Geo2.32

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

Let the point be (a,b)

Given it is equidistant from the two points,so

$\sqrt{(a+5)^2+(b-6)^2}=\sqrt{(a-3)^2+(b-2)^2}\,$

$a^2+25+10a+b^2+36-12b=a^2+9-6a+b^2+4-4b\,$

Simplifying we get,

$16a-8b=-48\,$

$2a-b=-6\,$

Since the point lies on the given equation,

$3a+b=-4\,$

Solving these two equations,

$a=-2,b=2\,$

Hence the point is (-2,2)