Question

Solve the following system of equations graphically : \(x + 3y = 6\) and \(2x - 3y = 12\). Also, find the area of the triangle formed by the lines \(x + 3y = 6\), \(x = 0\) and \(y = 0\).

Hint:

"Supplementary" means sum to 180°, while "Complementary" means sum to 90°. A quick way to remember: 'C' comes before 'S', and 90 comes before 180.

Answer 1

Step 1: Convert equations into slope-intercept form

Given equations:
x + 3y = 6
2x - 3y = 12

From x + 3y = 6
3y = 6 - x
y = (6 - x) / 3

From 2x - 3y = 12
-3y = 12 - 2x
3y = 2x - 12
y = (2x - 12) / 3

Step 2: Find points to draw the graphs

For x + 3y = 6
If x = 0 → y = 2
Point A = (0, 2)
If y = 0 → x = 6
Point B = (6, 0)

For 2x - 3y = 12
If x = 0 → y = -4
Point C = (0, -4)
If y = 0 → x = 6
Point D = (6, 0)

Plot these points on graph paper and draw both straight lines. The intersection point of the two lines gives the graphical solution.

Step 3: Find intersection point (verification)

x + 3y = 6
2x - 3y = 12

Add both equations:
3x = 18
x = 6

Substitute x = 6 into first equation:
6 + 3y = 6
3y = 0
y = 0

Graphical Solution: (6, 0)

Step 4: Area of triangle formed by x + 3y = 6, x = 0 and y = 0

Intercepts of x + 3y = 6:
x-intercept = 6
y-intercept = 2

Vertices of triangle:
(0,0), (6,0), (0,2)

Base = 6 units
Height = 2 units

Area = (1/2) × base × height
Area = (1/2) × 6 × 2
Area = 6 square units

Final Answer:
Intersection point = (6, 0)
Area of triangle = 6 square units