Solution

CH.7 COORDINATE GEOMETRY

Class 10 - Mathematics
Section A
1. (b) (2, 0)
Explanation: Let the required point be P(x, 0). Then,
2 2 2 2
PA = PB ⇒ (x + 1) = (x − 5)

2 2
⇒ x + 2x + 1 = x − 10x + 25

⇒ 12x = 24 ⇒ x = 2

So, the required point is P(2, 0).

2. (d) 2
Explanation: 2
3. (d) (3, 0)
Explanation: Let the required point be P(x, 0) then,
2 2 2 2 2 2
AP = BP ⇒ (x − 7) + (0 − 6) − (x + 3) + (0 − 4)

2 2
⇒ x − 14x + 85 = x + 6x + 25

−20x − 60 = x = 3

4. (a) (3, 0)
Explanation: The given point P lies on x-axis
Let the co-ordinates of P be (x, 0)
The point P lies on the perpendicular bisector of the line segment joining the points A(7, 6), B(-3, 4)
2 2
∴ PA = PB ⇒ PA = PB

2 2 2 2
⇒ (x − 7) + (0 − 6) = (x + 3) + (0 − 4)

2 2
⇒ x − 14x + 49 + 36 = x + 6x + 9 + 16

⇒ -14x + 85 = 6x + 25
⇒ 6x + 14x = 85 - 25 ⇒ 20x = 60
60
x = = 3
20

∴ co-ordinates of P will be (3, 0)

5. (b) -12
Explanation:
Given, P is the mid - point of the line segment joining the points Q and R
Where;
P=( a

3
, 4)

Q = ( - 6, 5)
R = ( - 2, 3)
Shown in the figure given below;

−6−2 5+3
∴ Mid - point of QR = P( 2
,
2
) = (-4, 4)
P = ( - 4, 4)
Since, midpoint of line segment having points (x1, y1) and (x2, y2);
( x1 + x2 ) (y + y )
= (
2
,
1

2
2
)

a
But given coordinates of mid - point P is ( 3
, 4) ;

∴ (
a

3
, 4) = ( - 4, 4)
On comparing the coordinates, we get
a

3
=-4

1/6
 ∴ a = - 12
Hence, the required value of a = - 12
6. (d) 10
Explanation: The distance of the point P( -6,8) from the origin (0, 0)
−−−−−−−−−
2 2
= √(−6) + 8

−−−−−−
= √36 + 64
−−−
= √100

= 10
7. (d) 3:1
Explanation: The point lies on y-axis
Its abscissa will be zero
Let the point divides the line segment joining the points (-3, -4) and (1, -2) in the ratio m:n
m x2 +n x1 m×1+n×(−3)
∴ 0 = ⇒ 0 =
m+n m+n
m−3n
⇒ = 0 ⇒ m − 3n = 0
m+n

m 3
⇒ m = 3n ⇒ =
n 1

∴ Ratio = 3:1
8. (b) 7
Explanation: P(2, 4), Q(0, 3), R(3, 6) and S(5, y) are the vertices of parallelogram PQRS

Join PR and QS which intersect at O

∴ O is mid-point of PR and QS

When O is mid point of PR then coordinates

2+3 4+6
of O will be = ( 2
,
2
)

5
= ( , 5)
2

O is mid-point of QS
y+3
∴ 5 = ⇒ y + 3 = 10
2

⇒ y = 10 - 3 = 7
9. (a) Both A and R are true and R is the correct explanation of A.
Explanation: Both A and R are true and R is the correct explanation of A.
10. (d) A is false but R is true.
5k+1 5k+1
Explanation: Let joining of (1, 1) and (5, 5) meet x-axis in k : 1. Now, x = k+1
and y = k+1

5k+1
But for x-axis y = 0. So, k+1
=0⇒k=− 1

⇒ 1 : 5 externally
11. (b) Both A and R are true but R is not the correct explanation of A.
−−−−−−−−−−−−−− − −−−− −−
Explanation: It will be √(a − 0) + (b − 0) = √a + b 2 2 2 2

Section B
12. Given, A (-2,-1), B (a, 0), C (4, b) and D (1, 2)
We know that the diagonals of a parallelogram bisect each other.
Therefore, the coordinates of the mid-point of AC are same as the coordinates of the mid-point of BD i.e.
−2+4 −1+b a+1 0+2
( , ) = ( , )
2 2 2 2

b−1 a+1
⇒ (1, ) = ( , 1)
2 2

a+1 b−1
⇒ = 1 and = 1
2 2

⇒a + 1 = 2 and b − 1 = 2

2/6
 ⇒a = 1 and b = 3

Hence, a = 1 and b = 3
13. Let A(-2,3), B(8,3) and C(6,7) be the vertices of a given triangle. Then,
(AB)2 = (8 + 2)2 + (3 - 3)2 = (10)2 + 02 = 100
(BC)2 = (8 - 6)2 + (3 - 7)2 = 22 + 42 = 4 + 16 = 20
(AC)2 = (-2 - 6)2 + (3 - 7)2 = 82 + 42 = 64 + 16 = 80
Clearly, (AB)2 = (AC)2 + (BC)2 [pythagoras theorem]
Therefore, A(-2,3), B(8,3) and C(6,7) are the vertices of right-angled triangle.

14.

Let P and Q trisect the line AB

i.e AP=PQ=QB
P divides AB in the ratio 1: 2
⇒ AP:PB = 1:2

By section formula, we get

1×0+2×−2 1×8+2×0 0−4 8+0 −4
Coordinates of P is ( 1+2
,
1+2
) =( 3
,
3
) =( 3
,
8

3
)

∴ and Q divides AB in the ratio 2: 1

i.e. AQ:QB= 2:1 ,
By section formula
2×0+1×(−2) 2×8+1×0 0−2 16+0 −2
Coordinates of Q are ( 2+1
,
2+1
) =( 3
,
3
) =( 3
,
16

3
)

15. Let A(6, -1) and B(2,3) be the given point and P(x,y) be the required point.
Since point P is equidistant from points A and B,therefore,
PA = PB ⇒ (PA)2 = (PB)2
⇒ (x - 6)2 + (y + 1)2 = (2 - x)2 + (3 - y)2
⇒ 36 + x2 - 12x + y2 + 1 +2y = 4 + x2 - 4x + 9 + y2 - 6y
⇒ -12x + 4x + 2y + 6y = 4 + 9 - 1 - 36

⇒ -8x + 8y = -24

-8 (x - y) = -24
⇒ x - y = 3

Hence, x - y = 3
16. According to the question, A(7, -3), B(5, 3) and C(3, -1).

AD and BE are medians of ΔABC.

D is the mid-point of BC and
E is the mid-point of AC
∴ Coordinates of D = ( 5+3

2
,
3−1

2
) = (
8

2
,
2

2
) = (4, 1)

7+3 −3−1 −4
Coordinates of E = ( 2
,
2
) = (
10

2
,
2
) = (5, −2)

−−−−−−−−−−−−−− − −−−−−−−−− − −−−−− −−

Now, median 2 2 2 2
AD = √(4 − 7) + (1 + 3) = √(−3) + (4) = √9 + 16 = √25 = 5 units
−−−−−−−−−−−−−−− − −−−− −−−− −−
And, median 2 2
BE = √(5 − 5) + (−2 − 3) = √0 + (−5)
2
= √25 = 5 units

Section C
17. Coordinates of points on a circle are A(2, 1), B(5, -8) and C(2, -9).
Let the coordinates of the centre of the circle be O(x, y).
OA = OB
−−−−−−−−−−−−−− − −−−−−−−−−−−−−− −
√(x − 2)2 + (y − 1)2 = √(x − 5)2 + (y + 8)2

(x - 2)2 + (y - 1)2 = (x - 5)2 + (y + 8)2

x2 + 4 - 4x + y2 + 1 - 2y = x2 + 25 - 10x + y2 + 64 + 16y
6x - 18y - 84 = 0
x - 3y - 14 = 0 ......... (i)

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 Similarly, OC = OB
−−−−−−−−−−−−−− − −−−−−−−−−−−−−− −
√(x − 2)2 + (y + 9)2 = √(x − 5)2 + (y + 8)2

(x - 2)2 + (y + 9)2 = (x - 5)2 + (y + 8)2

x2 + 4 - 4x + y2 + 81 + 18y = x2 + 25 - 10x + y2 + 64 + 16y
6x + 2y - 4 = 0
3x + y - 2 = 0 ....... (ii)
By solving (i) and (ii), we get,
x = 2 and y = -4
So, the coordinates of the centre of circle are (2, -4).
18. Here ( 24

11
, y) be any point on line formed by joining the points P and Q.Let it divide the line segment in the ratio of K:1.
Let PA : QA be K : 1

Where( 24

11
,y ) represent point A.
Using intersection formula,
m x2 +n x1
= x
m+n
3k+2 24
⟹ =
k+1 11

⟹ (3k+2)11 =24 (k+1)

⟹ 33k + 22 = 24k + 24
⟹ 33k-24k = 24-22
⟹ 9k =2
2
⟹ k =
9

Thus the ratio in which line is divided is equal to k:1=2:9

−18+14
Also y = = −
11
4

11

Thus y= − 4

11
which is coordinate of y.
−4
Therefore the ratio in which the line is divided by point ( 24

11
,
11
) is 2:9
19. P(a, b) is mid-point of AB, where A(10, - 6) and B(k, 4)
k+10 −6+4
P(a, b) = ( 2
,
2
)

k+10
a= 2
, b =-1
Now,we are given that , a - 2b = 18 or, a + 2 = 18 [as b=-1]
or, a =16
k+10
As a = ⟹
2
k+10

2
= 16 or, k = 22
Therefore, P(a, b) = (16, -1)
−−−−−−−−−−−−−−−− −
AB =√(22 − 10) + (4 + 6) 2 2

−−−−−−−−− −
= √(12)2 + (10)2
−−−
=√244
−−
= 2√61 units
20. If P (-4, 6) lies on the line segment joining A (k, 10) and B (3, -8), then P, A and B are collinear.

∴(-4 × 10 + k × -8 + 3 × 6) - (6k + 30 + -4 × -8) = 0

⇒ (-40 - 8k + 18) - (6k + 30 + 32) = 0

⇒ (-22 - 8k) - (6k + 62) = 0

⇒ -14k - 84 = 0

⇒ k = - 6

3λ−6 −8λ+10
Suppose P divides AB in the ratio λ : 1. Then, the coordinates of Pare ( λ+1
,
λ+1
) .But, the coordinates of P are (-4, 6).

4/6
 3λ−6 −8λ+10
∴ = −4 and = 6
λ+1 λ+1
2
⇒ λ =
7
2
Hence, P divides AB in the ratio 7
: 1 or 2: 7.

21.

We have P(p, -2) and Q ( 5

3
, q) are the points of trisection of the line segment joining A(3, -4) and B(1, 2)
We know AP : PB = 1 : 2
m x2 +n x1 m y +n y
By section formula [ m+n
], [
2

m+n
1
] coordinates of P are
1×1+2×3 1×2+2×(−4)
( , )
1+2 1+2

7
= ( , −2)
3

Hence, P =
7

Again we know that AQ : QB = 2 : 1

Therefore, Coordinates of Q are (using section formula )
2×1+1×3 2×2+1×(−4)
( , )
2+1 2+1

5
= ( , 0)
3

Hence, q = 0
Therefore, value of p and q is 7

3
and 0 respectively.
Section D
22. Position of Neena = (3, 6)
Position of Karan = (6, 5)
−−−−−−−−−−−−−− −
Distance between Neena and Karan = √(6 − 3) + (5 − 6) 2 2

−−−− −−−−
= √9 + (−1)2
−−
= √10
23. Co-ordinate of seat of Akash = 2, 3
24.
2+5 3+2
Co-ordinate of middle point = ( 2
,
2
)

= 3.5, 2.5
25. Binu = (5, 5); Karan = (6, 5)
−−−−−−−−−−−−−−−
2
Distance = √(6 − 5) 2
+ (5 − 2)

−−−−
= √1 + 9
−−
= √10

Section E
26. (d) ₹108
Explanation: Distance travelled by second bus = 7.2 km
∴ Total fare = 7.2 × 15 = ₹108

27. (b) none of these

−−−−−−−−−−−−−− −
Explanation: Required distance = √(2 + 2) 2
+ (3 + 3)
2

−− −−−− −−−−−− −−
= 2 2
√4 + 6 = √16 + 36 = 2√13 km ≈ 7.2 km
–
28. (d) 5√2 km
−−−−−−−−−−−−−− −
Explanation: Required distance = √(3 + 2) 2
+ (2 + 3)
2

− −−−−− –
= √5 + 5 = 5√2 km
2 2

29. (c) Second bus

Explanation: Distance between B and C
−−−−−−−−−−−−−− − −−− − –
= √(3 − 2) + (2 − 3) = √1 + 1 = √2 km
2 2

Thus, distance travelled by first bus to reach to B

– – –
= AC + CB = 5√2 + √2 = 6√2 km ≈ 8.48 km
and distance travelled by second bus to reach to B

5/6
 −−
= AB = 2√13 km = 7.2 km
∴ Distance of first bus is greater than distance of the cond bus, therefore second bus should be chosen.

30. (b) none of these

Explanation: Distance travelled by first bus = 8.48 km
∴ Total fare = 8.48 × 10 = ₹84.80

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