Question

Solve the following system of linear equations graphically:
$x-y+1=0,3x+2y-12=0.$
Calculate the area bounded 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 x − y + 1 = 0
x − y + 1 = 0
⇒ y = (x + 1) .........(i)
Putting x = −1, we get y = 0.
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.

x	−1	1	2
y	0	2	3

Now, plot the points A(-1, 0), 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 x − y + 1 = 0.

Graph of 3x + 2y − 12 = 0
3x + 2y − 12 = 0
⇒ 2y = (−3x + 12)
∴ $y=\frac{-3x+12}{2}$ ............(ii)
Putting x = 0, we get y = 6.
Putting x = 2, we get y = 3.
Putting x = 4, we get y = 0.
Thus, we have the following table for the equation 3x + 2y − 12 = 0.

x	0	2	4
y	6	3	0

Now, plot the points P(0, 6) and Q(4 , 0). The point C(2, 3) has already been plotted. Join PC and CQ to get the graph line PQ. Extend it on both ways.
Then, PQ is the graph of the equation 3x + 2y − 12 = 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 ΔACQ formed by these two lines and the x-axis are Q(4, 0), C(2, 3) and A(−1, 0).
Now, consider ΔACQ.
Here, height = 3 units and base (AQ) = 5 units
Area of ΔACQ = $\frac{1}{2}\times \mathrm{base}\times \mathrm{height}$ sq. units
= $\left(\frac{1}{2}\times 5\times 3\right)=7.5$ sq. units.