Question

In the given figure OQ is the perpendicular bisector of RS and UP is perpendicular to OQ. To prove $\frac{1}{OP}+\frac{1}{OQ}=?$

• A. $\frac{1}{OT}$
• B. $\frac{2}{OT}$
• C. $\frac{3}{OT}$
• D. $\frac{4}{OT}$

Answer 1

The correct option is B $\frac{2}{OT}$

In $△POU$ and $△QOR$

$\angle O=\angle O$ (same angle)

$\angle P=\angle Q$ [each ${90}^o$]

$\therefore △POU\sim △QOR...$ ($AA$ test of similarity)

$\therefore \frac{OP}{OQ}=\frac{PV}{QR}---\left(1\right)$

Also, in $△UPT$ and $△SQT$

$\angle P=\angle Q...$ {each ${90}^o$}

$\angle UTP=\angle STQ....$ (vertically opposite angles)

$\therefore △UPT\sim △STQ...$ ($AA$ test of similarity)

$\therefore \frac{UP}{SQ}=\frac{PT}{QT}$

but $SQ=RQ...$ ($Q$ is mid point)

$\therefore \frac{UP}{RQ}=\frac{PT}{QT}.....\left(2\right)$

$\therefore$ from $\left(1\right)$ and $\left(2\right)$

$\frac{PT}{QT}=\frac{OP}{OQ}$

$\frac{OT-OP}{OQ-OT}=\frac{OP}{OQ}$

$\therefore OT.OQ-OQ.OP=OP.OQ-OP.OT$

$\therefore \left(OQ\right).\left(OT\right)+\left(OP\right)\left(OT\right)=2\left(OP\right).\left(OQ\right)$

$\therefore OT(OQ+OP)=2\left(OP\right).\left(OQ\right)$

$\therefore \frac{OQ+OP}{\left(OP\right).\left(OQ\right)}=\frac{2}{OT}$

$\therefore \frac{1}{OP}+\frac{1}{OQ}=\frac{2}{OT}$