Thank you for reporting, we will resolve it shortly
Q.
If $O$ is the origin and $A$ and $B$ are points on the line $3x- 4y + 25 = 0$ such that $OA = OB = 13$. then the area of $\Delta\,OAB$ (in sq. units) is
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 $\Delta A O P]$
Area of $ \Delta O A B =\frac{1}{2} \times 24 \times 5 $
$=12 \times 5=60 $
