$\text{ The point on the line }3\mathrm{x}+4\mathrm{y}=5 \text{ which is equidistant from }(1,2)\text{ and (3,4) is}$

$\begin{array}{l}\mathrm{PA}=\mathrm{PB}\\ (\mathrm{x}-1{)}^2+(\mathrm{y}-2{)}^2=(\mathrm{x}-3{)}^2+(\mathrm{y}-4{)}^2\\ \mathrm{x}+\mathrm{y}-5=0\to \left(1\right)\\ \text{ Given }3\mathrm{x}+4\mathrm{y}-5=0\to \left(2\right)\\ \mathrm{Solve} 1 \mathrm{and} 2 \mathrm{we} \mathrm{get} \\ \mathrm{Solve} \left(1\right) \mathrm{and} \left(2\right) \mathrm{we} \mathrm{get}\\ \begin{array}{l}\mathrm{x}=15,\mathrm{y}=-10\\ (15,-10)\end{array}\end{array}$