On a graph paper, draw a horizontal line X'OX and vertical line YOY' as the x- axis and the y-axis respectively.
Graph of ` 4 x Z- 5y - 20=0`
` 4x - 5y - 20 = 0 rArr 5y = (4 x - 20 )`
` " "rArr y = (( 4 x - 20 ))/( 5) " " ` ... (i)
Putting ` x = 0 ` in (i), we get `y = -4 `
Putting ` x = 2 ` in (i), we get `y = -2.4`
Putting ` x = 5` in (i), we get `y = 0`
Thus, we have the following table for ` 4x - 5y -20 = 0`
Now, plot the points ` A(0, -4), B(2, -2.4)and C(5, 0)` on the graph paper.
Join AB and BC to get the graph line ABC. Extend it on both ways.
Thus, the line ABC is the graph of ` 4x - 5y-20 = 0`
Graph of ` 3x + 5y - 15 = 0`
` 3x + 5y - 15 = 0 rArr 5y = ( 15 - 3x )" "`... (ii)
Putting ` x = -5 ` in (ii), we get ` y = 6 `.
Putting ` x = 0 `in (ii), we get ` y = 3 `
Putting ` x = 5 `in (ii), we get `y = 0 `
Thus, we have the following table for ` 3x + 5y - 15 = 0`
On the same graph paper as above, plot the points `P(-5, 6) and Q(0, 3)`. The third point `C(5, 0)` has already been plotted.
Join PQ and QC to get the graph line PQC. Extend it on both ways.
Thus, the line ` PQC ` is the graph of ` 3x + 5y - 15=0 `.
The two graph lines intersect at the point `C(5, 0)`
` therefore x = 5, y = 0` is the solution of the given system of equations.
Clearly, the given equations are represented by the graph lines ABC and PQC respectively.
The verticles of ` DeltaAQC` formed by these lines and the y-axis are ` A(0,-4), Q (0, 3) and C(5, 0)`
