Q. Suppose $A,\ B$ are two points on $2x - y + 3 = 0$ and $P(1,\ 2),$ is such that $PA = PB$ Then, the mid point of $AB$ is
A) $\left( - \frac{1}{5},\frac{13}{5} \right)$
B) $\left( - \frac{7}{5},\frac{9}{5} \right)$
C) $\left( \frac{7}{5},\frac{- 9}{5} \right)$
D) $\left( \frac{- 7}{5},\frac{- 9}{5} \right)$
Explanation
(a)
Since, $AP = BP$ and $PM$ is perpendicular to the line
$2x - y + 3 = 0$ …(i)
Where, $M$ is the mid point $AB$
[image:10_image30.emf]
$Therefore,\ its\ slope\ is\ \left( - \frac{1}{2} \right)$
$\therefore Equation\ of\ line\ PM\ is\ y - 2 = - \frac{1}{2}(x - 1)$
$\Rightarrow \ 2y + x - 5 = 0\$ ….(ii)
Solving Eqs. (i) and (ii), we get the mid point of $AB$ is
$M\left( - \frac{1}{5},\frac{13}{5} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.