Chapter 7: Coordinate Geometry

Overview

This page provides comprehensive Chapter 7: Coordinate Geometry - Case Based Worksheet - SJMaths. Competency and Case Study based questions for Class 10 Coordinate Geometry. Practice for CBSE Board Exams.

Case Based / Competency Worksheet

Directions:

Read the following case studies carefully and answer the questions that follow.

  1. Case Study 1: Sports Day

    In a rectangular playground $ABCD$, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along $AD$. Niharika runs $\frac{1}{4}$th the distance $AD$ on the 2nd line and posts a green flag. Preet runs $\frac{1}{5}$th the distance $AD$ on the 8th line and posts a red flag.

    1. What is the distance between the green flag and the red flag?
      (A) $\sqrt{61}$ m
      (B) $\sqrt{101}$ m
      (C) $\sqrt{51}$ m
      (D) $\sqrt{41}$ m
      Solution: (A) $\sqrt{61}$ m
      Step 1: Green Flag (G): $x=2$, $y = \frac{1}{4} \times 100 = 25$. $G(2, 25)$.
      Step 2: Red Flag (R): $x=8$, $y = \frac{1}{5} \times 100 = 20$. $R(8, 20)$.
      Step 3: Distance $GR = \sqrt{(8-2)^2 + (20-25)^2} = \sqrt{36 + 25} = \sqrt{61}$ m.
    2. If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
      (A) $(5, 22.5)$
      (B) $(10, 22)$
      (C) $(2, 20)$
      (D) $(5, 20)$
      Solution: (A) $(5, 22.5)$
      Step 1: Midpoint formula: $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$.
      Step 2: $M = (\frac{2+8}{2}, \frac{25+20}{2}) = (5, 22.5)$.
  2. Case Study 2: Section Formula Application

    Point $P$ divides the line segment joining the points $A(2, 1)$ and $B(5, -8)$ such that $\frac{AP}{AB} = \frac{1}{3}$.

    1. What is the ratio in which $P$ divides $AB$?
      (A) 1:3
      (B) 1:2
      (C) 2:1
      (D) 2:3
      Solution: (B) 1:2
      Reason: Given $\frac{AP}{AB} = \frac{1}{3}$. Since $AB = AP + PB$, we have $\frac{AP}{AP+PB} = \frac{1}{3} \Rightarrow 3AP = AP + PB \Rightarrow 2AP = PB \Rightarrow \frac{AP}{PB} = \frac{1}{2}$.
    2. What are the coordinates of point $P$?
      (A) $(3, -2)$
      (B) $(3, 2)$
      (C) $(-3, -2)$
      (D) $(-3, 2)$
      Solution: (A) $(3, -2)$
      Step 1: Using section formula with ratio $1:2$.
      Step 2: $x = \frac{1(5) + 2(2)}{3} = \frac{9}{3} = 3$. $y = \frac{1(-8) + 2(1)}{3} = \frac{-6}{3} = -2$.
  3. Case Study 3: Class Room Seating

    In a classroom, 4 friends are seated at the points $A, B, C$ and $D$ as shown in a grid. The coordinates are $A(3, 4)$, $B(6, 7)$, $C(9, 4)$ and $D(6, 1)$.

    1. Find the distance between $A$ and $C$.
      (A) 6 units
      (B) $6\sqrt{2}$ units
      (C) 5 units
      (D) 4 units
      Solution: (A) 6 units
      Step 1: $A(3, 4)$ and $C(9, 4)$ have the same y-coordinate.
      Step 2: Distance $= |9 - 3| = 6$ units.
    2. What type of quadrilateral is $ABCD$?
      (A) Square
      (B) Rhombus
      (C) Rectangle
      (D) Parallelogram
      Solution: (A) Square
      Step 1: All sides are equal: $AB = BC = CD = DA = \sqrt{3^2+3^2} = 3\sqrt{2}$.
      Step 2: Diagonals are equal: $AC = 6$, $BD = \sqrt{(6-6)^2 + (1-7)^2} = 6$.
      Step 3: Equal sides and equal diagonals imply a Square.
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