Question

Solve the following system of equations graphically:
$4x-5y+16=0,2x+y-6=0.$
Determine the vertices of the triangle formed by these lines and the x-axis.

Answer 1

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 4x − 5y + 16 = 0

4x − 5y + 16 = 0
⇒ 5y = (4x +16)
∴ $y=\frac{4x+16}{5}$ ...........(i)
Putting x = 1, we get y = 4.
Putting x = −4, we get y = 0.
Putting x = 6, we get y = 8.
Thus, we have the following table for the equation 4x − 5y + 16 = 0.

x	1	−4	6
y	4	0	8

Now, plot the points A(1, 4), B(−4, 0) and C(6, 8) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 4x − 5y + 16 = 0.

Graph of 2x + y − 6 = 0
2x + y − 6 = 0
⇒ y = (−2x + 6) ...........(ii)
Putting x = 1, we get y = 4.
Putting x = 3, we get y = 0.
Putting x = 2, we get y = 2.
Thus, we have the following table for the equation 2x + y − 6 = 0.

x	1	3	2
y	4	0	2

Now, plot the points P(3, 0) and Q(2, 2). The point A(1, 4) has already been plotted. Join PQ and QA to get the graph line PA. Extend it on both ways.
Then, PA is the graph of the equation 2x + y − 6 = 0.

The two graph lines intersect at A(1, 4).
∴ The solution of the given system of equations is x = 1 and y = 4.
Hence, these lines form ΔBAP with the x-axis, whose vertices are B(−4, 0), A(1, 4) and P(3, 0).