This set of Class 10 Maths Chapter 7 Multiple Choice Questions & Answers (MCQs) focuses on “Coordinate Geometry – Distance Formula”.

1. The distance between the points (5, 7) and (8, -5) is ________

a) √153

b) √154

c) √13

d) √53

View Answer

Explanation: Using distance formula,

Distance between (5, 7) and (8, -5) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(8-5)^2 + (-5-7)^2}\)

= \( \sqrt {(3)^2 + (-12)^2} \)

= \( \sqrt {9 + 144}\)

= √153

2. The distance of the point (9, -12) from origin will be ___________

a) 13

b) 15

c) 14

d) 17

View Answer

Explanation: Distance between (9, -12) and (0, 0) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(0-9)^2 + (0 + 12)^2} \)

= \( \sqrt {(9)^2 + (-12)^2} \)

= \( \sqrt {81 + 144}\)

= √225 = 15

3. What will be the value of x, if the distance between the points (5, 11) and (2, x) is 10?

a) -11 + √91, -11 – √91

b) 11 + √91, 11 – √91

c) 11 + √91, 11 + √91

d) -11 + √91, 11 – √91

View Answer

Explanation: Distance between (5, 11) and (2, x) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(2-5)^2 + (x-11)^2} \)

= \( \sqrt {x^2-22x + 121 + (-3)^2} \)

= \( \sqrt {x^2-22x + 121 + 9} \)

= \( \sqrt {x^2-22x + 130} \)

The distance between (5, 11) and (2, x) is 10

∴ \( \sqrt {x^2-22x + 130} \) = 10

Squaring on both sides we get,

x

^{2}– 22x + 130 = 100

x

^{2}– 22x + 130 – 100 = 0

x

^{2}– 22x + 30 = 0

x = 11 + √91, 11 – √91

4. What will be the point of x-axis which will be equidistant from the points (9, 8) and (3, 2)?

a) (10, 0)

b) (13, 0)

c) (11, 0)

d) (12, 0)

View Answer

Explanation: Let the point on x-axis be (x, 0)

Distance between (9, 8) and (x, 0) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(x-9)^2 + (0-8)^2} \)

= \( \sqrt {x^2-18x + 81 + (-8)^2} \)

= \( \sqrt {x^2-18x + 81 + 64} \)

= \( \sqrt {x^2-18x + 145} \)

Distance between (3, 2) and (x, 0) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(x-3)^2 + (0-2)^2} \)

= \( \sqrt {x^2-6x + 9 + (-2)^2} \)

= \( \sqrt {x^2-6x + 9 + 4} \)

= \( \sqrt {x^2-6x + 13} \)

Since, the point ( x, 0) is equidistant to (3, 2) and (9, 8)

The distances will be equal

∴ \( \sqrt {x^2-18x + 145} = \sqrt {x^2-6x + 13} \)

Squaring on both sides we get,

x

^{2}– 18x + 145 = x

^{2}– 6x + 13

-18x + 145 = -6x + 13

-18x + 6x = -145 + 13

-12x = -132

x = \( \frac {132}{12} \) = 11

The point is (11, 0)

5. What will be the point of y-axis which will be equidistant from the points (-1, 0) and (3, 9)?

a) (5, \( \frac {89}{18} \))

b) (1, \( \frac {89}{18} \))

c) (16, \( \frac {89}{18} \))

d) (0, \( \frac {89}{18} \))

View Answer

Explanation: Let the point on y-axis be (0, y)

Distance between (-1, 0) and (0, y) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(0 + 1)^2 + (y-0)^2} \)

= \( \sqrt {y^2 + (1)^2} \)

= \( \sqrt {y^2 + 1} \)

Distance between (3, 9) and (0, y) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(0-3)^2 + (y-9)^2} \)

= \( \sqrt {y^2-18y + 81 + (-3)^2} \)

= \( \sqrt {y^2-18y + 81 + 9} \)

= \( \sqrt {y^2-18y + 90} \)

Since, the point (0, y) is equidistant from (-1, 0) and (3, 9)

The distances will be equal

∴ \( \sqrt {y^2 + 1} = \sqrt {y^2-18y + 90} \)

Squaring on both sides we get,

y

^{2}+ 1 = y

^{2}– 18y + 90

1 – 90 = -18y

-89 = -18y

y = \( \frac {89}{18} \)

The point is (0, \( \frac {89}{18} \))

6. If the point P(a, b) is equidistant from the points (3, 1) and (2, 0) then ____________

a) a + b = -3

b) a – b = -3

c) a + b = 3

d) a – b = 3

View Answer

Explanation: The point is (a, b)

Distance between (3, 1) and (a, b) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(a-3)^2 + (b-1)^2} \)

= \( \sqrt {a^2-6a + 9 + b^2-2b + 1} \)

= \( \sqrt {a^2-6a + 10 + b^2-2b} \)

Distance between (2, 0) and (a, b) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(a-2)^2 + (b-0)^2} \)

= \( \sqrt {a^2-4a + 4 + b^2 } \)

Since, the point (a, b) is equidistant from (-1, 0) and (3, 9)

The distances will be equal

∴ \( \sqrt {a^2-6a + 10 + b^2-2b} = \sqrt {a^2-4a + 4 + b^2 } \)

Squaring on both sides we get,

a

^{2}– 6a + 10 + b

^{2}– 2b = a

^{2}– 4a + 4 + b

^{2}

-6a + 10 – 2b = -4a + 4

-2a – 6 = 2b

-a – b = 3

a + b = -3

7. The point on y-axis which is at a distance 5 unit from the point (-5, 7) is ___________

a) (7, 0)

b) (0, 7)

c) (1, 7)

d) (7, 7)

View Answer

Explanation: Let the point on y-axis be (0, y)

Distance between (-5, 7) and (0, y) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(0 + 5)^2 + (y-7)^2} \)

= \( \sqrt {y^2-14y + 49 + 25} \)

= \( \sqrt {y^2-14y + 74} \)

The distance between (-5, 7) and (0, y) is 5

∴ \( \sqrt {y^2-14y + 74} \) = 5

Squaring on both sides, we get,

y

^{2}– 14y + 74 = 25

y

^{2}– 14y + 49 = 0

y = 7, 7

Hence, the point is (0, 7)

8. The point on x-axis which is at a distance 12 unit from the point (4, 6) is ___________

a) (-4 + \( \sqrt {11i} \), 0), (-4 – \( \sqrt {11i} \), 0)

b) (-4 – \( \sqrt {11i} \), 0), (4 – \( \sqrt {11i} \), 0)

c) (4 – \( \sqrt {11i} \), 0), (4 – \( \sqrt {11i} \), 0)

d) (4 + \( \sqrt {11i} \), 0), (4 – \( \sqrt {11i} \), 0)

View Answer

Explanation: Let the point on x-axis be (x, 0)

Distance between (4, 6) and (x, 0) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(x-4)^2 + (0-6)^2} \)

= \( \sqrt {x^2-8x + 16 + 36} \)

= \( \sqrt {x^2-8x + 52} \)

The distance between (4, 6) and (x, 0) is 12

∴ \( \sqrt {x^2-8x + 52} \) = 12

Squaring on both sides, we get,

x

^{2}– 8x + 52 = 25

x

^{2}– 8x + 27 = 0

x = 4 + \( \sqrt {11i} \), 4 – \( \sqrt {11i} \)

9. If A(0, 3), B(5, 0) and C(-5, 0) are the vertices of ∆ABC, then the triangle is __________

a) Right-angled

b) Isosceles

c) Scalene

d) Equilateral

View Answer

Explanation: Distance between (0, 3) and (5, 0) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(5-0)^2 + (0-3)^2} \)

= \( \sqrt {5^2 + -3^2 } \)

= \( \sqrt {25 + 9} \)

= √34

Distance between (5, 0) and (-5, 0) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(-5-5)^2 + (0-0)^2} \)

= \( \sqrt {-10^2} \)

= 10

Distance between (0, 3) and (-5, 0) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(-5-0)^2 + (0-3)^2} \)

= \( \sqrt {-5^2 + (-3)^2} \)

= \( \sqrt {25 + 9} \)

= √34

Since, the two sides of the triangle are equal.

Hence, the triangle will be isosceles triangle.

10. The area of the triangle if A (-1, -1), B(-1, 3) and C (2, -1) are the vertices of the triangle is ____________

a) 8 units

b) 4 units

c) 6 units

d) 5 units

View Answer

Explanation: Distance between A (-1, -1) and B (-1, 3) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(-1 + 1)^2 + (3 + 1)^2} \)

= \( \sqrt {4^2} \)

= √16

= 4

Distance between B (-1, 3) and C(2, -1) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(2 + 1)^2 + (-1-3)^2} \)

= \( \sqrt {3^2 + (-4)^2} \)

= \( \sqrt {9 + 16} \)

= 5

Distance between A (-1, -1) and C (2, -1) = \( \sqrt {(x_2-x_1)^2 + (y_2-y_1)^2} \)

= \( \sqrt {(2 + 1)^2 + (-1 + 1)^2} \)

= \( \sqrt {3^2 + 0^2} \)

= 3

Now, AC

^{2}+ AB

^{2}= 4

^{2}+ 3

^{2}= 16 + 9 = 25

BC

^{2}= 5

^{2}= 25

Hence, it is a right-angled triangle, right-angled at A.

Area of triangle = \( \frac {1}{2}\) × base × height = \( \frac {1}{2}\) × 4 × 3 = 6 units

**Sanfoundry Global Education & Learning Series – Mathematics – Class 10**.

To practice all chapters and topics of class 10 Mathematics, __ here is complete set of 1000+ Multiple Choice Questions and Answers__.

**Related Posts:**

- Practice Class 9 Mathematics MCQs
- Practice Class 8 Mathematics MCQs
- Check Class 10 - Mathematics Books