Question

Find area of triangle between the coordination axis of line $3x-4y-12=0$

Answer 1

Area of triangle between the coordinate axis of line $3x-4y-12=0$ is

$=\left|\frac{c^2}{ab}\right|$ sq.units

$=\left|\frac{(-12{)}^2}{\left(3\right)(-4)}\right|$

$=\left|\frac{144}{-12}\right|$

$=12$ sq.units