Question

Solve the following given system of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:

x – y + 1 = 0, 3x + 2y – 12 = 0

Answer 1

1. Table of values for graphing

To plot both equations graphically, we find at least two points for each line by rewriting them in terms of y:

Line 1: x – y + 1 = 0 ⇒ y = x + 1

If x = 0, y = 1 ⇒ (0, 1)

If x = –1, y = 0 ⇒ (–1, 0)

If x = 2, y = 3 ⇒ (2, 3)

Line 2: 3x + 2y – 12 = 0 ⇒ `y = (12 - 3x)/2`

If x = 0, y = 6 ⇒ (0, 6)

If x = 4, y = 0 ⇒ (4, 0)

If x = 2, y = 3 ⇒ (2, 3)

2. Graphical Representation

Plotting these coordinates on a Cartesian plane reveals that the two lines intersect at a common point, which represents the solution to the system.

3. Vertices of the triangle

The triangle is bounded by Line 1, Line 2 and the x-axis (y = 0). Its verties are:

A(2, 3) (Point of intersection of the two lines)

B(–1, 0) (Intersection of x – y + 1 = 0 with the x-axis)

C(4, 0) (Intersection of 3x + 2y – 12 = 0 with the x-axis)

4. Calculation of area

The area of a triangle is given by the standard formula:

Area = `1/2 xx "Base" xx "Height"`

Base (BC): Extends along the x-axis from x = –1 to x = 4.

Base = 4 – (–1) = 5 units

Height: The perpendicular distance from vertex A(2, 3) to the x-axis, which matches its y-coordinate.

Height = 3 units

Area = `1/2 xx 5 xx 3`

Area = 7.5 square units