Question

If $P.Q$ are two points on the line $3x+4y+15=0$ such that $OP=OQ=9$, then the area of $\Delta OPQ$ is

• A. $6\sqrt{2}$
• B. $9\sqrt{2}$
• C. $12\sqrt{2}$
• D. $18\sqrt{2}$

Answer 1

Answer: (D)

$18\sqrt{2}$

$OC=$ Length of the perpendicular from $(0,0)$ to the line $3x+4y+15=0$

$⟹OC=\frac{|0+0+15|}{\sqrt{3^2+4^2}}=\frac{15}{\sqrt{9+16}}=\frac{15}{5}=3$

In $△OPQ$,

$OP=OQ$

Hence the triangle is isoceles.

$⟹OC$ bisects base $PQ$. $C$ is the mid point of $PQ$

In right angle $△OCQ$

$O{Q}^2=O{C}^2+C{Q}^2$

$⟹C{Q}^2=9^2-3^2$

$⟹CQ=\sqrt{72}=6\sqrt{2}$

As $C$ is the mid point of $PQ$

$⟹PQ=2\times 6\sqrt{2}=12\sqrt{2}$

$Area=\frac{1}{2}\times base\times height=\frac{1}{2}\times PQ\times OC$

$=\frac{1}{2}\times 12\sqrt{2}\times 3$

$=18\sqrt{2}(unit{)}^2$ ------Option D