Question

Solve the following system of equations graphically and find the vertices and area of the
triangle formed by these lines and the x-axis:
$$\begin{gathered} x-y+3=0 \\ 2x+3y-4=0 \end{gathered}$$

Answer 1

From the first equation, write y in terms of x
$y=x+3.....\left(i\right)$
Substitute different values of x in (i) to get different values of y
$$\begin{gathered} \mathrm{For}x=-3,y=-3+3=0 \\ \mathrm{For}x=-1,y=-1+3=2 \\ \mathrm{For}x=1,y=1+3=4 \end{gathered}$$
Thus, the table for the first equation (x − y + 3 = 0) is

x	−3	−1	1
y	0	2	4

Now, plot the points A(−3,0), B(−1,2) and C(1,4) on a graph paper and join
A, B and C to get the graph of x − y + 3 = 0.
From the second equation, write y in terms of x
$y=\frac{4-2x}{3}.....\left(ii\right)$
Now, substitute different values of x in (ii) to get different values of y
$$\begin{gathered} \mathrm{For}x=-4,y=\frac{4+8}{3}=4 \\ \mathrm{For}x=-1,y=\frac{4+2}{3}=2 \\ \mathrm{For}x=2,y=\frac{4-4}{3}=0 \end{gathered}$$
So, the table for the second equation ( 2x + 3y − 4 = 0 ) is

x	−4	−1	2
y	4	2	0

Now, plot the points D(−4,4), E(−1,2) and F(2,0) on the same graph paper and join
D, E and F to get the graph of 2x − 3y − 4 = 0.



From the graph it is clear that, the given lines intersect at (−1,2).
So, the solution of the given system of equation is (−1,2).
The vertices of the triangle formed by the given lines and the x-axis are (−3,0), (−1,2) and (2,0).
Now, draw a perpendicular from the intersection point E on the x-axis. So,
$$\begin{gathered} \mathrm{Area}\left(∆EAF\right)=\frac{1}{2}\times AF\mathit{\times }EM \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}=\frac{1}{2}\times 5\mathit{\times }\mathit{2} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}=5\mathrm{sq}.\mathrm{units} \end{gathered}$$
Hence, the vertices of the triangle formed by the given lines and the x-axis are (−3,0), (−1,2)
and (2,0) and its area is 5 sq. units.