Question

Draw the graph of the following pair of linear equations :
$x+3y=6$ and $2x-3y=2$
Find the ratio of the areas of the two triangles formed by first line, $x=0,y=0$ and second line, $x=0,y=0$.

Answer 1

$\begin{array}{ccc}& \text{First Line}& \text{Second Line}\\ & x+3y=6& 2x-3y=12\\ \Rightarrow & x=6-3y& \Rightarrow 2x=12+3y\\ & & \Rightarrow x=\frac{12+3y}{2}\end{array}$

First line
$\begin{array}{cccc}x& 6& 3& 0\\ y& 0& 1& 2\end{array}$

Second line
$\begin{array}{cccc}x& 6& 3& 0\\ y& 0& -2& -4\end{array}$


$Areaofatriangle=\frac{1}{2}\times$ base $\times$ corresponding altitude

$\therefore \frac{\text{Area of}\Delta AOB}{\text{Area of}\Delta AOC}=\frac{\frac{1}{2}\times OA\times OB}{\frac{1}{2}\times OA\times OC}$

$\Rightarrow \frac{OB}{OC}=\frac{2}{4}=\frac{1}{2}$

$\therefore$ Required ratio = 1:2