Coordinate Geometry class 10 questions with solutions

Q 1 :
The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is

5 units
12 units
11 units
$(7 + \sqrt{5})$ units

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(2) 12 units

Q 2 :
Assertion (A): The value of y is 3, if the distance between the points P(2, -3) and Q(10, y) is 10.

Reason (R): Distance between two points is given by $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A)
Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of Assertion (A)
Assertion (A) is true but reason (R) is false.
Assertion (A) is false but reason (R) is true.

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(1)

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A)

Q 3 :
AD is a median of $∆$ ABC with vertices A(5, –6), B(6, 4) and C(O, O). Length AD is equal to :

$\sqrt{68}$ units
$2 \sqrt{15}$ units
$\sqrt{101}$ units
10 units

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(1)

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Using the mid point formula, the coordinates of mid-point of BC are

Co-ordinates of $D (x, y) = (\frac{6 + 0}{2}, \frac{4 + 0}{2}) = (3, 2)$

Now, length of $A D = \sqrt{{(5 - 3)}^{2} + {(- 6 - 2)}^{2}} = \sqrt{4 + 64} = \sqrt{68}$

Q 4 :
If the distance between the points (3, –5) and (x, –5) is 15 units, then the values of x are :

12, –18
–12, 18
18, 5
–9, –12

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(2)

$Here, (15)^{2} = {(3 - x)}^{2} + {(- 5 + 5)}^{2}$

$\Rightarrow 225 = 9 - 6 x + x^{2} \Rightarrow x^{2} - 6 x - 216 = 0$

$\Rightarrow x^{2} - 18 x + 12 x - 216 = 0 \Rightarrow x (x - 18) + 12 (x - 18) = 0$

$\Rightarrow (x - 18) (x + 12) = 0 \Rightarrow x = - 12, 18$

Q 5 :
If the vertices of a parallelogram PQRS taken in order are P(3, 4), Q(-2, 3) and R(-3, -2), then the coordinates of its fourth vertex S are

(-2, -1)
(-2, -3)
(2, -1)
(1, 2)

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(3)

We know that the diagonals of a parallelogram bisect each other.

Therefore, midpoint of QS = midpoint of PR. Let the coordinate of S is (x, y)

$\Rightarrow (\frac{x - 2}{2}, \frac{y + 3}{2}) = (\frac{3 - 3}{2}, \frac{4 - 2}{2})$

$\Rightarrow \frac{x - 2}{2} = \frac{3 - 3}{2} and \frac{y + 3}{2} = \frac{4 - 2}{2}$

$\Rightarrow x - 2 = 0 \Rightarrow x = 2$

$and y + 3 = 2 \Rightarrow y = - 1$

Hence, fourth vertex S are (2, -1).

Q 6 :
The centre of a circle is at (2, 0). If one end of a diameter is at (6, 0), then the other end is at :

(0, 0)
(4, 0)
(-2, 0)
(-6, 0)

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(3)

$(\frac{6 + x}{2}, \frac{0 + y}{2}) = (2, 0) \Rightarrow \frac{6 + x}{2} = 2, \frac{0 + y}{2} = 0$

⇒ x = –2 and y = 0

Q 7 :
Assertion (A): If the co-ordinates of the mid-points of the sides AB and AC of $ΔABC$ are D(3,5) and E(-3,-3) respectively, then BC = 20 units

Reason (R) : The line joining the mid points of two sides of a triangle is parallel to the third side and equal to half of it.

Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)
Assertion (A) is true but reason (R) is false.
Assertion (A) is false but reason (R) is true.

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(1)

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For assertion : Distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ $= \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

$∴$ Distance between $D E = \sqrt{{(- 3 - 3)}^{2} + {(- 3 - 5)}^{2}}$

$= \sqrt{36 + 64} = \sqrt{100} = 10 units$

By midpoint theorem, distance between $B C$ $= 2 \times distance between D E = 2 \times 10 = 20 units .$

So, assertion is true.

For reason: By midpoint theorem, the line joining the mid points of two sides of a triangle is parallel to the third side and equal to half of it. So, reason is also true.

Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)

Q 8 :
XOYZ is a rectangle with vertices X(–3, 0), O(0, 0), Y(0, 4) and Z(x, y). The length of its each diagonal is

$5$ units
$\sqrt{5}$ units
$x^{2} + y^{2}$ units
$4$ units

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(1)

We know that, Length of Diagonals are equal

In Rectangle, ZO = YX

$\Rightarrow Z O^{2} = Y X^{2} \Rightarrow {(x - 0)}^{2} + {(y - 0)}^{2} = {(0 + 3)}^{2} + {(4 - 0)}^{2}$

$\Rightarrow x^{2} + y^{2} = 25$

$\Rightarrow x^{2} + y^{2} = 5 Both diagonals are 5 units$

Q 9 :
The point on x-axis which is equidistant from the points (5, – 3) and (4, 2) is :

(4.5, 0)
(7, 0)
(0.5, 0)
(– 7, 0)

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(2) (7, 0)

Q 10 :
PQ is a diameter of a circle with centre O(2, – 4). If the coordinates of the point P are (– 4, 5), then the coordinates of the point Q will be :

(– 3, 4.5)
(– 1, 0.5)
(4, – 5)
(8, – 13)

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(4) (8, – 13)

Topic Question Set

Q 1 :
The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is

Q 2 :
Assertion (A): The value of y is 3, if the distance between the points P(2, -3) and Q(10, y) is 10.

Reason (R): Distance between two points is given by $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

Q 3 :
AD is a median of $∆$ ABC with vertices A(5, –6), B(6, 4) and C(O, O). Length AD is equal to :

Q 4 :
If the distance between the points (3, –5) and (x, –5) is 15 units, then the values of x are :

Q 5 :
If the vertices of a parallelogram PQRS taken in order are P(3, 4), Q(-2, 3) and R(-3, -2), then the coordinates of its fourth vertex S are

Q 6 :
The centre of a circle is at (2, 0). If one end of a diameter is at (6, 0), then the other end is at :

Q 7 :
Assertion (A): If the co-ordinates of the mid-points of the sides AB and AC of $ΔABC$ are D(3,5) and E(-3,-3) respectively, then BC = 20 units

Reason (R) : The line joining the mid points of two sides of a triangle is parallel to the third side and equal to half of it.

Q 8 :
XOYZ is a rectangle with vertices X(–3, 0), O(0, 0), Y(0, 4) and Z(x, y). The length of its each diagonal is

Q 9 :
The point on x-axis which is equidistant from the points (5, – 3) and (4, 2) is :

Q 10 :
PQ is a diameter of a circle with centre O(2, – 4). If the coordinates of the point P are (– 4, 5), then the coordinates of the point Q will be :

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Topic Question Set

Q 1 : The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is

Q 2 : Assertion (A): The value of y is 3, if the distance between the points P(2, -3) and Q(10, y) is 10. Reason (R): Distance between two points is given by (x2-x1)2+(y2-y1)2

Q 3 : AD is a median of ∆ABC with vertices A(5, –6), B(6, 4) and C(O, O). Length AD is equal to :

Q 4 : If the distance between the points (3, –5) and (x, –5) is 15 units, then the values of x are :

Q 5 : If the vertices of a parallelogram PQRS taken in order are P(3, 4), Q(-2, 3) and R(-3, -2), then the coordinates of its fourth vertex S are

Q 6 : The centre of a circle is at (2, 0). If one end of a diameter is at (6, 0), then the other end is at :

Q 7 : Assertion (A): If the co-ordinates of the mid-points of the sides AB and AC of ΔABC are D(3,5) and E(-3,-3) respectively, then BC = 20 units Reason (R) : The line joining the mid points of two sides of a triangle is parallel to the third side and equal to half of it.

Q 8 : XOYZ is a rectangle with vertices X(–3, 0), O(0, 0), Y(0, 4) and Z(x, y). The length of its each diagonal is

Q 9 : The point on x-axis which is equidistant from the points (5, – 3) and (4, 2) is :

Q 10 : PQ is a diameter of a circle with centre O(2, – 4). If the coordinates of the point P are (– 4, 5), then the coordinates of the point Q will be :

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Q 1 :
The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is

Q 2 :
Assertion (A): The value of y is 3, if the distance between the points P(2, -3) and Q(10, y) is 10.

Reason (R): Distance between two points is given by $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

Q 3 :
AD is a median of $∆$ ABC with vertices A(5, –6), B(6, 4) and C(O, O). Length AD is equal to :

Q 4 :
If the distance between the points (3, –5) and (x, –5) is 15 units, then the values of x are :

Q 5 :
If the vertices of a parallelogram PQRS taken in order are P(3, 4), Q(-2, 3) and R(-3, -2), then the coordinates of its fourth vertex S are

Q 6 :
The centre of a circle is at (2, 0). If one end of a diameter is at (6, 0), then the other end is at :

Q 7 :
Assertion (A): If the co-ordinates of the mid-points of the sides AB and AC of $ΔABC$ are D(3,5) and E(-3,-3) respectively, then BC = 20 units

Reason (R) : The line joining the mid points of two sides of a triangle is parallel to the third side and equal to half of it.

Q 8 :
XOYZ is a rectangle with vertices X(–3, 0), O(0, 0), Y(0, 4) and Z(x, y). The length of its each diagonal is

Q 9 :
The point on x-axis which is equidistant from the points (5, – 3) and (4, 2) is :

Q 10 :
PQ is a diameter of a circle with centre O(2, – 4). If the coordinates of the point P are (– 4, 5), then the coordinates of the point Q will be :