Question

Solve graphically the system of equations
$4x-5y-20=0,3x+5y-15=0.$
Find the coordinates of the vertices of the triangle formed by these two lines and the y-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 − 20 = 0
4x − 5y − 20 = 0
⇒ 5y = (4x − 20)
∴ $y=\frac{4x-20}{5}$............(i)
Putting x = 0, we get y = −4.
Putting x = 5, we get y = 0.
Putting x = 10, we get y = 4.
Thus, we have the following table for the equation 4x − 5y − 20 = 0.

x	0	5	10
y	−4	0	4

Now, plot the points A(0, −4), B( 5, 0) and C(10, 4) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of 4x − 5y − 20 = 0.

Graph of 3x + 5y − 15 = 0
3x + 5y − 15 = 0
⇒ 5y = (−3x + 15)
∴ $y=\frac{-3x+15}{5}$ ............(ii)
Putting x = 5, we get y = 0.
Putting x = 0, we get y = 3
Putting x = −5, we get y = 6.
Thus, we have the following table for the equation 3x + 5y − 15 = 0.

x	5	0	−5
y	0	3	6

Now, plot the points P(0, 3) and Q(−5 , 6). The point B(5, 0) has already been plotted. Join PQ and PB to get the graph line QB. Extend it on both ways.
Then, QB is the graph of the equation 3x + 5y − 15 = 0.

The two graph lines intersect at B(5, 0).
∴ The solution of the given system of equations is x = 5 and y = 0.
Clearly, the vertices of ΔPBA formed by these two lines and y-axis are P(0, 3), B(5, 0) and A(0, −4).