Chapter 7 · Class 10 CBSE · MCQ Test
Coordinate Geometry MCQ Test — Class 10 CBSE
Practice 10 multiple-choice questions with instant answer reveal and explanations.
Coordinate Geometry — MCQ Questions
1Find the distance between the points $(0, 0)$ and $(3, 4)$.
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Answer: 5
Hint: Use the distance formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
Solution:
$d = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{9 + 16}$ — = \sqrt{25} = 5
2The midpoint of the line segment joining $(2, 4)$ and $(6, 8)$ is:
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Answer: $(4, 6)$
Hint: Midpoint = $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.
Solution:
Midpoint $= \left(\frac{2+6}{2}, \frac{4+8}{2}\right)$ — = (4, 6)
3The distance of the point $(5, 12)$ from the origin is:
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Answer: 13
Hint: Distance from origin = $\sqrt{x^2 + y^2}$.
Solution:
$d = \sqrt{5^2 + 12^2} = \sqrt{25 + 144}$ — = \sqrt{169} = 13
4The coordinates of the point which divides the line segment joining $(1, 3)$ and $(5, 7)$ in the ratio $1:1$ are:
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Answer: $(3, 5)$
Hint: Ratio $1:1$ means the midpoint.
Solution:
Ratio $1:1$ is the midpoint.
$= \left(\frac{1+5}{2}, \frac{3+7}{2}\right)$ — = (3, 5)
5Find the distance between $(1, 1)$ and $(4, 5)$.
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Answer: 5
Hint: Use the distance formula.
Solution:
$d = \sqrt{(4-1)^2 + (5-1)^2} = \sqrt{9 + 16}$ — = 5
6The point $(0, 5)$ lies on which axis?
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Answer: y-axis
Hint: If the x-coordinate is 0, the point is on the y-axis.
Solution:
The x-coordinate is 0, so the point lies on the y-axis.
7The distance between the points $(3, 0)$ and $(0, 4)$ is:
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Answer: 5
Hint: Apply the distance formula.
Solution:
$d = \sqrt{(3-0)^2 + (0-4)^2} = \sqrt{9+16}$ — = 5
8Find the midpoint of the segment joining $(-2, 3)$ and $(4, -1)$.
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Answer: $(1, 1)$
Hint: Average the x-coordinates and y-coordinates.
Solution:
Midpoint $= \left(\frac{-2+4}{2}, \frac{3+(-1)}{2}\right)$ — = (1, 1)
9In which quadrant does the point $(-3, 4)$ lie?
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Answer: II
Hint: Negative x, positive y is the second quadrant.
Solution:
$x < 0$ and $y > 0$, so the point lies in Quadrant II.
10Find the distance between $(-1, -2)$ and $(2, 2)$.
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Answer: 5
Hint: Use the distance formula carefully with negative coordinates.
Solution:
$d = \sqrt{(2-(-1))^2 + (2-(-2))^2} = \sqrt{9 + 16}$ — = 5
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Tips for Coordinate Geometry MCQs
- 1Read each question carefully and identify what is being asked before looking at the options.
- 2Try to solve the problem mentally or on paper first, then match your answer with the options.
- 3Use elimination — rule out clearly wrong options to improve your chances even when unsure.
- 4Check units, signs, and edge cases — these are common traps in Coordinate Geometry MCQs.
- 5Review your mistakes after completing the test to build lasting understanding.
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