Question

Graphically , solve the following pair of equations:

$$\begin{gathered} 2x+y=6 \\ 2x-y+2=0 \end{gathered}$$

Find the ratio of the areas of the two triangles formed by the lines representing these equations with the x-axis and the lines with the y-axis.

Answer 1

The given linear equations are:
$$\begin{gathered} 2x+y=6.....\left(\mathrm{i}\right) \\ 2x-y+2=0.....\left(\mathrm{ii}\right) \end{gathered}$$
For (i), we have

x	0	3
y	6	0

For (ii), we have

x	0	−1
y	2	0

Thus, we plot the graph for these two equations and mark the point where these two lines intersect.


From the graph we see that the two lines intersect at point E(1, 4).
Now, the area of triangle CEB is
$A_1=\frac{1}{2}\times 4\times 4=8\mathrm{square}\mathrm{unit}$
The area of triangle AED is
$A_2=\frac{1}{2}\times 4\times 1=2\mathrm{square}\mathrm{unit}$
So, the ratio of the areas of the two triangles will be
$\frac{A_1}{A_2}=\frac{8}{2}=\frac{4}{1}$
Thus, the required ratio is 4 : 1.