question_answer

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, \[x=0,\text{ }y=0\] and second line, \[x=0,\text{ }y=0\].

Answer:

First Line
\[x+3y=6\]
\[\Rightarrow x=6-3y\]
x	6	3	0
y	0	1	2
\[\left( 6,\text{ }0 \right),\text{ }\left( 3,\text{ }1 \right),\text{ }\left( 0,\text{ }2 \right)\]
Second Line
\[2x-3y=12\]
\[\Rightarrow \,2x=12+3y\]
\[\Rightarrow \,x=\frac{12+3y}{2}\]
x	6	3	0
y	0	\[2\]	\[~-4\]
\[\left( 6,\text{ }0 \right),\text{ }\left( 3,\text{ }\text{ }2 \right),\text{ }\left( 0,\text{ }\text{ }4 \right)\]
Area of triangle \[=\frac{1}{2}\times \text{base}\times \text{corresponding altitude}\]
\[\therefore \frac{\text{Area}\,\,\text{of}\,\,\Delta \,AOB}{\text{Area}\,\,\text{of}\,\,\Delta \,AOC}=\frac{1/2\times OA\times OB}{1/2\times OA\times OC}\]
\[\Rightarrow \frac{OB}{OC}=\frac{2}{4}=\frac{1}{2}\]
\[\therefore \]  Required ratio \[=1:2\].