Class 10 Maths Previous Year Questions - Coordinate Geometry

Ans: (d)
Assertion says that point
divides the line joining the points A(1, 2) and B(–1, 1) in 1 : 2. 
∴ By section formula,

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)
Point P(0, 2) is the point of intersection of y-axis with line 3x + 2y = 4

Also, the distance of point P(0, 2) from x-axis is 2 units.

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

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 length of side = 400√2 units.
(iii) Here,

Using section formula, we have

Ans: (a)
Given, the equation of line is 4x- 3y = 9.
Putting x = 0, we get 4 x 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).
Given, PR = QR
⇒ PR² = QR²
⇒ (x - 5)² + (0 - 0)² = (x + 1)² + (0 - 0)²
⇒ x² - 10x + 25 = x² + 2x + 1
⇒ 12x =24
⇒ x = 2
Required 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
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

Hence, 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

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.

∴ Coordinates of P are

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 

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

Ans: (d)
Let coordinates of the point on the x-axis be P(x, 0). Let the given points be A(-4, 0) and B(10, 0). which also lie on x-axis.
Since P is equidistant from A and B.
∴ x = -4 + 10 / 2 = 6 / 2 = 3
So, P(3, 0) is equidistant from A(-4, 0) and B(10, 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 

Hence, the required point is  and 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.

∴ Coordinates of B =  (-5, -2)

Ans:  Given AB = 5 units

Ans: Let P(0, y) be the point on the y-axis which is equidistant from A(5, - 2) and B(-3, 2).

AP = BP ⇒ (AP)² = (BP)²
⇒ (5 - 0)² + (-2 - y)² = (-3 - 0)² + (2 - y)²
⇒ 25 + 4 + y² + 4y = 9 + 4 + y² - 4y
⇒ 8y = 9 - 25
⇒ y = -16/2 = -2
Hence, A and B are equidistant from (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 and4 = 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  [By Section Formula]
Since, y-coordinate of P is 0
.
Hence, the point P divides the line segment in the ratio 3 : 5.
Also, x-coordinate of P
=
∴ Coordinates of point P are (17/8, 0)

Ans: Given,


∴ Coordinates of R are

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 ration is 2: 7.

Now,

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

And the coordinates of the point where diagonals AC and BD intersect are M(2, –5). Let the coordinates of other two vertices i.e., C be (x₁, y₁) and D be (x₂, y₂). Since, diagonals of a parallelogram bisect each other, so, M is the mid-point of AC and BD. For line AC, where M is its mid-point.

Then,
i.e.
and
⇒ x₁ = 1 or y₁ = – 12
Similarly, M is the mid-point of BD. 
Then,

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,

Now,

Hence, the coordinates of the points of trisection are .

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 – 1 (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)

on squaring both sides]

But, C lies in the first quadrant.

Coordinates of C are

Hence, the coordinates of B and C are (5, 0) and  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.