# Class 10 Maths MCQ – Coordinate Geometry – Distance Formula

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

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

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

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,
x2 – 22x + 130 = 100
x2 – 22x + 130 – 100 = 0
x2 – 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)

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,
x2 – 18x + 145 = x2 – 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, $$\frac {89}{18}$$)

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,
y2 + 1 = y2 – 18y + 90
1 – 90 = -18y
-89 = -18y
y = $$\frac {89}{18}$$
The point is (0, $$\frac {89}{18}$$)
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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

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,
a2 – 6a + 10 + b2 – 2b = a2 – 4a + 4 + b2
-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)

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,
y2 – 14y + 74 = 25
y2 – 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)

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,
x2 – 8x + 52 = 25
x2 – 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

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

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, AC2 + AB2 = 42 + 32 = 16 + 9 = 25
BC2 = 52 = 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

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