Class 10 Maths Previous Year Questions - Coordinate Geometry

Ans: (d)

Assertion says that point ( -13 , 53 ) divides the line joining the points A(1,2) and B(-1,1) in 1:2.

∴ By section formula,

x = m₁x₂ + m₂x₁m₁ + m₂

= 1×(-1) + 2×13 = 13

y = m₁y₂ + m₂y₁m₁ + m₂

= 1×1 + 2×21+2

= 1 + 43

= 53

which is not equal to RHS i.e. 1/3

Ans:
Since, P(x, y) is equidistant from A(7, 1) and B(3, 5)
So, PA = PB
⇒ PA² = PB²
⇒ (x – 7)² + (y – 1)² = (x – 3)² + (y – 5)²
⇒ x² + 49 – 14x + y² + 1 – 2y = x² + 9 – 6x + y² + 25 – 10y
⇒ 6x – 14x + 50 – 34 + 10y – 2y = 0
⇒ – 8x + 8y + 16 = 0
⇒ 8x – 8y – 16 = 0
⇒ 8(x – y – 2) = 0
⇒ x – y – 2 = 0
⇒ x – y = 2

Ans: A (– 1, y); B(5, 7)
Since, AB is a diameter of circle and O is the centre of the circle.
OA = OB i.e., O divides AB in 1 : 1
So m₁ : m₂ = 1 : 1
So
⇒
⇒ – 6y = y + 7
⇒ – 7y = 7
⇒ y = – 1
Point O = (2, 3), A = (–1, – 1)
Now,


So, radius of circles = 5 units

Ans:
If y-axis divides points (5, 3) and (–1, 6) then coordinate of that point will be (0, y). Let P(0, y) divides A(5, 3) and B(–1, 6) in k : 1.
m₁ : m₂ = k : 1

⇒ 0 × (k + 1) = – k + 5
⇒ 0 = – k + 5
⇒ k = 5
So, m₁ : m₂ = 5 : 1

Ans: (b)
Distance from x-axis = y-coordinate of point = 7 units

Ans: (b)

• Assertion (A):To find the intersection of the y-axis with the line 3x + 2y = 4, set x = 0:

  3(0) + 2y = 4 ⇒ y = 2. So, the point of intersection is P(0,2).

  Assertion (A) is true.

• Reason (R):The distance of point P(0,2) from the x-axis is indeed 2 units.

  Reason (R) is true.

However, Reason (R) does not explain Assertion (A); it is just a separate true statement.
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

Ans: (d)
Distance of the point (-6, 8) from origin (0, 0)

= 10 Units

Ans: (b)
The points be A(-4, 0), B(4, 0) and C(0, 3).
Using distance formula

= 8 units

 = 5 units

 = 5 units
And, AB² ≠  BC² + CA²  [∵ BC = CA]
∴ ΔABC is an isosceles triangle.

Ans: Given centre of a circle is(2a, a - 7 )

Radius of the circle is 5√2 cm.
∴ Distance between centre (2a, a - 7) and (11, - 9 ) = radius of circle.

Ans: (d)
Let the point on the x-axis be (x, 0) which divides the line segment joining the points A(3, 6) and B(-12, -3) in the ratio k : 1

Using section formula, we have

(x, 0) = (-12)k + 3(1)k + 1 , (-3)k + 6(1)k + 1

⇒ -3k + 6k + 1 = 0

⇒ -3k + 6 = 0

⇒ k = 2

Hence, the required ratio is 2 : 1.

Ans: (i) We have. P = (-200, 0) and Q = (200, 0)

The coordinates of R and S are (200, 400) and (-200, 400).
(ii) (a) The length PQ = 200 + 200 = 400 units.
Area of square PQRS = 400  x 400 = 160000 sq. units.

(b) Length of diagonal PR = √2  x length of side = 400√2 units.
(iii) Here,

Using section formula, we have
∴ Kx₂ + x₁K + 1 , Ky₂ + y₁K + 1 = (-200, 400)

⇒ K(200) + (-600)K + 1 , K(800) + 0K + 1 = (-200, 400)

⇒ 200K - 600K + 1 , 800KK + 1 = (-200, 400)

∴ 800KK + 1 = 400

⇒ 800K = 400K + 400

⇒ 400K = 400

⇒ K = 1

Ans: (a)
Given, the equation of line is 4x- 3y = 9.
Putting x = 0, we get 4x(0) - 3y = 9 ⇒ y =  -3
So, the line 4x - 3y = 9 intersects the y-axis at (0, -3).

Ans: (a)
Let coordinates of the point on the x-axis be R (x, 0).

√((x - 5)² + (0 - 0)²) = √((x + 1)² + (0 - 0)²)

Simplify:

(x - 5)² = (x + 1)²

Expand:

x² - 10x + 25 = x² + 2x + 1

Solve for x:

-10x + 25 = 2x + 1

⇒ -12x = -24

⇒ x = 2

So, the point is (2, 0).

Ans: (d)
Let coordinate of point P= t
So, .(x-coordinate of point P = 2t  ∴ Point is P (2t, t).
Given, PQ = RP ⇒ PQ² = RP²
⇒ (2t - 2)² + (t + 5)² = (2t + 3)² + (t - 6)²  [By distance formula]
⇒ 4t² - 8t + 4 + t² + 10t + 25 = 4t²+ 12t + 9  +  t²- 12t + 36
⇒ 2t = 16
t = 8
P = 2t = 2 x 8 = 16
Coordinates of P are (16,  8).

Ans: (c)
Let point P(-4, 6) divides the line segment AB in the ratio m₁: m₂.

By section formula, we have

(-4,6) = 3m₁ - 6m₂m₁ + m₂ , -8m₁ + 10m₂m₁ + m₂

Now, -4 = 3m₁ - 6m₂m₁ + m₂

⇒ 3m₁ - 6m₂ = -4m₁ - 4m₂

⇒ 7m₁ = 2m₂ ∴ m₁ : m₂ = 2:7

Putting the value of m₁ : m₂ in the y-coordinate, we get

-8m₁ +10m₁ + m₂ = -8 × 27 + 10

-167 + 10 = 6

Hence, the required ratio is 2:7.

Ans: (c)
Coordinates of S are (-6, 4).

Ans: (b)
Coordinates of D are (-2, -4) and coordinates of H are (8, 2).
∴ Midpoint of DH =

Ans: (d)
Coordinates of A are (-2, 4) and coordinates of C are (4, -4).
Let (x, 0) divides the line segment joining the points A and C in the ratio m₁ : m₂
By section formula, we have

(x, 0) = 4m₁ - 2m₂m₁ + m₂ , -4m₁ + 4m₂m₁ + m₂

Now, 0 = -4m₁ + 4m₂m₁ + m₂

⇒ -4m₁ + 4m₂ = 0

⇒ m₁ : m₂ = 1:1

Ans: (c)
Coordinates of P are (-6, -4) and coordinates of G are (8, 6).

Ans: (b)
Coordinates of vertices of rectangle IJKL are respectively I(2, -2), J(2, -6), K(8, -6),L(8, -2).

Ans: (d)
From figure coordinates are A(2, 5), B(5, 7), C(8, 6) and D(6, 3)
Now,

Clearly, ABCD is a quadrilateral

Ans: (a)
Here, sports teacher is at O(0,0).
Now,

∴ OA is the minimum distance
∴ A is closest to sports teacher.

Ans: (a)
Required distance =
=

Ans: (c)
 Coordinates of mid-point of AC are

Ans: (b)
 Let point P(x, y) divides the line segment AD in the ration 1: 2.

∴ 1(6) + 2(2)1 + 2 , 1(3) + 2(5)1 + 2

⇒ x = 6 + 43 , y = 3 + 103

⇒ x = 103 , y = 133

∴ Coordinates of P are ( 103 , 133 )

Ans: (c)
Required distance

Ans: (d)
Required distance =

Ans: In rectangle AOBC. AB is a diagonal.

So,

= 5 Units

Ans: Let the given points be A(7, 10), B(-2, 5) and C(3, - 4].
Using distance Formula, we have 

Also, AB² + BC² = 106 + 106 = 212 = AC²
So. ABC is an isosceles right angled triangle with ∠B = 90°.

Ans: (d)

To find the point on the x-axis equidistant from (-4, 0) and (10, 0), let the point be (x, 0).

Using the distance formula:

√( (x + 4)² ) = √( (x - 10)² )

Square both sides:

(x + 4)² = (x - 10)²

Expand:

x² + 8x + 16 = x² - 20x + 100

Cancel x² and simplify:

8x + 16 = -20x + 100

28x = 84

x = 3

Final Answer:

The point is (3,0).

Ans: (d)
Since, the point P(k 0) divides the line segment joining A(2, -2) and B(-7, 4) in the ratio 1 : 2.

Ans: (c)
Let the coordinates of centre of the circle be (x, y) and AB be the given diameter.
By Using mid-point formula.

We have,

∴ Coordinates of C are

Ans: Let the point P(0, y) on y-axis divides the line segment joining the points A(6, -4) and B(-2, -7) in the ratio k : 1.

By section formula, we have

-2k + 6k + 1 = 0

⇒ -2k + 6 = 0 ⇒ k = 3

and

-7k - 4k + 1 = y

⇒ -7(3) - 43 + 1 = y

4y = -21 - 4 = -25

⇒ y = -254

Hence, the required point is (0, -254) and the required ratio is 3:1.

Ans: We have, A(2, 5), B(x, y) and C(-1, 2) and point C divides AB in the ratio 3 :4.

∴ -1 = 3x + 87 ⇒ -7 = 3x + 8 ⇒ x = -153 = -5

and

2 = 3y + 207 ⇒ 14 = 3y + 20 ⇒ y = -63 = -2

∴ Coordinates of B = (-5, -2)

Ans:  Given AB = 5 units

Ans:
Any point on the y-axis has coordinates of the form (0, y).

Set Up the Distance Equation:

The distance from (0, y) to (5, -2) should be equal to the distance from (0, y) to (-3, 2):

√((0 - 5)² + (y - (-2))²) = √((0 - (-3))² + (y - 2)²)

Simplify the equation:

√(25 + (y + 2)²) = √(9 + (y - 2)²)

Square Both Sides:

25 + (y + 2)² = 9 + (y - 2)²

Expand and Simplify:

Expand both sides:

25 + y² + 4y + 4 = 9 + y² - 4y + 4

Simplify:

29 + y² + 4y = 13 + y² - 4y

Subtract y² from both sides:

29 + 4y = 13 - 4y

Add 4y to both sides:

29 + 8y = 13

Subtract 29 from both sides:

8y = -16

Divide by 8:

y = -2

The point on the y-axis that is equidistant from (5,-2) and (-3,2) is (0,-2).

Ans: Let coordinates of the point A be (x, y) and O is the mid point of AB.

By using mid-point formula,
we have
⇒ -4 = x + 3 and 4 = y + 4
⇒ x = -7 and y = 0
∴ Coordinates of A are (-7, 0).

Ans: Let the point R(x, 0) on x-axis divides the line segment PQ in the ratio k: 1.

∴ By section formula, we have

∴ Required ratio is 2 : 1 .

Ans: Let the point P(x, 0) divides the segment joining the points A(1, -3) and B (4, 5) in the ratio k : 1

Coordinates of P are (4k + 1, 5k - 3)k + 1 (By Section Formula)

Since, y-coordinate of P is 0

5k - 3k + 1 = 0 ⇒ 5k - 3 = 0 ⇒ 5k = 3 ⇒ k = 35

Hence, the point P divides the line segment in the ratio 3:5.

Also, x-coordinate of P

= 4k + 1k + 1 = 43/5 + 1 = 178

∴ Coordinates of point P are (17/8, 0)

Ans: Given,

AR = 34 AB ⇒ ABAR = 43

⇒ ABAR - 1 = 43 - 1 = 1

⇒ AB - ARAR = 4 - 33 = 13

⇒ BRAR = 13 ⇒ AR : BR = 3:1

∴ Coordinates of R are

( 3(0) + 1(-4)3 + 1 , 3(6) + 1(0)3 + 1 )

= ( -44 , 184 ) = (-1, 92 )

Ans: Let the coordinates of A be (x, y). Here, 0(3, - 1] is the mid point of AB.

By using mid point formula, we have⇒ x = 4, y = - 8∴ Coordinates of A are (4, - 8).

Ans:
Let P(x₁, y₁) and Q(x₂, y₂) are the points of trisection of line segment AB.
∴ AP = PQ = QB
Now. point P divides AB internally in the ratio 1 : 2
∴ By section formula, we have  

Since, point P(3, -2) lies on the line 2x - y + k = 0
⇒ 6 + 2 + k = 0
⇒ k = - 8

Ans: Let point P(x₁, y₁) divides the line segment joining the points A(-2, -5) and B(6, 3) in the ratio k: 1

∴ Coordinates of P are

The point P lies on line x - 3y = 0

∴ Required ratio is 13 : 3.Now, coordinates of P are

Ans: Let the point P(-4, y) divides the line segment joining the points A and B in the ration k: 1

∴ By section formula, coordinates of P are

∴ Required ratio is 2: 7.

Now,

Ans: Consider a parallelogram ABCD with A(3, 2) and B(–1, 0).

Let two adjacent vertices of a parallelogram be A ≡ (3,2) and B ≡ (-1,0).

Let coordinates of the other two vertices be C (x₁, y₁) and D (x₂, y₂).

We know that diagonals of a parallelogram bisect each other.

∴ Midpoint of AC and Midpoint of BD are the same, i.e., point O(2, -5).

∴ 3 + x₁2 = 2 and 2 + y₁2 = -5

⇒ x₁ = 1 and y₁ = -12 ⇒ C ≡ (1, -12)

Also, x₂ - 12 = 2 and y₂ + 02 = -5

⇒ x₂ = 5 and y₂ = -10 ⇒ D ≡ (5, -10)

Hence, the remaining vertices are (1, -12) and (5, -10).

Ans: The given line segment is A(3, – 2) and B(–3, –4). 
Here, C(x, y) and C'(x’, y’) are the points of trisection of AB. 
Then, AC : CB = 1 : 2 and AC’ : C’B = 2 : 1

By section formula,

C(x, y) = mx₂ + nx₁m + n , my₂ + ny₁m + n

Here, m = 1, n = 2

x₁ = 3, y₁ = -2

x₂ = -3, y₂ = -4

∴ C(x, y) = 1(-3) + 2×31 + 2 , 1(-4) + 2(-2)1 + 2

= -3 + 63 , -4 - 43

= (1, -8/3)

Now,

C'(x', y') = m'x₂ + n'x₁m' + n' , m'y₂ + n'y₁m' + n'

Here, m' = 2, n' = 1

∴ C'(x', y') = 2(-3) + 1×32 + 1 , 2(-4) + 1×(-2)2 + 1

= -6 + 33 , -8 - 23

= (-1, -10/3)

Hence, the coordinates of the points of trisection are (1, -8/3) and (-1, -10/3).

Ans: Given an equilateral triangle ABC of side 3 units. 
Also, coordinates of vertex A are (2, 0). 
Let the coordinates of B = (x, 0) and C = (x', y'). 
Then, using the distance formula,

On squaring both sides, we get
9 = (x – 2)²
x² + 4 – 4x = 9 
x² – 4x – 5 = 0 
x² – 5x + x – 5 = 0
(x – 5) (x + 1) = 0 
x = 5, – 1 But x = – 1 (Neglected , since it lies on positive x-axis) 
Coordinates of B are (5, 0). 
Now, AC = BC [Q Sides of an equilateral triangle are equal] 
or AC² = BC²
By using the distance formula, 
√(x' – 2)² + (y' – 0)² = √(x' – 5)² + (y' – 0)²
x' ² + 4 – 4x' + y' ² = x' ² + 25 – 10x' + y' ² 
6x' = 21 
x' = 7 / 2 

Also, AC = 3 units. (given)

√((x - 2)² + (y' - 0)²) = 3

On squaring both sides:

x² + 4 - 4x + y'² = 9

494 + 4 - 4 × 72 + y'² = 9 [∵ x' = 72]

⇒ y'² = 9 - 94 = 274

⇒ y' = √274 = ± 3√32

But, C lies in the first quadrant.

y' = 3√32

Coordinates of C are ( 72 , 3√32 )

Hence, the coordinates of B and C are (5, 0) and ( 72 , 3√32 ) respectively.

Ans: Given: ∆ABC with vertices A(–2, 0), B(2, 0), C(0, 2) and ∆PQR with vertices P(−4, 0) Q(4, 0) and R(0, 4). 
In ∆ABC, using distance formula,

Similarly, in ∆PQR, using distance formula,

Now,

and

So,

Since, the corresponding sides of ΔABC and ΔPQR are proportional,
∴ ∆ABC ~ ∆PQR
Hence, proved.