CBSE

Class 10 Mathematics
NCERT Exemplar Solutions
Chapter 7 COORDINATE GEOMETRY

EXERCISE 7.1

Choose the correct answer from the given four options in the following questions:

1. The distance of the point P(2, 3) from x–axis

(a) 2

(b) 3

(c) 1

(d) 5

Sol. (b): The perpendicular distance of P(2, 3) from x–axis is equal to the y coordinate so, it is
3 units. verifies ans. (b).

2. The distance between the points A(0, 6) and B(0, –2) is

(a) 6

(b) 8

(c) 4

(d) 2

Sol. (b): AB =
⇒ AB = 8 units

Hence, verifies Ans (b).

3. The distance of the point P(–6, 8) from the origin is

(a) 8

(b)

(c) 10

(d) 6

Sol. (c): Coordinates of origin are O(0, 0) and P(–6, 8)

∴ (OP)2 = (x2 – x1)2 + (y2 – y​1​ )2

= (–6 – 0)2 + (8 – 0)2 = 36 + 64

OP =

⇒ OP = 10 units. verifies ans. (c).

4. The distance between the points (0, 5) and (–5, 0) is

(a) 5

(b)

(c)

(d) 10

Sol. (b): Let A(0, 5) and B(–5, 0) are the two points.
Then, AB2 = (x2 – x1)2 + (y2 – y1)2

= (–5 – 0)2 + (0 – 5)2 = 25 + 25

⇒ AB2 = 50

⇒ AB = units. verifies ans. (b).

5. AOBC is a rectangle whose three vertices are A(0, 3), O(0, 0) and B(5, 0). The length of
its diagonal is

(a) 5

(b) 3

(c)

(d) 4

Sol. (c): A(0, 3) and B(5, 0)

The length of diagonal = AB

AB2 = (x2 – x1)2 + (y2 – y1)2

= (5 – 0)2 + (0 – 3)2

= 25 + 9

⇒ AB = verifies Ans (c).

6. The perimeter of a triangle with vertices (0, 4), (0, 0), and (3, 0) is

(a) 5

(b) 12

(c) 11

(d) 7 +

Sol. (b): Perimeter of ΔABC = AB + BC + AC

Let A(0, 4), B(0, 0), C(3, 0) be the three vertices of ΔABC.

AB2 = (x2 – x1)2 + (y2 – y1)2

= (0 – 0)2 + (0 – 4)2 = 0 + 16

⇒ AB = = 4 cm

AC2 = (3 – 0)2 + (0 – 4)2 = 9 + 16

⇒ AC2 = 25

⇒ AC = 5 cm

BC2 = (3 – 0)2 + (0 – 0)2 = 9 + 0

⇒ BC2 = 9

⇒ BC = 3 cm

∴ Perimeter = 4 cm + 5 cm + 3cm = 12

Hence, verifies Ans. (b).

7. The area of triangle with vertices A(3, 0), B(7, 0) and C(8, 4) is

(a) 14
(b) 28

(c) 8

(d) 6

Sol. (c): Area (A) of ΔABC whose vertices are A(3, 0), B(7, 0) and C(8, 4) is given by

Area of ΔABC = [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

= [3(0 – 4) + 7(4 – 0) + 8(0 – 0)]

= [–12 + 28 + 0] = [16] = 8 sq. units

Hence, verifies the Ans. (c).

8. The points (–4, 0), (4, 0) and (0, 3) are the vertices of a

(a) right triangle

(b) isosceles triangle

(c) equilateral triangle

(d) scalene triangle

Sol. (b): Let the vertices of ΔABC are A(–4, 0), B(4, 0) and C(0, 3).

AB2 = (x2 – x1)2 + (y2 – y1)2

⇒ AB2 = [4 – (–4)]2 + (0 – 0)2 = 64 + 0 = 64

⇒ AB = 8 cm

AC2 = [0 – (–4)]2 + (3 – 0)2 = 16 + 9 = 25

⇒ AC2 = 25

⇒ AC = 5 cm

BC2 = (0 – 4)2 + (3 – 0)2 = 16 + 9 = 25

⇒ BC2 = 25

⇒ BC = 5 cm

∴ AC = BC = 5 cm and AB = 8 cm

Hence, the triangle is an isosceles triangle. So, verifies ans. (b).

9. The point which divides the line segment joining the points (7, –6) and (3, 4) in ratio 1
: 2 internally lies in the

(a) Ist quadrant

(b) IInd quadrant

(c) IIIrd quadrant

(d) IVth quadrant

Sol. (d):

⇒
⇒

verifies the Ans. (d).

10. The point which lies on the perpendicular bisector of the line segment joining the
points A(–2, –5) and B(2, 5) is

(a) (0, 0)

(b) (0, 2)

(c) (2, 0)

(d) (–2, 0)

Sol. (a): The perpendicualr bisector of AB will pass through the mid–point of Ab. Mid–point

of A(x1, y1) and B(x2, y2) is given by .

= (0, 0)

So, the perpendicular bisector passes through (0, 0).

11. The fourth vertex D of a parallelogram ABCD whose three vertices are A(–2, 3), B(6,
7) and C(8, 3) is

(a) (0, 1)

(b) (0, –1)

(c) (–1, 0)

(d) (1, 0)

Sol. (b): We know that the diagonals AC and BD of parallelogram ABCD bisect each other.
OR

[The mid point of diagonal AC] = [Mid point of diagonal BD]

⇒ (3, 3)

Comparing both sides, we have

and

⇒ x4 + 6 = 6

⇒ y4 + 7 = 6

⇒ x4 = 0

⇒ y4 = 6 – 7 = –1

∴ The fourth vertex of parallelogram is (0, –1) verifies ans. (b).

12. If the point P(2, 1) lies on the line segment joining points A(4, 2) and B(8, 4), then

(a) AP = AB
(b) AP = PB

(c) PB = AB

(d) AP = AB

Sol. (d):

⇒ 8k + 4 = 2k + 2 4k + 2 = k + 1

⇒ 6k = –2 3k = –1

⇒ k = k =

Verification:

⇒ AP = –1 i.e., 1 part outside AB

and PB = 2
∴ AP = 1x unit

and AB = 3x – 1x = 2x units

So, AP = AB

⇒ 1 = × 2

⇒ 1 = 1, which is true

Hence, verifies the ans. (d).

13. If is the mid point of the line segment joining the points

(a) –4

(b) –12

(c) 12

(d) –6

Sol. (b): P(x, y) is mid–point of QR then

⇒ a = –4 × 3 = –12

Verifies the ans. (b).

14. The perpendicualr bisector of the line segment joining the points A(1, 5) and B(4,6)
cuts y–axis at

(a) (0, 13)

(b) (0, –13)

(c) (0, 12)

(d) (13, 0)

Sol. (a): The given points are A(1, 5) and B(4, 6).

The perpendicualr bisector of the line segment joinin g the points A(1, 5) and B(4, 6) cuts the
y–axis at P(0, y).

Now, AP = BP ⇒ AP2 = BP2

∴ 1 + (y – 5)2 = 16 + (y – 6)2

⇒ 1 + y2 – 10y + 25 = 16 + y2 – 12y + 36

⇒ –10y + 26 = –12y + 52

⇒ 12y – 10y = 52 – 26

⇒ 2y = 26

⇒ y = 26 ÷ 2 = 13

So, the reqwuirted point is (0, 13).

Hence, (a) is the correct answer.

15. The coordinates of the point which is equidistant from the three vertices of the
ΔAOB as shown in the figure is
(a) (x, y)

(b) (y, x)

(c)

(d)

Sol. (a): In a right triangle, the mid–point of the hypotenuse is equidistant from the three
vertices of triangle.

Mid–point of A(2x, 0) and B(0, 2y) is

Hence, (a) is the correct answer.

16. A circle drawn with origin as the centre passes through a point (13/2,0). The point
which does not lie in the interior of the circle is

(a)

(b)
(c)

(d)

Sol. (d): Radius of circle = 6.5 units

(a) Distance point from (0, 0) is

= 1.25 units

The distance 1.25 < 6.5. SO, the point lies in the interior of the circle.

(b) Distance of point from (0, 0) is

= 3.0731 < 6.25

So, the point lies in the interior of the circle.

(c) Distance of point from (0, 0) is

= 5.0249 < 6.5

So, the point lies in the interior of the circle.

(d) Distance of point from (0, 0) is

= 6.5 units

So, lies on the circle. It does not lie in the interior of the circle.

Hence, (d) is the correct answer.

17. A line intersects the y–axis at points P and Q respectively. If (2, –5) is the mid–point
of PQ, then co–ordinates of P and Q are respectively.

(a) (0, –5) and (2, 0)

(b) (0, 10) and (–4, 0)

(c) (0, 4) and (–10, 0)

(d) (0, –10) and (4, 0)

Sol. (d): P lies on y–axis so co–ordinates of P are (0, y).

Similaraly, co–ordinates of Q lies on x–axis = Q(x, 0)

Mid–point of PQ is

= M(2, –5), which is given

⇒ = M(2, –5)

⇒ = M(2, –5)

Comparing both side, we get

 and

⇒ x = 4 and y = –10

Hence, the co–ordinates of P(0, –10) and Q(4, 0) verifies ans. (d).

18. The area of the triangle with vertices (a, b + c) (b, c + a) and (c, a +b) is

(a) (a + b + c)2

(b) 0

(c) a + b + c

(d) abc

Sol. (b): If the vertices of ΔABC are

A(x1​, y1) = A(a, b + c)

B(x2, y2​) = B(b, c + a)

C(x3, y3) = C(c, a + b)

Then, Area of ΔABC = [x1(y2 – y3) + x2(y3 – y1) + x3=(y1 – y2)]

⇒ Area of ΔABC = [a{c + a – (a + b)} + b{a + b – (b + c)} + c{b + c – (c + a)}]

= [a(c – b) + b(a – c) + c(b – a)]

⇒ Area of ΔABC

= [ac – ab + ab – bc + bc – ac]

⇒ Area of ΔABC = 0 So, verifies the option (b).

19. If the distance between the point (4, p) and (1, 0) is 5, then the value of p is

(a) 4 only

(b) ± 4

(c) –4 only

(d) 0

Sol. (b): According to the question, the distance between A(4, p) and B(1, 0) is 5 units.

∴ AB = 5 units

⇒ (AB)2 = (5)2

⇒ (4 – 1)2 + (p – 0)2 = 25

⇒ (3)2 + (p)2 = 25

⇒ p2 = 25 – 9

⇒ p2 = 16

⇒ p = ± 4

Hence, verifies the ans. (b).

20. If the points A(1, 2), O(0, 0) and C(a, b) are collinear, then

(a) a = b

(b) a = 2b

(c) 2a = b

(d) a = –b

Sol. (c): Points A(x1, y1), B(x2, y2) and C(x3, y3) will be collinear if the area of ΔABC is zero so,
A(1, 2), B(0, 0), C(a, b) will collinear if area ΔABC = 0

or [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0

⇒ [1(0 – b) + 0(b – 2) + a(2 – 0)] = 0

⇒ (–b + 2a) = 0

⇒ + a = 0

⇒ –b + 2a = 0

⇒ 2a = b

Hence, verifies the ans. (c).

 EXERCISE 7.2

State whether the following statemts are true or false. Justify your answer.

1. ΔABC with vertices A(–2,0), B(2,0) and C(0,2) is similar to ΔDEF with vertices D(–4, 0),
E(4, 0) and F(0, 4).

Sol. True: ΔABC ~ ΔDEF is

In ΔABC,

AB2 = [2 – (–2)]2 + [0 + (0)]2 = (4)2 + 0 = (4)2

⇒ AB = 4 units

BC2 = (0 – 2)2 + (2 – 0)2 = 4 + 4 = 8

⇒ BC = units

AC2 = [0 – (–2)]2 + (2 – 0)2 = 22 + 22 = 4 + 4 = 8

⇒ AC = units

In ΔDEF,

DE2 = [4 – (–4)]2 + (0 – 0)2 = (8)2

⇒ DE = 8 units

EF2 = (0 – 4)2 + (4 – 0)2 = 42 + 42 = 16 + 16 = 32

⇒ EF = units

DF2 = [0 – (–4)]2 + (4 – 0)2

= 16 + 16 = 32

⇒ DE = units

Now,

Hence, ΔABC ~ ΔDEF.

2. Point P(–4, 2) lies on the line segment joining the points A(–4, 6) and B(–4, –6).

Sol. True: We observed that x–coordiante is same i.e., equal to (–4) so line is parallel to y–
axis. y–coordinate of P i.e., 2 lies between 6 and –6 of A and B respectively. Hence, P lies
between A and B on AB.

OR

Point P(–4, 2) will lie on the line AB if area of ΔABP is zero.

∴ i.e., ar(ΔABP) = 0

⇒ [x1(y2 – y1) + x2(y3 – y1) + x3(y1 – y2) = 0

⇒ [–4(–6 – 2) –4(2 – 6) –4(6 + 6)] = 0

⇒ [–4(–8) – 4(–4) –4(12)] = 0

⇒ 32 + 16 – 48 = 0

⇒ 48 – 48 = 0, which is true.

Hence, point P lies on the line joining A and B.

3. The points (0, 5), (0, –9) and (3, 6) are collinear.

Sol. False: Three points A, B and C wil be collinear if the

area ΔABC = 0

⇒ [0(–9 – 6) + 0(6 – 5) + 3(5 – (–9)] = 0

⇒ 0 + 0 + 3(14) = 0

⇒ 42 ≠ 0, which is false.

Hence, the given points are not collinear.

4. Point P(0, 2) is the point of intersection of y–axis and perpendicular bisector of line
segment joining the points A(–1, 1) and B(3, 3).

Sol. False: As the point P(0, 2) is the point of intersection of y–axis and perpendicular bisector
of the line joining the points A(–1, 1) and B(3, 3), then point P must be equidistant from A and
B. So, we must write PA = PB.

PA = units

PA = units

∴ PA ≠ PB

Hence, the given statement is false.

5. Points A(3, 1), B(12, –2) and C(0, 2) cannot be the vertices of a triangle.
Sol. True: Points A, B, C can from a triangle if the sum of any two sides is greater than the
third side.

AB2 = (x2 – x1)2 + (y2 – y1)2

⇒ AB2 = (12 – 3)2 + (–2 – 1) = 81 + 9 = 90

⇒ AB = units

BC2 = (0 – 12)2 + [2 – (–2)]2 = 144 + 16 = 160

⇒ BC = units

AC2 = (0 – 3)2 + (2 – 1)2 9 + 1 = 10

⇒ AC = units

∴ AC = units, AB = units and BC = units

Now, AB + AC = + = units = BC

So, A, B, C points cannot form a Δ.

6. Points A(4, 3), B(6, 4), C(5, –6) and D(–3, 5) are the vertices of a parallelogram.

Sol. False: The diagonals of parallelogram bisect each other so, ABCD will be a parallelogram
if

mid–point of diagonal AC = mid–point of diagonal BD

⇒
⇒

Hence, ABCD is not a parallelogram.

7. A circle has its centre at the origin and a point P(5, 0) lies on it. The point Q(6, 8) lies
outside the circle.

Sol. True: If the distance of Q from the centre O(0, 0) is greater than the radius then point Q
lies in the exterior or the circle. Point P(5, 0) lies on the circle and centre is at O(0, 0) so radius
= OP

OP2 = (x2 – x1)2 + (y2 – y1)2

= (5 – 0)2 + (0 – 0)2

⇒ OP2 = 52

⇒ OP = 5 units

Now, OQ2 = (6 – 0)2 + (8 – 0)2 = 36 + 64 = 100

⇒ OQ = 10 units

∴ OQ > OP (radius)

So, point Q lies exterior to circle.

8. The point A(2,7) lies on the perpendicular bisector of line segment joining the points
P(6,5) and Q(0,–4).

Sol. False: Any point (A) on perpendicular bisector will be equidistant from P and Q so

PA = QA

or PA2 = QA2
⇒ (2 – 6)2 + [7 – (5)]2 = (2 – 0)2 + [7 – (–4)]2

⇒ (–4)2 + (2)2 = 22 + (11)2

⇒ 16 + 4 = 4 + 121

⇒ 20 ≠ 125

So, a does not lie on the perpendicular bisector of PQ.

9. Point P(5, –3) is one of the two points of trisection of the line segment joining the
points A(7, –2) and B(1, –5).

Sol. True

Let pointP divides the line Ab in ratio k : 1 then

and

⇒ ,

⇒ ,

⇒ 5k + 5 = k + 7, –5k – 2 = –3k – 3

⇒ 4k = 7 – 5, –2k = –3 + 2

⇒ ,

So, P divides AB in 1 : 2 ratio.

Hence, P is one point of trisection of AB.

10. Points A(–6, 10), B(–4, 6) and C(3, –8) are collinear such that AB = AC.

Sol. True: Points A, B and C will be colinear if ar (ΔABC) = 0

ar ΔABC = [x1(y2​ – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0

⇒ [–6{6 – (–8)} –4(–8 –10) + 3(10 – 6)] = 0

⇒ –6(14) – 4(–18) + 3(4) = 0

⇒ –84 + 72 + 12 = 0

⇒ –84 + 84 = 0, which is true

So, points A, B and C are collinear.

AC2 = (x2 – x1)2 + (y2 – y1)2

= (3 + 6)2 + (–8 – 10)2 = 81 + 324

⇒ AC = = units

AB2 = [–4 – (–6)]2 + (6 – 10)2

= (–4 + 6)2 + (–4)2

= (2)2 + (–4)2 = 4 + 16

⇒ AB2 = 20

⇒ AB = units
Now, AB = AC

R.H.S.

= AB

Hence, AB = AC is true.

11. The point P(–2, 4) lies on a cirle of radius 6 and centre (3, 5).

Sol. False: The point P(–2, 4) lies on a circle if distance between P and centre is equal to the
radius so distance of P from centre O(3, 5) wil be

OP2 = (–2 – 3)2 + (4 – 5)2

⇒ OP2 = 25 + (–1)2

⇒ OP = ≠ radius 6

So, P does not lie on the circle. It will lie inside the circle.

12. The points A(–1, –2), B(4, 3), C(2, 5) and D(–3, 0) in that order form a recangle.

Sol. True: ABCD will form a recangle if

(i) it is a parallelogram. (ii) diagonals are equal.

For parallelogram: Diagonals bisect each other.

i.e., Mid point of AC = Mid point of BD is

i.e.,
⇒

Hence, ABCD is a parallelogram.

Now, Diagonal AC =

⇒ AC = units

and Diagonal BD =

⇒ BD = units

⇒ BD = units

∴ Diagonal AC = Diagonal BD

Hence, ABCD is a rectangle.

 EXERCISE 7.3

1. Name the type of triangle formed by the points A(–5, 6), B(–4, –2) and C(7, 5).

Sol. A(–5, 6), B(–4, –2), C(7, 5)

AB2 = (x2 – x1)2 + (y2 – y1)2

⇒ AB2 = (–4 + 5)2 + (–2 – 6)2

= (1)2 + (–8)2 = 1 + 64 = 65

⇒ AB = units

AC2 = (7 + 5)2 + (5 – 6)2

⇒ AC2 = (12)2 + (–1)2

⇒ AC2 = 144 + 1

⇒ AC = units

BC2 = (7 + 4)2 + (5 + 2)2 = 112 + 72 = 121 + 49

⇒ BC = units

As AB ≠ BC ≠ AC so scalene triangle.

AC2 + AB2 = 145 + 65 = 210 ≠ BC2, soit is not a right angled Δ

So, a scalene Δ will be formed.

2. Find the point on the x–axis which are at a distance of from point (7, –4). How
many such points are there?

Sol. Let point P(x, 0) be a point on x–axis, and A be the point (7, –4).

So, AP = [Given]

⇒ AP2 = 4 × 5 = 20

⇒ (x – 7)2 + [0 – (–4)]2 = 20

⇒ x2 + 49 – 14x + 16 = 20

⇒ x2 – 14x – 20 + 65 = 0

⇒ x2 – 14x + 45 = 0

⇒ x2 – 9x – 5x + 45 = 0

⇒ x(x – 9) –5(x – 9) = 0

⇒ (x – 9)(x – 5) = 0

⇒ x – 9 = 0 or x – 5 = 0

⇒ x = 9 or x = 5

Hence, there are two such points on x–axis whose distance from (7, –4) is . Hence,

required points are (9, 0), (5, 0).

3. What type of quadrilateral do the points A(2, –2), B(7, 3), C(11, –1) and D(6, –6) taken
in that order form?

Sol. (i) A quadrilateral is a parallelogram, if mid points of diagonals AC and BD are same.

(ii) A parallelogram is not a rectangle, if diagonals AC ≠ BD.

(iii) A parallelogram may be a rhombus if AB = BC.

(iv) If in a parallelogram diagonals are equal, then it is rectangle.

In a rectangle if the sides AB = BC, then the rectangle is a square.

For parallelogram with vertices A(2, –2), B(7, 3), C(11, –1), D(6, –6).

mid point of AC = mid point of BD

⇒ , which is true.

Hence, ABCD is a parallelogram.

How, we will check whether AC = BD

or AC2 = BD2

⇒ (11 – 2)2 + (–1 + 2)2 = (6 – 7)2 + (–6 – 3)2

⇒ (9)2 + (1)2 = (–1)2 + (–9)2

⇒ 81 + 1 = 1 + 81

⇒ 82 = 82, which is true.

As the diagonals are equal so it is a rectangle or square.

Now, we will check whether adjacent sides AB = BC

or AB2 = BC2

⇒ (7 – 2)2 + (3 + 2)2 = (11 – 7)2 + (–1 – 3)2

⇒ 52 + 52 = (4)2 + (–4)2

⇒ 25 + 25 = 16 + 16

⇒ 50 ≠ 32, which is false.

So, ABCD is not a square. Hence ABCD is a rectangle.

4. Find the value of a, if the distance between the points A(–3, –14) and B(a, –5) is 9 units.

Sol. Consider A(–3, –14) and B(a, –5).

According to the question, AB = 9

⇒ AB2 = 81

⇒ (a + 3)2 + (–5 + 14)2 = 81

⇒ a2 + 9 + 6a + (9)2 = 81

⇒ a2 + 6a + 9 = 81 – 81

⇒ (a + 3)2 = 0

⇒ a + 3 = 0

⇒ a = –3

5. Find a point which is equidistant from the points A(–5, 4) and B(–1, 6). How many such
points are there?

Sol. Let P(x, y) is equidistant from A(–5, 4) and B(–1, 6) then

PA = PB

⇒ PA2 = PB2

⇒ (x + 5)2 + (y – 4)2 = (x + 1)2 + (y – 6)2

⇒ x2 + 25 + 10x + y2 + 16 – 8y = x2 + 1 + 2x + y2 + 36 – 12y

⇒ 41 + 10x – 8y = 37 + 2x – 12y

⇒ 8x + 4y + 4 = 0

⇒ 2x + 1y + 1 = 0
The above equation shows that infinite points are equidistant from AB, because all the points
on perpendicular bisector of AB will be equidistant from AB.

⇒ One such point which is equidistant from A and B is the mid–point M of AB i.e.,

M(–3, 5)

So, (–3, 5) is equidistant from points A and B.

6. Find the coordinates of the points Q on the x–axis which lies on the perpendicular
bisector of the line segment joining the points A(–5, –2) and B(4, –2). Name the type of
triangle formed by the points Q, A and B.

Sol. Let Q(x, 0) be a point on x–axis which lies on the perpendicular bisector of AB.

∴ QA = QB

⇒ QA2 = QB2

⇒ (–5 – x)2 + (–2 – 0)2 = (4 – x)2 + (–2 – 0)2

⇒ (x + 5)2 + (–2)2 = (4 – x)2 + (–2)2

⇒ x2 + 25 + 10x + 4 = 16 + x2 – 8x + 4

⇒ 10x + 8x = 16 – 25

⇒ 18x = –9

⇒ x =
Hence, the point Q is .

Now, QA2 = + [–2 – 0]2

⇒ QA2 =

⇒ QA = units

Now, QB2 = + (–2 – 0)2 = + (–2)2

⇒ QB2 =

⇒ QB = units

and AB = = 9 units

⇒ AB = 9 units

As QA = QB

So, ΔQAB is an isosceles Δ.

7. Find the value of m if the points (5, 1), (–2, –3) and (8, 2m) are collinear.

Sol. Points A, B, C will be collinear if the area of ΔABC = 0.

i.e., [x1(y2​ – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0

⇒ [5(–3 – 2m) –2(2m – 1) + 8(1 + 3) = 0

⇒ –15 – 10m – 4m +2 + 32 = 0
⇒ –14m – 15 + 34 = 0

⇒ –14m + 19 = 0

⇒ –14m = –19

⇒ m =

Hence, the required value of m = .

8. If the point A(2, –4) is equidistant from P(3, 8) and Q(–10, y) then find the values of y.
Also find distance PQ.

Sol. According to the question,

PA = QA

⇒ PA2 = QA2

⇒ (3 – 2)2 + (8 + 4)2 = (–10 – 2)2 + (y + 4)2

⇒ 12 + 122 = (–12)2 + y2 + 16 + 8y

⇒ y2 + 8y + 16 – 1 = 0

⇒ y2 + 8y + 15 = 0

⇒ y2 + 5y + 3y + 15 = 0

⇒ y(y + 5) + 3(y + 5) = 0

⇒ (y + 5) (y + 3) = 0

⇒ y + 5 = 0 or y + 3 = 0

⇒ y = –5 or y = –3

So, the co–ordinates are P(3, 8), Q1(–10, –3), Q2(–10, –5).
Now, = (3 + 10)2 + (8 + 3)2 = 132 + 112

⇒ = 169 + 121

⇒ units

and = (3 + 10)2 + (8 + 5)2 = 132 + 132

= 132[1 + 1]

⇒ = 132 × 2

⇒ = units

Hence, y = –3, –5 and PQ = units and units.

9. Find the area of the triangle whose vertices are (–8, 4), (–6, 6) and (–3, 9).

Sol. Vertices of ΔABC are A(–8, 4), B(–6, 6) and C(–3, 9).

∴ Area of ΔABC = [x1(y2​ – y3) + x2(y3 – y1) + x3(y1 – y2)]

⇒ Area of Δ = [–8(6 – 9) –6(9 – 4) –3(4 – 6)]

= [–8(–3) – 6(5) – 3(–2)]

= [24 – 30 +6] = 0

Hence, the area of given triangle is zero.

10. In what ratio does the x–axis divides the line segment joining the points (–4, –6) and
(–1, 7)? Find the coordinates of the points of division.

Sol. Point P(x, 0) on x–axis intersects the line joining the points A(–4, 6) and B(–1, 7). Let P
divides the line in the ratio k : 1.
Using the section formula, we have

(I)

⇒ 7k – 6 = 0

⇒ m1 = 6 and m2 = 7

Again, using the section formula, we have

Now, [From (I)]

∴ Hence, the required point of intersection .

11. Find the ratio of the point divides the line segment joining the points

and B(2, –5).

Sol. Let points P divides the line segment AB in the ratio k : 1, then
The coordinates of P, by section formula are

⇒ 8k + 2 = 3k + 3

⇒ 8k – 3k = 3 – 3

⇒ 5k = 1

⇒ k =

⇒ m1 = 1 and m2 = 5

Now,

∴ y–coordinate of P is

⇒
y–coordinate of P is .

Hence, P divides AB in ratio 1 : 5.

12. If point P(9a – 2, –b) divides the line segment joining the points A(3a + 1, –3) and B(8a,
5) in the ratio 3 : 1, then find the values of a and b.

Sol. Point P(9a – 2, –b) divides the line segment joining the points A(3a + 1, –3) and B(8a, 5) in
the ratio 3 : 1 But, the coordinates of P are (9a – 2, –b).

Using section formula, we have

9a – 2 =

⇒ 36a – 8 = 27a + 1

⇒ 36a – 27a = 8 + 1

⇒ 9a = 9

⇒ a = = 1

–b =

⇒ –b =

⇒ b = –3

Hence, a = +1 and b = –3

13. If (a, b) is mid–point of the line segment joining points A(10, –6) and B(k, 4) and a –
2b = 18, then find the value of k and the distance AB.
Sol. Let P(a, b) is the mid–point of the line–segment joining the points A(10, –6) and B(k, 4).
Therefore, P(a, b) divides the line segment joining the points A(10, –6) and B(k, 4) in the ratio
1 : 1.

⇒ ... (I)

and

⇒ b = –1 .....(II)

But, a – 2b = 18 ......(III) [Given]

⇒ a – 2(–1) = 18 [Using (II)]
⇒ a = 18 – 2
⇒ a = 16

But, [From (I)]

⇒ 10 + k = 32

⇒ k = 32 – 10

⇒ k = 22

Now, the co–ordinates of A and B are given by A(10, –6) and B(22, 4)

∴ AB2 = (22 – 10)2 + (4 + 6)2

= 122 + 102 = 144 + 100

⇒ AB2 = 244

⇒ AB = units

Hence, the required value of k = 22, a = 15, b = –1 and AB = units.

14. If the centre of circle is (2a, a – 7) then find the values of a if he circle passes through
the point (11, –9) and has diameter units.

Sol. Let C(2a, a – 7) be the centre of the circle and it passes through the point P(11, –9).

∴ PQ =

⇒ CP =

⇒ CP2 = = 50

⇒ (2a – 11)2 + (a – 7 + 9)2 = 50

⇒ (2a)2 + (11)2 – 2(2a) (11) + (a + 2)2 = 50

⇒ 4a2 + 121 – 44a + (a)2 + (2)2 + 2(a)(2) = 50

⇒ 5a2 – 40a + 125 = 50

⇒ a2 – 8a + 25 = 10

⇒ a2 – 8a + 25 – 10 = 0

⇒ a2 – 8a + 15 = 0

⇒ a2 – 5a – 3a + 15 = 0

⇒ a(a – 5) –3(a – 5) = 0

⇒ (a – 5) (a – 3) = 0

⇒ a – 5 = 0 or a – 3 = 0

⇒ a = 5 or a = 3
Hence, the required values of a are 5 and 3.

15. The line segment joining the points A(3, 2) and B(5, 1) is divided at the point P in the
ratio of 1 : 2 and it lies on the line 3x – 18y + k = 0. Find the value of k.

Sol.

P divides AB in the ratio 1 : 2 Then, the coordinates of P(x, y) are given by

and

But, P(x, y) lies on the line 3x – 18y + k = 0

⇒ 33 – 90 + 3k = 0
⇒ 3k = 90 – 33
⇒ 3k = 57

⇒ k =

⇒ k = 19
Hence, the required value of k = 19.

16. If , E(7, 3) and are the mid–points of sides of ΔABC, find the

area of ΔABC.

Sol. In ΔABC, D is mid point BC, E is mid point of AC, and F is mid point of AB.

∴ ΔDEF ≅ ΔAFE ≅ ΔFBD ≅ ΔEDC

So, area of ΔABC = 4 (area of ΔDEF)

The mid–points of sides of ΔABC are

given by ,E(7, 3), and .

∴ Area ΔDEF = [x1(y2​ – y3) + x2(y3 – y1) + x3(y1 – y2)]

or
∴ Area of ΔABC = 4 × Area ΔDEF

= 11 square units
Hence, the required area of ΔABC is 11 square units.

17. The points A(2, 9), B(a, 5) and C(5, 5) are the verticles of a ΔABC right angled at B.
Find the values of a and hence the area of ΔABC.

Sol. ΔABC is right angled at B.

∴ By Pythagoras theorem,

AB2 + BC2 = AC2 …(I)

AB2 = (x2 – x1)2 + (y2 – y1)2

⇒ AB2 = (a – 2)2 + (5 – 9)2

⇒ AB2 = (a)2 + (2)2 – 2(a) (2) + (–4)2

= a2 + 4 – 4a + 16

⇒ AB2 = a2 – 4a + 20

BC2 = (a – 5)2 + (5 – 5)2

=(a)2 + (5)2 – 2(a)(5) + 02

⇒ BC2 = a2 + 25 – 10a

AC2 = (5 – 2)2 + (5 – 9)2

= 32 + (–4)2

= 9 + 16 = 25

⇒ AC = = 5 units

∴ a2 – 4a + 20 + a2 + 25 – 10a = (5)2 [From (I)]

⇒ 2a2 – 14a + 45 – 25 = 0

⇒ 2a2 – 14a + 20 = 0

⇒ a2 – 7a + 10 = 0

⇒ a2 – 5a – 2a + 10 = 0

⇒ a(a – 5) –2 (a – 5) = 0

⇒ (a – 5)(a – 2) = 0

⇒ a – 5 = 0 or a – 2 = 0

⇒ a = 5 or a = 2

If a = 5 then B(5, 5) and C(5, 5) and BC = 0, which is not possible.

Hence, a = 2.

Now, AB2 = a2 – 4a + 20

= (2)2 –4(2) + 20

= 4 – 8 + 20

⇒ AB2 = 24 – 8
⇒ AB2 = 16

⇒ AB = 4 units

And, BC2 = a2 + 25 – 10a

= (2)2 + 25 – 10(2) [ a = 2]

= 4 + 25 – 20 = 29 – 20 = 9

⇒ BC2 = 9

⇒ BC = 3 units

∴ Area of right angled triangle ABC = base × altitude

= BC × AB

= × 3 × 4

= 6 square units

Hence, the value of a = 2 and area of ΔABC is 6 sq. units.

18. Find the coordinates of the point R on the line segment joining the points P(–1, 3)

and Q(2, 5) such that PR = PQ.

Sol. PR = PQ

⇒
⇒

or or PR : QR = 3 : 2

∴ m1 = 3 and m2 = 2

Now, the coordinates of point R are given by

and

Hence, the required coordinates of R are .

19. Find the value of k if the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are
collinear.

Sol. Points A, B and C will be collinear if area of ΔABC = 0

⇒ [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2​) = 0

⇒ [(k + 1) {2k + 3 – 5k} + 3k{5k – 2k} + (5k – 1){2k – (2k + 3)}] = 0

⇒ (k + 1)(–3k + 3) + 3k(3k) + (5k – 1)(2k – 2k – 3) = 0

⇒ –3(k + 1)(k – 1) + 3(3k2) – 3(5k – 1) = 0

Divide by 3 on both sides, we have

[(k + 1)(–k + 1) + 3k2 + (5k – 1)(–1)] = 0

⇒ 1 – k2 + 3k2 – 5k + 1 = 0

⇒ 2k2 – 5k + 2 = 0

⇒ 2k2 – 4k – 1k + 2 = 0

⇒ 2k(k – 2) –1 (k – 2) = 0

⇒ (k – 2)(2k – 1) = 0

⇒ k – 2 = 0 or 2k – 1 = 0

⇒ k = 2 or 2k = 1

⇒ k = 2 or k =

Hence, the required value of k are 2 and .

20. Find the ratio in which the line 2x + 3y – 5 =0 divides the line segment joining the
points (8, –9) and (2, 1). Also find the coordinates of the point of division.

Sol.

Let the line given be equation I divides AB at P(x, y) in the ratio k : 1,

Then, using the section formula, the coordinates of P are given by

 and

⇒ and

⇒ and

⇒ P(x, y) = lies on line I so P must satisfy equation (I)

So substitute and in equation I

On multiplying by (k + 1) in above equation both sides, we get

2(2k + 8) +3(k – 9) –5(k + 1) = 0

⇒ 4k + 16 + 3k – 27 – 5k – 5 = 0

⇒ 2k – 16 = 0

⇒ k = = 8

∴ Point of intersection is given by

Hence, line of equation (I) divides AB in ratio 8 : 1 at .

 EXERCISE 7.4

1. If (–4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of
the third vertex, given that the origin lies in the interior of the triangle.

Sol. Let A(–4, 3), B(4, 3) and C(x, y) are the three vertices of ΔABC.

As the triangle is equilateral,

so AC = BC = AB

or AC2 = BC2 = AB2 (I)

Now, AB2 = (4 + 4)2 + (3 – 3)2

⇒ AB2 = (8)2 = 64

⇒ AB = 8 units (II)

AC2 = (x + 4)2 + (y – 3)2

= (x)2 + (4)2 + 2(x)(4) + (y)2 + (3)2 – 2(y)(3)

= x2 + y2 + 8x – 6y + 16 + 9

⇒ AC2 = x2 + y2 + 8x – 6y + 25 (III)

BC2 = (x – 4)2 + (y – 3)2

=(x)2 + (4)2 – 2(x)(4) + (y)2 + (3)2 – 2(y) (3)

= x2 + y2 – 8x – 6y + 16 + 9

⇒ BC2 = x2 + y2 – 8x – 6y + 25 (IV)
Now, AC2 = AB2 [From (I)]

⇒ x2 + y2 + 8x – 6y + 25 = 64 [From (III), (II)]

⇒ x2 + y2 + 8x – 6y = 64 – 25

⇒ x2 + y2 + 8x – 6y = 39 (V)

Again, BC2 = AB2 [From (I)]

⇒ x2 + y2 – 8x – 6y + 25 = 64 [From (II), (IV)]

⇒ x2 + y2 – 8x – 6y = 64 – 25

⇒ x2 + y2 – 8x – 6y = 39 (VI)

Substracting (V) from (VI), we have

⇒ x = 0

Putting x = 0 in (V), we have

(0)2 + y2 + 8(0) – 6y = 39

⇒ y2 – 6y – 39 = 0

D = b2 – 4ac (a = 1, b = –6, c = –39)

= (–6)2 –4(1) (–39) = 36 + 156

⇒ D = 192

⇒
⇒

⇒ y1 = 3 + and y2 = 3 –

Hence, the third vertex of ΔABC maybe C(0, 3 + ) and C’(0, 3 – ).

Now, C(0, 3 + )

= C(0, 3 + 4 × 1.732)

= C(0, 3 + 6.9)

= C(0, 9.9)

and C’(0, 3 – )

= C’(0, 3 –4 × 1.732)

= C’(0, 3 –6.9)

= C’(0, –3.9)

So, the required point so that origin lies inside it is (0, 3 – ).

2. A(6, 1), B(8, 2) and C(9, 4) are three vertices of a parallelogram ABCD. If E is the mid
point of DC, then find the area of ΔADE.
Sol. ABCD is a parallelogram so

[Mid point of diagonal BD] = [Mid point of diagonal AC]

∴ Mid point of BD = and Mid point of AC =

⇒ and

⇒ x4 = 15 – 8 and y4 = 5 – 2

⇒ x4 = 7 and y4 = 3

∴ D = (7, 3)

Mid point of DC is

Now, Area of ΔADE

 sq units = sq.units [In magnitude]

Hence, the area of ΔADE is sq. units.

3. The points A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of ΔABC.

(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of points P on AD such that AP : PD = 2 : 1.

(iii) Find the coordinates opf points Q and R on medians BE and CF respectively such
that BQ : QE = 2 : 1 and CR : RF = 2 : 1.

(iv) What are the coordinates of the centroid of the ΔABC?

Sol. (i) Median from A meets BC at D i.e., D is the mid–point of BC.

So, the coordinates of D are given by

(ii)

The coordinates of the point P on AD such that AP : PD = 2 : 1 are given by

⇒

∴ is the required point.

(iii) (a) Median BE meets the side AC at its mid–point E.

∴ Coordinates of E are .

Now the coordinates of Q such that BE is median and BQ : QE = 2 : 1 are given by

∴ The coordinates of point Q on median BE such at QB : QE = 2 : 1 are

(b) Median CF meets the side AB at its mid–point F.

∴ Coordinate of F are .
Now, the coordinates of R such that CF is median and CR : RF = 2 : 1 are given by

So, the coordinates of point R on the median CF such that CR : RF = 2 : 1 are

(iv) Coordinates of centroid G of ΔABC are

It is observed that coordinates of P, Q, R and G are same.

Hence, the medians intersect at the same point i.e., centroid which divides the medians in
the ratio 2 : 1.

4. If the points A(1, –2), B(2, 3), C(a, 2) and D(–4, –3) from a parallelogram, find the value
of a and height of the parallelgoram taking AB a base.

Sol. As ABCD is a parallelogram and diagonals of prallelogram bisect each other.

The mid–point of diagonals of parallelogram will coincide i.e.,

Mid–point of diagonal AC = Mid–point of diagonal BD

⇒
⇒

⇒ a = –2 –1 = –3

Hence, the value of a is –3.

Now, Area of ΔABD = base × altitude

⇒ units
Hence, the perpendicular distance between parallel sides AB and CD is units.

5. Student of a school are standing in rows and column in their playground for a drill
practice. A, B, C, D are the positions of four students as shown in the figure. Is it
possible to place Jaspal in the drill in such a way that he is equidistant from each of the
four students A, B, C and D? If so, what should be his position?

Sol. Coordinates of A, B, C and D from graph are A(3, 5), B(7, 9), C(11, 5), and D(7, 1).

To find the shape of ABCD:

AB2 = (7 – 3)2 + (9 – 5)2 = 42 + 42 = 42(1 + 1)

⇒ AB = units

BC2 = (11 – 7)2 + (5 – 9)2 = (4)2 + (–4)2 = 42(1 + 1)

⇒ BC = units

CD2 = (7 – 11)2 + (1 – 5)2 = (–4)2 + (–4)2 = 42 + 42

⇒ CD = units

DA2 = (7 – 3)2 + (1 – 5)2 = 42 = + (–4)2 = 42 + 42

⇒ DA = = units

∴ AB = BC = CD = DA = units.
So, ABCD will be either square or rhombus.

Now, Diagonal AC =

⇒ AC

⇒ AC = 8 units

and diagonal BD =

⇒ BD = 8 units

∴ Diagonal AC = Diagonal BD

So, the given quadrilateral ABCD is square. The point which is equidistant from point A, B, C,
D of a square ABCD will be at the intersecting point of diagonals and diagonals bisect each
other.

Hence, the required point O equidistant from A, B, C, D is mid point of any diagonal =

= (7, 5).

Hence, the required point is (7, 5).

6. Ayush start walking from his house to office. Instead of going to the office directly, he
goes to a bank first, from there to his daughter’s school and then reaches the office.
What is the extra distance travelled by Ayush in reaching his office? (Assume that all
distances covered are in straight lines.) If the house is situated at (2, 4) bank at (5, 8)
school at (13, 14) and office at (13, 26) and coordinates are in km.

Sol. Consider the coordinates of house H(2, 4) bank B(5, 8) school S(13, 14) and office O(13,
26).

Distance HB2 = (5 – 2)2 + (8 – 4)2 = 32 + 42 = 9 + 16 = 25

⇒ HB = 5 km
Distance BS2 = (13 – 5)2 + (14 – 8)2 = 82 + 62 = 64 + 36

⇒ BS2 = 100

⇒ BS = 10 km

Distance SO2 = (13 – 13)2 + (26 – 14)2 = 02 + 122 = 122

⇒ SO = 12 km

Total distance travelled by Ayush from house to bank to school and then to office

= HB + BS + SO

= 5 + 10 + 12 = 27

Direct distance from house to office = HO

⇒ HO2 = (13 – 2)2 + (26 – 4)2 = (11)2 + (22)2

⇒ HO2 = 121 + 484

⇒ HO = = 24.6 km

So, extra distance travelled by Ayush = 27 km – 24.6 km = 2.4 km.

Hence, extra distance travelled by Ayush = 2.4 km