Question

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

Answer 1

The area of the triangle between the co-ordinate axis of line $ax+by+c=0$ is $\left|\frac{c^2}{ab}\right|$ sq.units.

The area of the triangle between the co-ordinate axis of line $3x-4y-12=0$ is

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

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

$=12$ sq.units. (Ans)