Find the point on the line 3x + y + 4 = 0 which is equidistant to the points (–5, 6) and (3, 2).

Given A = (–5, 6), B = (3, 2)
$\mathrm{L}\equiv 3\mathrm{x}+\mathrm{y}+4=0\dots \left(1\right)$
P(x, y) be a point on the line L = 0
$$\begin{gathered} \text{ gvien }\mathrm{PA}=\mathrm{PB} \\ \text{S.O.B.S } \\ {\mathrm{PA}}^2={\mathrm{PB}}^2 \end{gathered}$$
$\begin{array}{l}(\mathrm{x}+5{)}^2+(\mathrm{y}-6{)}^2=(\mathrm{x}-3{)}^2+(\mathrm{y}-2{)}^2\\ {\mathrm{x}}^2+25+10\mathrm{x}+{\mathrm{y}}^2+36-12\mathrm{y}\\ =x^2+9-6\mathrm{x}+{\mathrm{y}}^2+4-4\mathrm{y}\\ 16\mathrm{x}-8\mathrm{y}+48=0\\ 2\mathrm{x}-\mathrm{y}+6=0\dots \left(2\right)\end{array}$
Solving equation (1) & (2)
$\begin{array}{llll} \mathrm{x}& & \mathrm{y} 1& \\ 1& 4& 3& 1\\ -1 & 6& 2& -1\end{array}$
$\begin{array}{l}\frac{\mathrm{x}}{6+4}=\frac{\mathrm{y}}{8-18}=\frac{1}{-3-2}\\ \frac{\mathrm{x}}{10}=\frac{\mathrm{y}}{-10}=\frac{1}{-5}\\ \Rightarrow \frac{\mathrm{x}}{10}=\frac{1}{-5}\\ \Rightarrow \mathrm{x}=\frac{10}{5}=-2\\ \frac{\mathrm{y}}{-10}=\frac{1}{-5}\Rightarrow \mathrm{y}=\frac{10}{5}=2\\ \therefore (\mathrm{x},\mathrm{y})=(-2,2)\end{array}$