Plot the points A(2, 5), B(–2, 2) and C(4, 2) on a graph paper. Join AB, BC and AC. Calculate the area of ∆ABC.
This question comes from Chapter 3: Coordinate Geometry, Class 9 Mathematics (NCERT). It tests two skills in one shot — the ability to plot ordered pairs accurately on a Cartesian plane and the ability to calculate the area of a triangle when its vertices are given as coordinates.
At first glance, the calculation looks tricky because A, B, and C are not arranged in an obvious shape. But once you plot them and notice that B(–2, 2) and C(4, 2) lie on the same horizontal line (y = 2), the problem becomes straightforward. You can treat BC as the base and drop a perpendicular from A to that line to get the height.This method — identifying a horizontal or vertical side as the base — is one of the fastest tricks for coordinate geometry problems in Class 9 board exams. Students preparing with eSaral's NCERT-aligned practice sets, taught by faculty like Saransh Gupta (IIT Bombay, AIR-41), consistently use this approach to solve such problems in under 90 seconds.
For more NCERT-based worked examples, visit NCERT Solutions for Class 9 & 10 Maths.Plot the points A(2, 5), B(–2, 2) and C(4, 2) on a graph paper. Join AB, BC, and AC. Calculate the area of ∆ABC.
Abscissa of D = Abscissa of A = 2
Ordinate of D = Ordinate of B = 2
Now,
BC = (2 + 4) units = 6 units
AD = (5 – 2) units = 3 units
Area of $\Delta A B C=\frac{1}{2} \times$ Base $\times$ Height
$=\frac{1}{2} \times B C \times A D$
$=\frac{1}{2} \times 6 \times 3$
$=9$
Hence, area of ∆ABC is 9 square units.
How to Plot the Points on Graph Paper
Step-by-Step Plotting Instructions
Follow these steps on your graph paper:
1. Draw the x-axis and y-axis. Mark the origin O(0, 0) where they intersect.
2. Plot A(2, 5): Move 2 units right on the x-axis, then 5 units up. Mark and label A.
3. Plot B(–2, 2): Move 2 units left on the x-axis (negative direction), then 2 units up. Mark and label B.
4. Plot C(4, 2): Move 4 units right on the x-axis, then 2 units up. Mark and label C.
5. Join A to B, B to C, and C to A using a ruler to form triangle ABC.
6. Draw D(2, 2): From A, drop a vertical line down to the line BC. It meets BC at D(2, 2). Draw this perpendicular AD with a dotted line.
Always use a sharp pencil and a ruler when plotting on graph paper in board exams. Marks are awarded for accuracy of plotting, not just the final numerical answer. Each plotted point correctly labelled can carry partial credit in CBSE Class 9 exams.
What the Graph Should Look Like
- B(–2, 2) and C(4, 2) sit on the same horizontal grid line.
- A(2, 5) sits above the line BC, directly above D(2, 2).
- The triangle is a non-equilateral, non-right triangle with one horizontal base.
- The dotted perpendicular AD divides the triangle into two smaller right triangles: △ADB and △ADC.
Common Mistakes Students Make in This Problem
Mistake 1: Calculating BC Incorrectly
Many students compute BC as:
$$BC = 4 - 2 = 2 \quad \text{(WRONG)}$$
They forget that B is at x = –2, not x = 2. The correct calculation accounts for the negative x-coordinate:
$$BC = 4 - (-2) = 6 \quad \text{(CORRECT)}$$
Mistake 2: Confusing D with an Existing Vertex
Point D(2, 2) is a construction point drawn only to measure height. Some students mistakenly label it as a fourth vertex or confuse it with B or C. Always mark D with a different symbol (e.g., a small cross) and note it as "foot of altitude."
Mistake 3: Using the Wrong Formula
The coordinate formula for area of a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is:
$$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$
Using this formula for verification:
$$= \frac{1}{2} |2(2-2) + (-2)(2-5) + 4(5-2)|$$
$$= \frac{1}{2} |2(0) + (-2)(-3) + 4(3)|$$
$$= \frac{1}{2} |0 + 6 + 12|$$
$$= \frac{1}{2} \times 18 = \mathbf{9 \text{ sq. units}} ✓$$
Both methods confirm the answer: 9 square units.