If $O$ is the origin and $A$ and $B$ are points on the line $3 x-4 y+25=0$ such that $\mathbf{O A}=\mathbf{O…
- 30
- 120
- 60
- 65
Solution

Required distance $O P=\left|\frac{0+0+25}{\sqrt{3^2+4^2}}\right|=\left|\frac{25}{5}\right|=5$ So, $A P=P B=12$ [By Pythagoras theorem in $\triangle A O P$ ] Area of $ \begin{aligned} \triangle O A B & =\frac{1}{2} \times 24 \times 5 \\ & =12 \times 5=60 \end{aligned} $
Asked in: AP EAMCET 2019 (21 Apr Shift 1)