Chapter 7: Coordinate Geometry

Overview

This page provides comprehensive Chapter 7: Coordinate Geometry - Assertion Reason Worksheet - SJMaths. Assertion and Reason type questions for Class 10 Coordinate Geometry. Practice for CBSE Board Exams.

Assertion-Reason Worksheet

Directions:

In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:

  • (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  • (B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
  • (C) Assertion (A) is true but Reason (R) is false.
  • (D) Assertion (A) is false but Reason (R) is true.
  1. Question 1:
    Assertion (A): The point $(0, 4)$ lies on the y-axis.
    Reason (R): The x-coordinate of a point on the y-axis is zero.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Since the x-coordinate is 0, the point lies on the y-axis. A is true.
    Step 2: R is the correct condition for a point to lie on the y-axis. R explains A.
  2. Question 2:
    Assertion (A): The distance of point $P(2, 3)$ from the x-axis is 3.
    Reason (R): The distance of point $(x, y)$ from the y-axis is $|x|$.
    (A)
    (B)
    (C)
    (D)
    Solution: (B)
    Step 1: Distance from x-axis is $|y| = |3| = 3$. A is true.
    Step 2: R is a true statement, but it explains distance from y-axis, not x-axis. So R does not explain A.
  3. Question 3:
    Assertion (A): The point $P(-2, 3)$ lies in the third quadrant.
    Reason (R): The signs of coordinates in the third quadrant are $(- , -)$.
    (A)
    (B)
    (C)
    (D)
    Solution: (D)
    Step 1: Point $(-2, 3)$ has $x < 0, y > 0$, so it lies in the II quadrant. A is false.
    Step 2: R is true.
  4. Question 4:
    Assertion (A): The point $(2, 2)$ is equidistant from $(1, 1)$ and $(3, 3)$.
    Reason (R): The origin has coordinates $(0, 0)$.
    (A)
    (B)
    (C)
    (D)
    Solution: (B)
    Step 1: Distance from $(1,1)$ is $\sqrt{1^2+1^2}=\sqrt{2}$. Distance from $(3,3)$ is $\sqrt{1^2+1^2}=\sqrt{2}$. A is true.
    Step 2: R is true but irrelevant to A.
  5. Question 5:
    Assertion (A): The mid-point of the line segment joining the points $P(-2, 8)$ and $Q(-6, -4)$ is $(-4, 2)$.
    Reason (R): The mid-point of line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Midpoint $= (\frac{-2-6}{2}, \frac{8-4}{2}) = (\frac{-8}{2}, \frac{4}{2}) = (-4, 2)$. A is true.
    Step 2: R is the correct formula used.
  6. Question 6:
    Assertion (A): The points $(1, 2), (2, 3)$ and $(3, 4)$ are collinear.
    Reason (R): Three points are collinear if they form an isosceles triangle.
    (A)
    (B)
    (C)
    (D)
    Solution: (C)
    Step 1: Slopes: $\frac{3-2}{2-1} = 1$ and $\frac{4-3}{3-2} = 1$. Slopes equal $\Rightarrow$ collinear. A is true.
    Step 2: Collinear points cannot form a triangle. R is false.
  7. Question 7:
    Assertion (A): The centroid of the triangle formed by the points $(0, 0), (3, 0)$ and $(0, 4)$ is $(1, 4/3)$.
    Reason (R): The centroid of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Centroid $= (\frac{0+3+0}{3}, \frac{0+0+4}{3}) = (1, 4/3)$. A is true.
    Step 2: R is the correct formula.
  8. Question 8:
    Assertion (A): The ratio in which the y-axis divides the segment joining $(5, -6)$ and $(-1, -4)$ is $5:1$.
    Reason (R): The x-coordinate of a point on the y-axis is zero.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Let ratio be $k:1$. Point on y-axis is $(0, y)$.
    Step 2: $x = \frac{k(-1) + 1(5)}{k+1} = 0 \Rightarrow -k+5=0 \Rightarrow k=5$. Ratio 5:1. A is true.
    Step 3: R is the property used to set $x=0$.
  9. Question 9:
    Assertion (A): The distance of the point $(3, 4)$ from the origin is 7.
    Reason (R): Distance from origin is $\sqrt{x^2+y^2}$.
    (A)
    (B)
    (C)
    (D)
    Solution: (D)
    Step 1: Distance $= \sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5$. A is false.
    Step 2: R is true.
  10. Question 10:
    Assertion (A): If the points $(1, 2), (4, y), (x, 6)$ and $(3, 5)$ are vertices of a parallelogram taken in order, then $x = 6$ and $y = 3$.
    Reason (R): Diagonals of a parallelogram bisect each other.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Midpoint of diagonals coincide. Midpoint of $(1,2)$ & $(x,6)$ is $(\frac{1+x}{2}, 4)$.
    Step 2: Midpoint of $(4,y)$ & $(3,5)$ is $(3.5, \frac{y+5}{2})$.
    Step 3: $\frac{1+x}{2} = 3.5 \Rightarrow x=6$. $4 = \frac{y+5}{2} \Rightarrow y=3$. A is true.
    Step 4: R is the property used.
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