The point on the line 3x + 4y = 5 which is equidistance from (1, 2) and (3, 4) is

: Let A (1, 2) and B (3, 4) and  be the point on 3x + 4y - 5 = 0

$\begin{array}{l}\Rightarrow 3\alpha +4\beta -5=0--\left(1\right)\\ GivenPA=PB\\ S.B.S\\ \Rightarrow P{A}^2=P{B}^2\end{array}$

$\begin{array}{l}\therefore {\left(\alpha -1\right)}^2+{\left(\beta -2\right)}^2={\left(\alpha -3\right)}^2+{\left(\beta -4\right)}^2\\ \Rightarrow {\alpha }^2+1-2\alpha +{\beta }^2+4-4\beta ={\alpha }^2+9-6\alpha +{\beta }^2+16-8\beta \\ \Rightarrow -2\alpha -4\beta +5+6\alpha +8\beta -25=0\\ \Rightarrow 4\alpha +4\beta -20\Rightarrow \alpha +\beta -5=0--\left(2\right)\\ Solving\left(1\right)\&\left(2\right)\\ \alpha \beta 1\\ \begin{array}{cccc}4& -5& 3& 4\\ 1& -5& 1& 1\end{array}\end{array}$

$\begin{array}{l}\Rightarrow \frac{\alpha }{-20+5}=\frac{\beta }{-5+15}=\frac{1}{3-4}\\ \Rightarrow \frac{\alpha }{-15}=\frac{\beta }{10}=\frac{1}{-1}\\ \therefore \alpha =\frac{-15}{-1},\beta =\frac{10}{-1}\\ =15,=-10\\ \therefore P\left(\alpha ,\beta \right)=\left(15,-10\right)\end{array}$