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

Let the point be $(x,y)$ as it lies on $3x+y+4=0$

$\Rightarrow y=-3x-4$

$\Rightarrow$ Point $=(x,-3x-4)$

Distance of this is equidistant from $(-5,6)$ and $(3,2)$

$\Rightarrow \sqrt{(x+5{)}^2+(6+3x+4{)}^2}=\sqrt{(x-3{)}^2+(2+3x+4{)}^2}$

$\Rightarrow (x+5{)}^2+(3x+10{)}^2=(x-3{)}^2+(3x+6{)}^2$

$\Rightarrow x^2+25+10x+9x^2+100+60x=x^2+9-6x+9x^2+36-36x$

$\Rightarrow 70x+125=30x+45$

$\Rightarrow 40x=-80$

$\Rightarrow x=-2$

$\Rightarrow y=-3(-2)-4=2$

$\Rightarrow$ Point $=(-2,2)$.