Question

Find graphically the vertices of triangle whose sides are 3x + 4y = 12, y − 6 = 0 and y = 2x − 8. Find the area of the triangle.

Answer 1

Given:

3x + 4y = 12

y − 6 = 0

y = 6

y = 2x − 8

Step 1: Find the vertices (points of intersection)

Intersection of 3x + 4y = 12 and y = 6

Substitute y = 6 into the first equation

3x + 4(6) = 12

3x + 24 = 12

3x = −12

x = −4

Point A = (−4, 6)

Intersection of y = 6 and y = 2x − 8

6 = 2x − 8

2x = 14

x = 7

Point B = (7, 6)

Intersection of 3x + 4y = 12 and y = 2x − 8

Substitute y = 2x − 8 into 3x + 4y = 12

3x + 4(2x − 8) = 12

3x + 8x − 32 = 12

11x − 32 = 12

11x = 12 + 32 = 44

x = `44/11`

x = 4

Point C = (4, 0)

Step 2: Use Area Formula for 3 points

A = (−4, 6)

B = (7, 6)

C = (4, 0)

Area = `1/2 |x_1(y_2 − y3) + x_2(y_3 − y_1) + x_3(y_1 − y_2)|`   ... [use the area of triangle method]

= `1/2 |−4(6 − 0) + 7(0 − 6) +4(6 − 6)|`

= `1/2 |−24 − 42 + 0|`

= `1/2 xx 66`

Area = 33 square units