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 5x - y = 7
5x - y = 7
⇒ y = (5x – 7) …(i)
Putting x = 0, we get y = -7.
Putting x = 1, we get y = -2.
Putting x = 2, we get y = 3.
Thus, we have the following table for the equation 5x - y = 7.
Now, plot the points A(0, -7), B(1, -2) and C(2, 3) 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 5x - y = 7.
Graph of x - y + 1 = 0
x - y + 1 = 0
⇒ y = x + 1 ……..(ii)
Putting x = 0, we get y = 1.
Putting x = 1, we get y = 2.
Putting x = 2, we get y = 3.
Thus, we have the following table for the equation x - y + 1 = 0.
Now, plot the points P(0, 1) and Q(1, 2). The point C(2, 3) has already been plotted.
Join PQ and QC to get the graph line PC. Extend it on both ways.
Then, PC is the graph of the equation x - y + 1 = 0.
The two graph lines intersect at C(2, 3).
∴ The solution of the given system of equations is x = 2 and y = 3.
Clearly, the vertices of ∆APC formed by these two lines and the y-axis are P(0, 1), C(2, 3) and A(0, -7).
Now, consider ∆APC.
Here, height = 2 units and base (AP) = 8 units
∴ Area ∆APC = \(\frac{1}2\) × base × height sq. units
= \(\frac{1}2\) × 8 × 2
= 8 sq. units.
