Question

Find the coordinates of the vertices and the area of the triangle enclosed by the y-axis and the graphs of x + 3y = 12 and x − 3y = 0.

Answer 1

Given Equations:

x + 3y = 12

x − 3y = 0

y-axis (i.e., x = 0)

Step 1: Find coordinates of the vertices

Intersection with y-axis (i.e., x = 0)

For x + 3y = 12

0 + 3y = 12

Y = 4

Point (0, 4)

For x − 3y = 0

0 − 3y = 0

y = 0

Point (0, 0)

Intersection of the two lines 

x + 3y = 12   ...(i)

x − 3y = 0   ...(ii)

(x + 3y) + (x − 3y) = 12 + 0    ...[Adding (i) & (ii) Equation]

x + x + 3y − 3y = 12
2x + 0 = 12

2x = 12

x = `12/2 = 6`

Step 2: Substitute value of x into one of the original equations

x − 3y = 0

x = 6

6 − 3y = 0

3y = 6

y = `6/3`

y = 2

Point (6, 2)

Step 3: 

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|0(4 − 2) + 0(2 − 0) + 6(0 − 4)|`

= `1/2|0 + 0 − 24|`

= `1/2 xx 24`

= 12 square units.