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Chapter 7: Coordinate Geometry

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This page provides comprehensive Chapter 7: Coordinate Geometry - MCQ Worksheet - SJMaths. Multiple Choice Questions (MCQ) worksheet for Class 10 Coordinate Geometry. Practice for CBSE Board Exams.

MCQ Worksheet

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  1. Question 1: The distance of the point $P(2, 3)$ from the x-axis is:
    (A) 2
    (B) 3
    (C) 1
    (D) 5
    Solution: (B) 3
    Reason: The distance of a point $(x, y)$ from the x-axis is given by $|y|$. Here, $|3| = 3$.
  2. Question 2: The distance between the points $A(0, 6)$ and $B(0, -2)$ is:
    (A) 6
    (B) 8
    (C) 4
    (D) 2
    Solution: (B) 8
    Step 1: Distance formula: $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
    Step 2: $\sqrt{(0-0)^2 + (-2-6)^2} = \sqrt{(-8)^2} = \sqrt{64} = 8$.
  3. Question 3: The midpoint of the line segment joining the points $P(-2, 8)$ and $Q(-6, -4)$ is:
    (A) $(-4, 2)$
    (B) $(4, 2)$
    (C) $(2, 6)$
    (D) $(-4, -2)$
    Solution: (A) $(-4, 2)$
    Step 1: Midpoint formula: $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$.
    Step 2: $(\frac{-2+(-6)}{2}, \frac{8+(-4)}{2}) = $(\frac{-8}{2}, \frac{4}{2}) = (-4, 2)$.
  4. Question 4: The point which divides the line segment joining the points $(7, -6)$ and $(3, 4)$ in ratio 1:2 internally lies in the:
    (A) I quadrant
    (B) II quadrant
    (C) III quadrant
    (D) IV quadrant
    Solution: (D) IV quadrant
    Step 1: Using section formula: $x = \frac{1(3) + 2(7)}{1+2} = \frac{17}{3}$, $y = \frac{1(4) + 2(-6)}{1+2} = \frac{-8}{3}$.
    Step 2: Since $x > 0$ and $y < 0$, the point lies in the IV quadrant.
  5. Question 5: The distance of the point $P(-6, 8)$ from the origin is:
    (A) 8
    (B) $2\sqrt{7}$
    (C) 10
    (D) 6
    Solution: (C) 10
    Step 1: Distance from origin: $\sqrt{x^2 + y^2}$.
    Step 2: $\sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.
  6. Question 6: If the distance between the points $(4, p)$ and $(1, 0)$ is 5, then the value of $p$ is:
    (A) 4 only
    (B) $\pm 4$
    (C) -4 only
    (D) 0
    Solution: (B) $\pm 4$
    Step 1: $\sqrt{(4-1)^2 + (p-0)^2} = 5$.
    Step 2: Squaring both sides: $3^2 + p^2 = 25 \Rightarrow 9 + p^2 = 25 \Rightarrow p^2 = 16 \Rightarrow p = \pm 4$.
  7. Question 7: The perimeter of a triangle with vertices $(0, 4)$, $(0, 0)$ and $(3, 0)$ is:
    (A) 5
    (B) 12
    (C) 11
    (D) $7 + \sqrt{5}$
    Solution: (B) 12
    Step 1: Vertices form a right-angled triangle at origin.
    Step 2: Sides are 4 (on y-axis), 3 (on x-axis). Hypotenuse = $\sqrt{3^2+4^2} = 5$.
    Step 3: Perimeter = $3 + 4 + 5 = 12$.
  8. Question 8: If the points $A(1, 2)$, $O(0, 0)$ and $C(a, b)$ are collinear, then:
    (A) $a = b$
    (B) $a = 2b$
    (C) $2a = b$
    (D) $a = -b$
    Solution: (C) $2a = b$
    Step 1: For collinear points, slopes must be equal. Slope of OA = $\frac{2-0}{1-0} = 2$.
    Step 2: Slope of OC = $\frac{b-0}{a-0} = \frac{b}{a}$.
    Step 3: $\frac{b}{a} = 2 \Rightarrow b = 2a$.
  9. Question 9: The ratio in which the x-axis divides the segment joining $(3, 6)$ and $(12, -3)$ is:
    (A) 2:1
    (B) 1:2
    (C) -2:1
    (D) 1:-2
    Solution: (A) 2:1
    Step 1: Let ratio be $k:1$. The y-coordinate of the point on x-axis is 0.
    Step 2: $y = \frac{k(-3) + 1(6)}{k+1} = 0 \Rightarrow -3k + 6 = 0 \Rightarrow k = 2$. Ratio is 2:1.
  10. Question 10: If $(a/3, 4)$ is the midpoint of the segment joining the points $P(-6, 5)$ and $R(-2, 3)$, then the value of $a$ is:
    (A) -4
    (B) -12
    (C) 12
    (D) -6
    Solution: (B) -12
    Step 1: Midpoint x-coordinate = $\frac{-6 + (-2)}{2} = \frac{-8}{2} = -4$.
    Step 2: Given x-coordinate is $a/3$. So, $a/3 = -4 \Rightarrow a = -12$.
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