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

Given $P(1, 2)=(x, y)$ and line $2x-y+3=0$

If $PA=PB$ then the mid point of AB is Foot of $\perp r$ from p to $2x-y+3=0$

$\therefore$ Let $Q\left(h, k\right)$ be the foot of $\perp r$

$$\begin{gathered} \frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{-\left(a{x}_1+b{y}_1+c\right)}{a^2+b^2} \\ \Rightarrow \frac{h-1}{2}=\frac{k-2}{-1}=\frac{-\left(2\left(1\right)-2+3\right)}{2^2+{\left(-1\right)}^2} \\ \Rightarrow \frac{h-1}{2}=\frac{k-2}{-1}=\frac{-3}{5} \end{gathered}$$

$\begin{array}{l}\frac{h-1}{2}=\frac{-3}{5},\frac{k-2}{-1}=\frac{-3}{5}\\ h-1=\frac{-6}{5},k-2=\frac{3}{5}\\ h=\frac{-6}{5}+1,k=\frac{3}{5}+2\\ h=\frac{-1}{5}\text{\hspace{0.17em},k\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{13}{5}\\ \therefore \left(h,k\right)=\left(\frac{-1}{5},\frac{13}{5}\right)\end{array}$

$\therefore$ Mid point of AB = $\left(\frac{-1}{5},\frac{13}{5}\right)$