If O is the origin and A and B are points on | vedclass

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(C) Let $P$ be the foot of the perpendicular from the origin $O(0, 0)$ to the line $3x - 4y + 25 = 0$.The perpendicular distance $OP$ is given by:$OP

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) Let $P$ be the foot of the perpendicular from the origin $O(0, 0)$ to the line $3x - 4y + 25 = 0$.The perpendicular distance $OP$ is given by:$OP = \left| \frac{3(0) - 4(0) + 25}{\sqrt{3^2 + (-4)^2}} ight| = \left| \frac{25}{\sqrt{9 + 16}} ight| = \left| \frac{25}{5} ight| = 5$.In the right-angled triangle $	riangle OAP$,by the Pythagoras theorem:$AP^2 + OP^2 = OA^2$$AP^2 + 5^2 = 13^2$$AP^2 = 169 - 25 = 144$$AP = 12$.Since $OP \perp AB$,$P$ is the midpoint of $AB$,so $AB = 2 	imes AP = 2 	imes 12 = 24$.The area of $	riangle OAB = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes AB 	imes OP = \frac{1}{2} 	imes 24 	imes 5 = 60$ sq units.

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 $\triangle OAB$ (in sq units) is

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