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Chapter 7: Coordinate Geometry

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This page provides comprehensive Chapter 7: Coordinate Geometry - Standard Worksheet - SJMaths. Standard level practice worksheet for Class 10 Coordinate Geometry. Practice for CBSE Board Exams.

Standard Level Worksheet

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  1. Question 1: Find the point on the x-axis which is equidistant from $(2, -5)$ and $(-2, 9)$.
    Solution:
    Step 1: Let point be $P(x, 0)$. $PA = PB \Rightarrow PA^2 = PB^2$.
    Step 2: $(x-2)^2 + (0+5)^2 = (x+2)^2 + (0-9)^2$.
    Step 3: $x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81$.
    Step 4: $-4x + 29 = 4x + 85 \Rightarrow -8x = 56 \Rightarrow x = -7$.
    Answer: $(-7, 0)$.
  2. Question 2: Find the ratio in which the y-axis divides the line segment joining the points $(5, -6)$ and $(-1, -4)$. Also find the point of intersection.
    Solution:
    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 is $5:1$.
    Step 3: $y = \frac{5(-4) + 1(-6)}{5+1} = \frac{-20-6}{6} = \frac{-26}{6} = \frac{-13}{3}$.
    Answer: Ratio 5:1, Point $(0, -13/3)$.
  3. Question 3: If the points $A(6, 1)$, $B(8, 2)$, $C(9, 4)$ and $D(p, 3)$ are the vertices of a parallelogram, taken in order, find the value of $p$.
    Solution:
    Step 1: Diagonals of a parallelogram bisect each other. Midpoint of AC = Midpoint of BD.
    Step 2: Midpoint AC: $(\frac{6+9}{2}, \frac{1+4}{2}) = (7.5, 2.5)$.
    Step 3: Midpoint BD: $(\frac{8+p}{2}, \frac{2+3}{2}) = (\frac{8+p}{2}, 2.5)$.
    Step 4: $\frac{8+p}{2} = 7.5 \Rightarrow 8+p = 15 \Rightarrow p = 7$.
  4. Question 4: Find the coordinates of the points of trisection of the line segment joining $(4, -1)$ and $(-2, -3)$.
    Solution:
    Step 1: Let points be $P$ and $Q$. $P$ divides in 1:2, $Q$ in 2:1.
    Step 2: $P = (\frac{1(-2)+2(4)}{3}, \frac{1(-3)+2(-1)}{3}) = (\frac{6}{3}, \frac{-5}{3}) = (2, -5/3)$.
    Step 3: $Q = (\frac{2(-2)+1(4)}{3}, \frac{2(-3)+1(-1)}{3}) = (\frac{0}{3}, \frac{-7}{3}) = (0, -7/3)$.
  5. Question 5: Find the ratio in which the line segment joining $A(1, -5)$ and $B(-4, 5)$ is divided by the x-axis. Also find the coordinates of the point of division.
    Solution:
    Step 1: Let ratio be $k:1$. Point on x-axis is $(x, 0)$.
    Step 2: $y = \frac{k(5) + 1(-5)}{k+1} = 0 \Rightarrow 5k - 5 = 0 \Rightarrow k = 1$. Ratio 1:1.
    Step 3: $x = \frac{1(-4) + 1(1)}{2} = -3/2$.
    Answer: Ratio 1:1, Point $(-3/2, 0)$.
  6. Question 6: If $(1, 2)$, $(4, y)$, $(x, 6)$ and $(3, 5)$ are the vertices of a parallelogram taken in order, find $x$ and $y$.
    Solution:
    Step 1: Midpoint of diagonals coincide.
    Step 2: $\frac{1+x}{2} = \frac{4+3}{2} \Rightarrow 1+x = 7 \Rightarrow x = 6$.
    Step 3: $\frac{2+6}{2} = \frac{y+5}{2} \Rightarrow 8 = y+5 \Rightarrow y = 3$.
    Answer: $x=6, y=3$.
  7. Question 7: Find the coordinates of a point $A$, where $AB$ is the diameter of a circle whose centre is $(2, -3)$ and $B$ is $(1, 4)$.
    Solution:
    Step 1: Let $A$ be $(x, y)$. Centre $(2, -3)$ is the midpoint of $AB$.
    Step 2: $\frac{x+1}{2} = 2 \Rightarrow x+1 = 4 \Rightarrow x = 3$.
    Step 3: $\frac{y+4}{2} = -3 \Rightarrow y+4 = -6 \Rightarrow y = -10$.
    Answer: $A(3, -10)$.
  8. Question 8: If $A$ and $B$ are $(-2, -2)$ and $(2, -4)$, respectively, find the coordinates of $P$ such that $AP = \frac{3}{7} AB$ and $P$ lies on the line segment $AB$.
    Solution:
    Step 1: $AP = \frac{3}{7} AB \Rightarrow \frac{AP}{AB} = \frac{3}{7} \Rightarrow AP:PB = 3:4$.
    Step 2: Using section formula with ratio 3:4.
    Step 3: $x = \frac{3(2) + 4(-2)}{7} = \frac{6-8}{7} = -2/7$.
    Step 4: $y = \frac{3(-4) + 4(-2)}{7} = \frac{-12-8}{7} = -20/7$.
    Answer: $P(-2/7, -20/7)$.
  9. Question 9: Find the area of a rhombus if its vertices are $(3, 0)$, $(4, 5)$, $(-1, 4)$ and $(-2, -1)$ taken in order.
    Solution:
    Step 1: Area of rhombus = $\frac{1}{2} \times d_1 \times d_2$.
    Step 2: $d_1 = AC = \sqrt{(-1-3)^2 + (4-0)^2} = \sqrt{16+16} = 4\sqrt{2}$.
    Step 3: $d_2 = BD = \sqrt{(-2-4)^2 + (-1-5)^2} = \sqrt{36+36} = 6\sqrt{2}$.
    Step 4: Area $= \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = \frac{1}{2} \times 24 \times 2 = 24$ sq units.
  10. Question 10: Find the values of $y$ for which the distance between the points $P(2, -3)$ and $Q(10, y)$ is 10 units.
    Solution:
    Step 1: $PQ = 10 \Rightarrow PQ^2 = 100$.
    Step 2: $(10-2)^2 + (y+3)^2 = 100 \Rightarrow 64 + (y+3)^2 = 100$.
    Step 3: $(y+3)^2 = 36 \Rightarrow y+3 = \pm 6$.
    Step 4: $y = 3$ or $y = -9$.
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