Question

A point is equidistant from (-2,4) and (3,5). It is also equidistant from (-2,-3) and (6,-1). Find the point



Please I need Help with this

Answer 1

The point must be on the right-bisector of the line segment from (-2,4) to (3,5)

slope of line segment is 1/5, so slope of right-bisector is -5
midpoint of line segment is (1/2 , 9/2)

equation of right-bisector of line segment is
y - 9/2 = -5(x - 1/2)
2y - 9 = -10x + 5
10x + 2y = 14
5x + y = 7 or y = -5x+7

2nd line segment:
slope of line segment = 2/8 = 1/4
slope of right-bisector = -4
midpoint = (2, -2)
equation of right-bisector
y+2 = -4(x-2)
y +2 = -4x+8
y = -4x + 6

then:
-4x+6 = -5x+7
x = 1
which makes y = -4(1)+6 = 2

the point is (1,2)