x1, y1) B(0, b) = (x2, y2)
m1 : m2 = 1:2 (2, –2) = (x, y)

By using section formula

m1 = x – x1 and m1 = y – y1
m2 x2 – x m2 y2 – y

or, 21 = 2 – a or, 1 = –2 – 0
0 – 2 2 b+2

or, 1 = 2–a or, 1 = –2
2 –2 2 b+2

or, –1 = 2 – a or, b + 2 = – 4

a = 3 b=–6

Now, the equation of straight line is

x + y = 1
a b

or, x + y = 1
3 –6

or, x – y = 1
3 6

or, 2x6– y = 1

or, 2x – y = 6 is the required equation of the line.

130 Infinity Optional Mathematics Book - 9

5. Find the equation of the straight lines which passes through the point

(2, 2) and makes intercepts on the axes, the sum of whose length is 9.

Solution: Here, Y

Let, AB be a straight line which cuts x-axis Q B(0, b)
b (2, 2)
and y-axis at A and B such that (OA) = a and

OB = b. Then the coordinates of A and B are (a,

0) and (0, b) respectively. The line AB passes

through the point (2, 2).

Now, a + b = 9 X' Oa A(a,0) X

b = 9 – a .............. (i)

The equation of line is, x + y = 1 Y'
a b

or, x + 9 y a = 1
a –

or, (9 –a(a9) x + ay = 1
– a)

or, (9 – a) x + ay = a(9 – a) ......... (i)

Since, the line passes through the point (2, 2) it should satisfy equation (i)

Now, put x = 2 and y = 2 then

(9 – a) × 2 + a × 2 = 9a – a2

or, 18 – 2a + 2a + a2 – 9a = 0

or, a2 – 9a + 18 = 0

or, a2 – (6 + 3) a + 18 = 0

or, a2 – 6a – 3a + 18 = 0

or, a(a – 6) – 3(a – 6) = 0

or, (a – 6) (a – 3) = 0

Either, a – 6 = 0 ⇒ a = 6 OR, a – 3 = 0 ⇒ a = 3



Case I: When a = 6 then from equation (i)

(9 – 6) x + 6y = 6(9 – 6)

or, 3x + 6y = 18

or, 3 (x + 2y) = 18

or, x + 2y = 6 .......... (ii)



Case II: When a = 3 then from equation (i) we get

(9 – 3) x + 3y = 3(9 – 3)

or, 6x + 3y = 18

or, 3(2x + y) = 18

or, 2x + y = 6 ........... (iii)

Hence, equation (ii) and (iii) represent the required equations of straight lines.

Infinity Optional Mathematics Book - 9 131

Exercise 4.4

Section 'A'
1. (a) What do you mean by x-intercept and y-intercept of a straight line?
(b) Write the equation of a straight line in double intercept form.
2. Find the equation of the straight line in the following condition.
(i) x– intercept (a) = 5, y – intercept (b) = 10
(ii) x – intercept = 4, y – intercept = – 8
(iii) x – intercept = –3, y–intercept = 7

Section 'B'

3. Find the equation of the straight line whose intercepts on the axes of
x and y are 4 and –3 respectively. Also, prove that the line are passes
through the point (–6, –4).

4. (a) Find the equation of the straight line which passes through

(i) the point (4, 1) and makes equal intercepts on both the axes.

(ii) the point (–3, –4) and makes equal intercepts on the axes.

(iii) the point (3, –2) and makes intercepts equal in magnitude but
opposite in sign.

(iv) the point (6, –5) and cuts off equal intercepts in magnitude but
opposite in sign.

(b) Find the equation of the straight line which passes through

(i) the point (3, 2) and makes an intercept on the x-axis is twice as
long as that on the y-axis.

(ii) the point (–1, 3) whose intercept on the y-axis is thrice that on
x-axis.
Section 'C'

5. (a) Find the equation of a straight line, a portion of whose length intercepted
between the axes is bisected at the point (3, 4).

(b) Obtain the equation of a straight line whose portion of the line intercepted
between the axes is bisected at the point (–4, –6).

(c) Prove that the equation of straight line in which the point (m, n) bisects
x y
the portion of the line between the axes is 2m + 2n =1

6. (a) Find the equation of the straight line which passes through the point (2,

3) and is such that the portion of it between the axes is divided by the

point in the ratio 1:2.

(b) Obtain the equation of a straight line the portion of which intercepted
between the axes is divided by the point (–5, –6) in the ratio 3 : 4.

132 Infinity Optional Mathematics Book - 9

7. (a ) Determine 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.

(b) Obtain the equation of straight line which passes through the point (2,
1) and makes intercepts on the axes whose product is 8.

(c) 'a' and 'b' are the intercepts made by a line on the coordinate axes. If 2a
+ b = 6 and the line passes through the point (1, 1), find the equation of
the straight line.

8. (a) Find the equation of the straight lines which have slope –3 and form
with the coordinate axes a triangle of area 24 sq. units. 4

(b) A straight line passes through the point (0, –6) and cuts off a triangle of
area 12 sq. units from the axes of coordinates, find the equation of the
straight lines.

(iii) Normal form (Perpendicular form) (x cosa + y sin a = p)

To find the equation of a straight line in Y
terms of the length of the perpendicular 'p' B
from origin upon it and the angle 'a' which

makes by perpendicular with the positive

direction of the x-axis. M

Let, AB be a straight line which meets x -axis p Q(x,y)
at A so that x -intercept = OA and y-axis at AX

B so that y-intercept = OB. Let, OM = p be X' a
the length of perpendicular drawn from O O

on AB and ∠AOM = a. Y'
Let, Q(x, y) be any point on AB.

In right angled triangle OAM.

cos a = OM = p
OA OA
p
or, OA = cosa .............. (i)

We have, ∠MOB = 90° – a

In ∆OMB

Cos(90° – a) = OM
OB
p
or, sina = OB

or, OB = p ............... (ii)
sina

Infinity Optional Mathematics Book - 9 133

According to double intercepts from, the equation of AB is

x + y =1
OA OB

or, px + y =1
p

cosa sina

or, xcopsa + ysina = 1
p

\ xcosa + ysina = p is the required equation of straight line in the perpendicular
form.

Can you derive xcosa + ysina = p by any other (alternative) method? Teachers
or facilitators are requested to encourage the student to do this by other
method.

Note: The length of perpendicular line drawn from origin to the line (P) is always
positive and the value of a will from 0° to 360°.

WORKED OUT EXAMPLES

1. The length of the perpendicular from the origin upon a line is 3 and

the perpendicular makes an angle of 135° with the positive x-axis.

Find the equation of the line. Also, show that it passes through

(–2 2, 2). YB
Solution: Here,

Length of perpendicular (P) = 3

Angle made by perpendicular with x-axis (a) = 135°

We have, M
The equation of line is
xcosa + ysina = P X' A 3 X
or, xcos135° + ysin135° = 3 135°

O

or, x – 1 +y 1 =3 Y'
2 2

or, –x + y = 3
2

or, –x + y = 3 2

or, x – y + 3 2 = 0 ....... (i) is the required equation of straight line. If it

134 Infinity Optional Mathematics Book - 9

passes through (–2 2, 2) it should satisfy equation (i)

Now, put, x = –2 2 and y = 2 then from equation (i), we get

–2 2 – 2 + 3 2 = 0

or, –3 2 + 3 2 = 0

0 = 0 which is true.

Hence, this shows that the line passes through the point (–2 2, 2).

2. A straight line makes an intercept of 4 units of length of the x-axis

and the perpendicular from the origin on it is inclined at 60° to the

x-axis. Find the equation of the line. Y

Solution: Here, B

x- intercept (OA) = a = 4 units

Angle made by perpendicular with x-axis (a) = 60° M
In right angled ∆OMA p
60°
cos60° = OM X' O4 AX
OA
Y'
12 = p
4

or, p = 2

Again, the equation of line is
xcosa + ysina = p

or, xcos60° + ysin60° = 2

or, x × 1 + y × 3 =2
2 2

or, x + 3y = 4 is the required equation of line.

Infinity Optional Mathematics Book - 9 135

Exercise 4.5

Section 'A'

1. (a) Write the equation of straight line in perpendicular form.
(b) What does p and a represent in xcosa + ysina = p?
(c) Write the range of angle made by the perpendicular with x-axis (a).

Section 'B'

2. Find the equation of a straight line if

(i) Length of perpendicular from the origin on the line (p) = 4 units.

Angle made by the perpendicular with x-axis (a) = 30°

(ii) p = 9 units, a = 60°

(iii) p = 8 units a = 150°
5
(iv) p = 17 units, a = 135°
(v) p = units a = 60°

Section 'C'

3. (a) The length of the perpendicular from the origin on a straight line is 3
units and the perpendicular is inclined at 120° with the x-axis. Find the
equation of the line and show that it passes through (–3, 3).

(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 straight line and prove that the line passes through the point
(2, 6 3).

4. Find the equation of the straight line PQ in the following figures.

(i) ∠XPQ = 60° and OM = 3 units (ii) ∠XOM = 30° and OM = 3 units

YY

Q X' P O X
M 30°

3 units 3
M
X' 60° O X
P Q

Y' Y'
5. (a) Find the equation of a straight line which cuts off an intercept –2 from

the x-axis and whose perpendicular distance from origin is 1 units.

136 Infinity Optional Mathematics Book - 9

(b) Find the equation of the straight lines whose perpendicular distance
from origin is 2 units and passes through the point (4, 0).

Y

6. In the adjoining figure, a straight line AB cuts the B (0, 4)

x-axis at A and y-axis at B. The coordinates of B M

is (0, 4). OM⊥AB and ∠AOM = 30°. Find the the

equation of line AB. 30°
O
X' AX

Y'

IV General Equation of first Degree
What is called first degree ? Discuss on it.
The equation of the form Ax + By + C = 0, where A, B and C are constants and

x and y are variable of degree one is known as general equation of first degree.
First degree equation is also called linear equation. The linear equation
always represents a straight line . We can reduce this linear equation in three
standard forms.
Reduction of the linear equation Ax + By + C = 0 in three standard forms.
(a) In slope intercept form (y = mx + b)
The given equation of a straight line is
Ax + By + C = 0
or, By = –Ax – C

or, y = – BAx – C
B

or, y = – A x+ – C ............ (i)
B B

Equation (i) is in the form of y = mx + b .......... (ii)

Comparing equation (i) and (ii), we get

Slope (m) = – A = – Coefficient of x
B Coefficient of y

and y – intercept (b) = – C =– Constant term
B Coefficient of y

Is the value of 'B' becomes zero? (How?)

(b) In double intercept from x + x = 1
a b

The given equation of a straight line is

Ax + By + C = 0

Infinity Optional Mathematics Book - 9 137

or, Ax + By = –C

Dividing both sides by –C, we get

Ax–+CBy = – C
– C

or, –AC x+ B y=1
–C

or, x + y = 1 ............ (i)]
–C –C

AB

which is in the form of x + x = 1 .......... (ii)
a b

Comparing equation (i) and (ii), we get

x-intercept (a) = –C = – Constant term
A Coefficient of x

y-intercept (b) = –C = – Constant term
B Coefficient of y

What is the nature of line if A = 0 and B = 0?
(c) In perpendicular (normal) form (x cosa + ysina = p)
The given equation of a line is
Ax + By + C = 0 ..............(i)
The equation of a line in the perpendicular form is
xcosa + ysina = p
or, xcosa + ysina – p = 0 ............... (ii)
Equation (i) and (ii) will be identical if

cosa = sina = –p = k(suppose)
A B C

∴ Cosa = Ak, sina = Bk and p = –Ck

We know that,

cos2a + sin2a = 1

or, A2k2 + B2k2 = 1

or, k2(A2 + B2) = 1

or, k2 = 1
A2 + B2

138 Infinity Optional Mathematics Book - 9

or, k=± 1
A2 + B2

∴ cosa = ± A B2, sina = ± B B2 and p = ± C
A2 + A2 + A2 + B2

Substituting the values of cosa, sina and p in equation (ii) we get

± A B2 x + B y=± C B2 .............. (iii)
A2 + A2 + B2 A2 +

∴ The equation (iii) is in the form of x cos a + y sin a = p.
The sign of the value of k should be chosen such that p is always positive.

Note: To reduce the equation to the perpendicular form.

(i) Divide both sides of the given equation by

A2 + B2 i.e. (coeff. of x)2 + (coeff. of y)2
(ii) Transpose the constant term to the right hand side and make it positive if

necessary as term of R.H.S must be positive to get p.

WORKED OUT EXAMPLES

1. Reduce the equation 2y – 10x = 8 to the slope intercept form and find
slope and the y-intercept.

Solution: Here,

the given equation of line is
2y – 10x = 8
Now,
2y = 10x + 8

or, y = 10 x + 8
2 2

or, y = 5x + 4 ............ (i)

∴ Equation (i) is in the form of y = mx + b ......... (ii)
By comparing equation (i) and (ii), we get
Slope (m) = 5
and y-intercept (b) = 4

Infinity Optional Mathematics Book - 9 139

2. Reduce the equation 4x + 7y + 5 = 0 to the double intercept form. Also,
find x-intercept and y - intercept.

Solution: Here, the given equation of straight line is 4x + 7y + 5 = 0
Now, 4x + 7y = – 5

or, 4 x + 7 y = –5 [ Dividing both sides by –5]
–5 –5 –5

or, x + y =1 ............... (i)
–5 –5

47

∴ Equation (i) is in the form of x + y = 1 .............. (ii)
a b

By comparing equation (i) and (ii),we get

x-intercept (a) = –5
4

and y-intercept (b) = –5
7

3. Reduce the equation 3x – y + 2 = 0 to the perpendicular form. Also,
find the length of the perpendicular from the origin on the line and
the angle made by the perpendicular with the x-axis.

Solution : Here, the given equation of line is 3x – y + 2 = 0 .............(i)
Now,

(Coefficient of x)2 + (Coefficient of y)2

= ( 3)2 + (–1)2

= 3 + 1 = 2
Dividing equation (i) on both sides by 2 we get

3 x – 1 y + 2 = 0
2 2 2 2

or, 23 x – 1 y + 1 = 0
2

or, 23 x – 1 y =–1
2

or, – 3 x+ 1 y = 1 [ Multiplying both sides by –1]
2 2

or, x – 3 +y 1 = 1 ............. (ii)
2 2

∴ Equation (ii) is in the form of xcosa + ysina = p ............ (iii)

140 Infinity Optional Mathematics Book - 9

By comparing equation (ii) and (iii), we get CAST Rule

cosa =– 23, sina = 1 and p =1 S A
2 II θ
Since, the value of cosa is negative and sina 180° – θ I

is positive, so a lies in the second quadrant. T C
Principle angle (P.A) = 30° 180° + θ 360°θ
∴ a = 180° – P.A.
= 180° – 30° = 150° III IV

Where θ is the acute
angle of the 1st quadrant

Hence, a = 150° & p =1.

4. The line 2x + 3y + m = 0 cuts the x-axis and y-axis at the points P and
Q respectively. If the area of ∆POQ is 27 square units, find the value
of 'm'.

Solution: Here, the given equation of line is 2x + 3y + m = 0 ............ (i)
Y
Area of ∆POQ = 27 sq. units

Now, by formula, Q

x-intercept (OP) = a = – Constant term =0
Coefficient of x 2x + 3y + m

= – m
2

y-intercept (OQ) = b = – Constant term
Coefficient of y
X' O PX
= – m
3
Y'
Since, ∆POQ is a right angled triangle then by formula

Area of ∆POQ = 1 × OP × OQ
2

or, 27 sq. unit = 1 × –m × –m sq. units..
2 2 3

or, 27 × 12 = m2

or, m2 = 324

∴ m = ± 18

Hence, m = ± 18.

Infinity Optional Mathematics Book - 9 141

Exercise 4.6

Section 'A'

1. (a) What are the three standard forms of first degree equation? Write them.

(b) Write slope and y-intercept of the equation y = –4x – 3

(c) What are the values of x-intercept and y-intercept of the equation x + y
= 1? Write them. 2 3

2. (a) Write the general form of first degree equation,

(b) What are called 'a', 'b' and 'c' in general equation of first degree

ax + by + c = 0-

(c) Reducing the equation 3x + 4y = 10 into perpendicular from is

3x 42 + 4y 42 = 10 42 then find the value of cos a, sin a and p.
32 + 32 + 32 +

Section 'B'

3. Find the slope and y-intercept of the following lines by reducing the

equations into slope intercept form.

(i) 3y + x = 2 3 (ii) 2x + 3y – 6 = 0

(iii) 7x – 8y + 15 = 0 (iv) 3x – 2y = 5 2

4. Find the intercepts (x-intercept and y-intercept) of the following

lines by reducing the equations into double - intercept form.

(i) 4x – 3x – 12 = 0 (ii) 8x – 5y = –60
(iii) 3x + y – 2 3 = 0 5x 3y
(iv) 6 + 5 = 15

5. Reduce the following given equations into the perpendicular form

and find a and p.

(i) 3x – y – 4 = 0 (ii) x – y + 4 = 0
(iii) y – x = 2 2 (iv) x – 3y + 8 = 0

6. What intercepts does the line x – y = 1 make on the axes? What is the gradient
3 5

of this line?

142 Infinity Optional Mathematics Book - 9

Section 'C'

7. (a) Reduce the equation 3x – y – 4 = 0 into three standard forms.
(b) Reduce the equation 3x + y + 4 = 0 into the form of xcosa + sina = p.

Also, find the value of and p.

8. (a) Convert the equation y = mx + b into xcosa + ysina = p and prove that

m2 + 1 = 1
b2 b2 p2
x y
(b) Reduce the equation a + b =1 into the form of xcosa + ysina = p. Also,

prove that 1 + 1 = p12.
a2 b2

9. (a) The line 2x + 3y + 18 = 0 meets the x-axis and y-axis at the points A and

B respectively. Find the area of triangle AOB.

(b) The line 4x + 5y = 20 cuts the x-axis and y-axis at the points P and T

respectively. Find the area of triangle POT.

(c) Find the value of 'h' so that the line whose equation is 2x + 3y + h = 0

will form a triangle with the coordinate axes whose area is 48 sq. units.

10. Make a report by doing the following activities and present in the class.

(a) Write down five equations of straight lines.

(b) Draw the lines on the graph of those equations.

(c) Convert those equations into three standard form of equations.

Equation of a straight lines in special cases :

We can obtain the equation of a straight q B
lines if A Y

☻ It's slope and a point through which P(x, y)
it passes are known. D C(x1, y1)

☻ Two points through which it passes X
are known. Y'

(i) Point slope form
To find the equation of a straight line X'

passing through a given point (x1, y1)
with a given slope (m).

Infinity Optional Mathematics Book - 9 143

Let, θ be the angle of inclination of the straight line AB which passes through

the point C(x1, y1).
Therefore, slope of line AB is 'm' = tanθ. If P(x, y) be any point on the line AB

then slope of PC = y2 – y1 = y – y1
x2 – x1 x – x1

Since, A, D, C, P and B are collinear so slope of AB = Slope of PC

m = y – y1
x – x1

or, y – y1 = m(x – x1)

which is the required equation of a straight line.

If the above line AB passes through origin then what would be its equation?

Find it. Y

(ii) Two points form: B
To find the equation of a straight line passing N(x2, y2)
P(x, y)
through two given points (x1, y1) and (x2, y2).
Let, AB be a straight line passing through the

points M(x1, y1) and N(x2, y2). P(x, y) be any X' O M(x1, y1)
A
point on the line AB. X

Since, the point M, N and P lies on the same Y'

straight line.

Slope of MP = Slope of MN

or, xy –– y1 = y2 – y1 [∴ Slope (m) = y2 – y1 ]
x1 x2 – x1 x2 – x1

∴ y – y1 = y2 – y1 (x – x1) which is the required equation of a straight
x2 – x1

line.

Can you derive the equation of straight lines (i) y – y1 = m(x – x1) and

(ii) y – y1 = y2 – y1 (x – x1) by any alternative methods? Try to do them by
x2 – x1

other methods.

144 Infinity Optional Mathematics Book - 9

WORKED OUT EXAMPLES

1. Find the equation of a straight line passing through the point (2, 3)
and makes an angle 60° with the x-axis.

Solution: Here,

The angle of inclination of the line (θ) = 60°

Passes point (2, 3) = (x1, y1)
Now, by formula,

Slope (m) = tanθ = tan60° = 3

The equation of the line is

y – y1 = m(x – x1)

or, y – 3 = 3(x – 2)

or, y – 3 = 3x – 2 3

or, 0 = 3x – y + 3 – 2 3

or, 3x – y + (3 – 2 3) = 0 is the required equation of line.

2. Find the equation of a straight line passing through the points (–3, 4)
and (2, 7).

Solution: Here, a straight line passing through the points

Let, (–3, 4) = (x1, y1) (2, 7) = (x2, y2)

The equation of line in two point form is

y – y1 = y2 – y1 (x – x1)
or, x2 – x1
or,
y–4= 7 – 4 (x + 3)
2 + 3

y – 4 = 3 (x + 3)
5

or, 5y – 20 = 3x + 9

or, 3x – 5y + 29 = 0 is the required equation of straight line.

Infinity Optional Mathematics Book - 9 145

3. If the points (1, –4), (2, 5) and (x, 14) are collinear, find the value of x.
Also, find its equation.

Solution: Here,
The points (1, –4), (2, 5) and (x, 14) are collinear.

Let, (1, –4) = (x1, y1)

(2, 5) = (x2, y2)

The equation of line is

y – y1 = y2 – y1 (x – x1)
or, x2 – x1
or,
y + 4 = 5 + 4 (x – 1)
2 – 1

y+4= 9 (x – 1)
1

or, y + 4 = 9x – 9

or, 9x – y = 13 ......... (i)

Since, the points are collinear, so the remaining point (x, 14) should satisfy

equation (i)

Now, put x = x and y = 14 then from equation (i), we get

or, 9 × x – 14 = 13

or, 9x = 13 + 14
27
or, x= 9 = 3

∴ x = 3

Hence, x = 3 and equation (i) 9x – y = 13 is the required equation of a straight
line.

4. Find the equation of the straight line passing through the point

(–3, 4 ) and the centroid of triangle whose vertices are (5, –4), (3, 5)

and (–2, 2). P(5, –4)

Solution: Here, B

Let, P(5, – 4), Q(3, 5) and R(–2, 2) be the vertices of (–3, 4) G

∆PQR. G be the centroid of DPQR. A T R(–2 2)
Q(3, 5)

By centroid formula, the coordinates of centroid

146 Infinity Optional Mathematics Book - 9

x = x1 + x2 + x3 and y = y1 + y2 + y3
3 3

= 25 + 33 – 2 y == –4 +5 + 2
= 1 3

∴ The coordinates of centroid G(2, 1).

Again, the equation of line AB passing through (–3, 4) and (2, 1) is

y – y1 = y2 – y1 (x – x1)
or, x2 – x1
or,
y– 4= 1 – 4 (x + 3)
2 + 3

y–4= –3 (x + 3)
5

or, 5y – 20 = – 3x – 9

or, 3x + 5y = –9 + 20

or, 3x + 5y = 11 which is the required equation of line AB.

5. Find the equation of the straight line joining the point (3, –2) and

(5, –4). At what point does it cut y-axis.

Solution: Here,

Let, the line joining the point (3, –2) and (5, –4) cut y-axis at point P.

Let, (3, –2) = (x1, y1)

(5, –4) = (x2, y2) Y

The equation of line is P

y – y1 = y2 – y2 (x – x1) X' O X
x2 – x1
or, (5, –4)
y + 2 = –4 + 2 (x – 3) (3, – 2)
or, 5–3

or, y + 2 = –2 (x – 3)
or, 2
or,
or, y + 2 = – 1(x – 3) Y'

y+2=–x+3

x+y=3–2

x + y = 1 ....... (i) is the required equation of line.

Infinity Optional Mathematics Book - 9 147

Since, the line cuts on y-axis at P i.e. P lies on y-axis. So its x-coordinate
(x) = 0, so, substituting x = 0 in equation (i), we get

0 + y = 1
or, y = 1

∴ The straight line cuts on y-axis at (0, 1).

6. If the points M(h, 2k) lies on the line x – 2y – 3 = 0 and the point

N(2k, h) lies on the lines y = 11 – 2x. Find the equation of MN and the

length of MN.

Solution: Here, M (h, 2k)x –2y – 3 = 0
The point M(h, 2k) lies on line x – 2y – 3 = 0

∴ h – 2(2k) – 3 = 0

or, h = 4k + 3 ........... (i) N (2k, h) y = 11 – 2x

Similarly, the point N(2k, h) lies on the line y = 11 – 2x

∴ h = 11 – 2 × 2k

h = 11 – 4k ........... (ii)

Equating equation (i) and (ii), we get

4k + 3 = 11 – 4k

or, 4k + 4k = 11 – 3

or, 8k = 8
∴ k = 1

Substituting k = 1 in equation (i), we get

h = 4 × 1 + 3

or, h = 7

∴ The coordinates of M = (h, 2k) = (7, 2 × 1) = (7, 2) = (x1, y1) and, the

coordinates of N = (2k, h) = (2 × 1, 7) = (2, 7) = (x2, y2).

The equation of AB is

y – y1 = y2 – y1 (x – x1)
or, x2 – x1
or,
y–2= 7 – 2 (x – 7)
2 – 7

y–2= 5 ( x – 7)
–5

or, y – 2 = –1(x – 7)

148 Infinity Optional Mathematics Book - 9

or, y – 2 = –x + 7

\ x + y = 9 is the required equation of line.

For the length of MN, by distance formula

MN = (x2 – x1)2 + (y2 – y1)2

MN = (2 – 7)2 + (7 – 2)2 = 25 + 25 = 5 2 units.

7. In what ratio does the line joining the points (1, 3) and (2, 7) divided
by the line joining the points (0, 9) and (1, 6).

Solution: Here,
Let, the line joining A(1, 3) and B(2, 7) be divided by the line joining C(0, 9)

and D(1, 6) at P(x, y) in the ratio m1: m2.

The equation of line CD (divider) is C (0,9)

y – y1 = y2 – y1 (x – x1) m2
or, y – 9= 6x–2 x1 – 0) P (x, y)
or, – m1
1– 9 (x A (1,3)
0
–3 B (2,7)
y–9= 1 (x)

or, 3x + y = 9 ............ (i)

For the straight line AB, by using section formula, D (1,6)

x = m1x2 + m2x1 and y = m1y2 + m2y1
m1 + m2 m1 + m2

= m1 ×2 + m2 × 1 y = m1 ×7 + m2 × 3
m1 + m2 m1 + m2

x = 2mm11++mm22 y = 7m1 + 3m2
m1 + m2

Since, P(x, y) lies on CD so it must satisfy equation (i)

or, 3 2m1 + m2 + 7m1 + 3m2 = 9
m1 + m2 m1 + m2

or, 6m1 + 3m2 + 7m1 + 3m2 = 9
m1 + m2

or, 13m1 + 6m2 = 9m1 + 9m2

or, 13m1 – 9m1 = 9m2 – 6m2

Infinity Optional Mathematics Book - 9 149

or, 4m1 = 3m2

or, mm12 = 3
4

∴ m1 : m2 = 3: 4 is the required ratio.

Exercise 4.7

Section 'A'

1. (a) Write the equation of straight line in a point slope form.

(b) What is the equation of straight line passing through the points (x1, y1)
and (x2, y2) ? Write it.

2. (a) What is the equation of straight line passing through the point (h, k) and

having slope 'g'?

(b) Find the equation of a straight line passing through (0, 0) and having

slope 1.

Section 'B'

3. (a) Find the equation of a straight line passing through the point.

(i) (4, 1) and having slope = –1

(ii) (–1, 2) and having slope (m) = 2
3

(b) Find the equation of a straight line passing through

(i) the point (2, –2) and making angle 60° with x-axis.

(ii) the point (4, 5) and making angle 135° with x-axis.

(iii) the point (1, 2) and making angle 150° with x-axis.
(iv) The point (–3, –2) and making angle tan-1 1 with x-axis.

3

(c) Find the equation of a straight line passing through (2, 1) and making
an angle of 45° with x-axis. Also, show that the line passes through
(0, –1).

150 Infinity Optional Mathematics Book - 9