Question

A point $P$ is taken on ${}^{\prime }{L}^{\prime }$ such that $\frac{2}{OP}=\frac{1}{OA}+\frac{1}{OB}.$
Then the locus of $P$ is :

• A. $3x+3y=40$
• B. $3x+3y+40=0$
• C. $3x-3y=40$
• D. $3y-3x=40$

Answer 1

The correct option is D $3y-3x=40$

Let the variable line be $y=mx$
Therefore the coordinates of point p is $(h,mh)$
Therefore A being the coordinates of the intersection of $y=mx$ with $y-x-10=0$ implies
$A=(\frac{10}{m-1},\frac{10m}{m-1})$
Similarly $B$ being the coordinates of the intersection of $y=mx$ with $y-x-20=0$ implies
$B=(\frac{20}{m-1},\frac{20m}{m-1})$
Hence, $OP=h\sqrt{1+m^2}$
$\Rightarrow OA=\frac{10\sqrt{1+m^2}}{m-1}$
$\Rightarrow OB=\frac{20\sqrt{1+m^2}}{m-1}$
Since $\frac{2}{OP}=\frac{1}{OA}+\frac{1}{OB}$
Hence, $\frac{2}{h\left(\sqrt{1+m^2}\right)}=\frac{m-1}{10\left(\sqrt{1+m^2}\right)}+\frac{m-1}{20\left(\sqrt{1+m^2}\right)}$
$\Rightarrow \frac{2}{h}=\frac{3(m-1)}{2}$
$\Rightarrow 4=3mh-3h$
$\Rightarrow 3\left(y\right)-3\left(x\right)=40$
$\Rightarrow 3x-3y+40=0$