Question

Suppose A, B are two points on $2x-y+3=0$ and $P(1,2)$ is such that $PA=PB$. Then the midpoint of $AB$ is

• A. $(\frac{-1}{5},\frac{13}{5})$
• B. $(\frac{-7}{5},\frac{9}{5})$
• C. $(\frac{7}{5},\frac{-9}{5})$
• D. $(\frac{-7}{5},\frac{-9}{5})$

Answer 1

The correct option is A $(\frac{-1}{5},\frac{13}{5})$

The points of line $2x-y+3=0$ can be of the form

$y=2x+3$

Now $P$ is not lying on line and $PA=PB$

so mid point of $AB$ will be perpendicular from $P$ to line $2x-y+3=0$

$y=2x+3$ ...$\left(1\right)$

compare with $y=mx+c$

so slope of this line $m=2$

so slope of line perpendicular to it $=\frac{-1}{m}=\frac{-1}{2}$

Now equation of line passing through $(1,2)$ and having slope $\frac{-1}{2}$ is , $y-y_1=m(x-x_1)$

$y-2=\frac{-1}{2}(x-1)$

$2y-4=-x+1$

$x+2y=5$ ......$\left(2\right)$

put $y$ from equation $\left(1\right)$ to $\left(2\right)$

$x+4x+6=5$

$5x=-1$

$x=\frac{-1}{5}$

So, $y=2x+3=\frac{-2}{5}+3=\frac{13}{5}$

So coordinate of mid point is $=(\frac{-1}{5},\frac{13}{5})$