Question

Find the area of the triangle whose sides are along the lines $x=0,y=0$ and $4x+5y=20$.

Answer 1

Given the equation of the straight line is $4x+5y=20$ can be written as $\frac{x}{5}+\frac{y}{4}=1$.

From this it is clear that the $x-$ and $y-$ intercepts are $5$ and $4$ units respectively.

So the area of the triangle formed by the $x-$, $y-$ and the given line is

$=\frac{1}{2}\times 4\times 5$

$=10$ sq. units.