Question

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

Answer 1

Let the point on the line 3x+y+4=0 equidistant from points A (-5,6)and B (3,2) be C (x,y)
AC = BC..........(i)
Ac =
$\sqrt{(x + 5) ^{2} + (y - 6) ^{2} }$
Also, BC =
$\sqrt{(x - 3) ^{2} + (y - 2) ^{2} }$

(x+5)^2+(y-6)^2 = (x-3)^2 + y-2)^2
or, x^2+10x+25+y^2-12y+36=x^2-6x+9+y^2-4y+4
or, 10x +61 - 12y = -6x-4y+13
or, 16x-8y=-48
or, 2x -y = -6
y = 2x+6..........

now, putting the value of y in the equation,
3x+y+4=0
or, 3x+2x+6+4=0
x = -10/5
x = -2#
then,
y =2 (-2)+6
= 2 #
Therefore,
the point is, C (-2,2),,

Hope u got the answer