Coordinate Geometry Distance Formula - Class 10 Maths MCQ

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

Answer: a
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

Answer: b
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

Answer: b
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)
View Answer

Answer: c
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, \( \frac {89}{18} \))
View Answer

Answer: d
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
View Answer

Answer: a
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)
View Answer

Answer: b
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)
View Answer

Answer: d
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
View Answer

Answer: b
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

Answer: c
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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