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Class 10 Maths Chapter 7: Coordinate Geometry MCQs with answers

10 multiple-choice questions on Coordinate Geometry, each with its answer and the reason, written from the NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.

Question 1
The distance of the point P(−6, 4) from the y-axis is
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Answer: (a) 6 units

The distance from the y-axis is the size of the abscissa: |−6| = 6 units. (4 is the distance from the x-axis; a distance is never −6; 2√13 is OP.)

Question 2
A kite K is flying at the point (5, −8) on a grid whose x-axis runs along a straight canal bank. The distance of K from the x-axis is
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Answer: (b) 8 units

The distance from the x-axis is the size of the ordinate: |−8| = 8 units. (5 is the distance from the y-axis; √89 is the distance from the origin.)

Question 3
Which of the following points lies on the x-axis?
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Answer: (b) (−3, 0)

A point on the x-axis has ordinate 0, so it has the form (x, 0): (−3, 0). (0, −3) lies on the y-axis.

Question 4
The distance between the points A(1, −2) and B(7, 6) is
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Answer: (a) 10 units

AB = √[(7 − 1)² + (6 − (−2))²] = √(36 + 64) = √100 = 10. (14 adds the gaps 6 + 8; 4√5 comes from adding the coordinates instead of subtracting them.)

Question 5
The distance of the point (−9, 12) from the origin is
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Answer: (c) 15 units

OP = √(x² + y²) = √(81 + 144) = √225 = 15. (21 and 3 add and subtract 9 and 12; 3√7 = √63 subtracts the squares.)

Question 6
The distance between P(−3, 5) and Q(2, −7) is
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Answer: (b) 13 units

PQ = √[(2 − (−3))² + (−7 − 5)²] = √(25 + 144) = √169 = 13. (√5 comes from adding the coordinates; 17 adds the gaps 5 + 12.)

Question 7
While finding the distance between two points, Aarav gets PQ² = 49. The distance PQ is
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Answer: (c) 7 units

A distance is never negative, so only the positive square root is taken: PQ = 7 units.

Question 8
A straight canal runs along the x-axis. A pump house is to be built on the canal so that it is equidistant from two villages at A(−1, 4) and B(5, 2). The pump house is at
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Answer: (c) (1, 0)

Let the point be (x, 0). (x + 1)² + 16 = (x − 5)² + 4 gives 2x + 17 = −10x + 29, so 12x = 12 and x = 1. The point is (1, 0); both distances are √20.

Question 9
The points A(1, −1), B(3, 3) and C(4, 5) are
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Answer: (c) collinear

AB = √(4 + 16) = 2√5, BC = √(1 + 4) = √5, AC = √(9 + 36) = 3√5. Since AB + BC = 3√5 = AC, the three points are collinear.

Question 10
In the triangle with vertices A(2, 3), B(6, 6) and C(−1, 7), the right angle is at
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Answer: (a) A

AB² = 16 + 9 = 25, AC² = 9 + 16 = 25, BC² = 49 + 1 = 50. AB² + AC² = BC², so by the converse of Pythagoras the angle opposite BC, that is ∠A, is 90°.

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