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 \[=1:2\].
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