The point on the line 4x – y – 2 = 0 which is equidistant from the points ( -5, 6)and (3, 2) is

 $\begin{array}{l}Letp\left(\alpha ,\beta \right)bethepo\mathrm{int}on4x-y-2=0\\ \Rightarrow 4\alpha -\beta -2=0--\left(i\right)\end{array}$

And given points A ( -5, 6) B (3 ,  2)

Given  PA=PB

$\begin{array}{l}\Rightarrow P{A}^2=P{B}^2\\ \Rightarrow {\left(\alpha +5\right)}^2+{\left(\beta -6\right)}^2={\left(\alpha -3\right)}^2+{\left(\beta -2\right)}^2\\ \Rightarrow {\alpha }^2+25+10\alpha +{\beta }^2+36-12\beta ={\alpha }^2+9-6\alpha +{\beta }^2+4-4\beta \\ \Rightarrow 16\alpha -8\beta +48=0\\ \Rightarrow 2\alpha -\beta +6=0--\left(2\right)\\ \left(1\right)\to \left(2\right)\\ 2\alpha -8=0\Rightarrow \alpha =\raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=4\\ From\left(1\right)\beta =4\alpha -2=16-2=14\\ \therefore \left(\alpha ,\beta \right)=\left(4,14\right)\end{array}$