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 :

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)

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 3 :

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 4 :

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 5 :

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 6 :

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 7 :

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 8 :

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)

Q 9 :

Point P divides the line segment joining the points A(4, –5) and B(1, 2) in the ratio 5 : 2. Coordinates of point P are

$(\frac{5}{2}, \frac{- 3}{2})$
$(\frac{11}{2}, 0)$
$(\frac{13}{7}, 0)$
$(0, \frac{13}{7})$

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(3) $(\frac{13}{7}, 0)$

$x = \frac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}} = \frac{5 \times 1 + 2 \times 4}{5 + 2} = \frac{5 + 8}{7} = \frac{13}{7}$

$y = \frac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}} = \frac{5 \times 2 + 2 \times (- 5)}{5 + 2} = \frac{10 - 10}{7} = 0$

Q 10 :

If end points of a diameter of a circle are $(- 5, 4) a n d (1, 0),$ then the radius of the circle is:

$2 \sqrt{13} u n i t s$
$\sqrt{13} u n i t s$
$4 \sqrt{2} u n i t s$
$2 \sqrt{2} u n i t s$

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(2)

Let given end points of diameter be A(-5,4) and B(1,0).

∴ Diameter has length,

$A B = \sqrt{(1 + 5)^{2} + (0 - 4)^{2}} = \sqrt{36 + 16} = \sqrt{52} = 2 \sqrt{13} units Radiusofcircle = \frac{2 \sqrt{13}}{2} = \sqrt{13} units .$

Topic Question Set

Q 1 :

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

Q 2 :

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

Q 3 :

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

Q 4 :

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 5 :

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 6 :

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 7 :

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

Q 8 :

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 :

Q 9 :

Point P divides the line segment joining the points A(4, –5) and B(1, 2) in the ratio 5 : 2. Coordinates of point P are

Q 10 :

If end points of a diameter of a circle are $(- 5, 4) a n d (1, 0),$ then the radius of the circle is:

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Topic Question Set

Q 1 : The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

Want Us to Email you About Special Offers & Updates?

Q 1 :

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