Question

Solve the following system of linear equations graphically :
4x - 5y - 20 = 0
3x + 3y - 15 = 0
Determine the vertices of the triangle formed by the lines, represented by the above equations and the y-axis.

Answer 1

4x - 5y - 20 = 0   ...(1)
3x + 3y - 15 = 0  ...(2)

4x - 5y - 20 = 0
⇒ 4x = 5y + 20
Corresponding values of x and y can be tabulated as :

x	0	-5	5
y	-4	-8	0

Plotting points (0, -4), (-5, -8), (5, 0) and joining them, we get a line l₁ which is the graph of equation (1).

Again, 3x + 3y - 15 = 0
⇒ x + y - 5 = 0
⇒ x+ y = 5
Corresponding values of x and y can be tabulated as :

x	0	-5	5
y	-4	-8	0

Plotting points (0, 5), (5, 0), (2, 3) and joining them, we get a line l₂ which is the graph of equation (2).

The lines l₁ and l₂ intersect at (5, 0). Thus, the solution of equations (1) and (2) is x = 5 and y = 0.
Now, it can be seen that ΔABC is formed by the two lines l₁ and l₂ and the y-axis.
The vertices of ΔABC is A(0, 5), B(5, 0) and C(0, -4).