Find the point equidistant from (-6, -1) and (-1, 2) and at a distance 5 units from (-2, 7).

Question

Thank you for those who will help.

Answer 1

(-6,-1) and (-1,2) have a point half way between them at (-7/2, 1/2)

A line through those points has slope = (2+1)/(-1+6) =3/5

A perpendicular line has slope -5/3

y-1/2 = -5/3(x+7/2)

y= -5x/3 -35/6+3/6= -5x/3 -32/6

y=-5x/3 -16/3 is the equation of all points equidistant between (-6,-1) and (-1,2)

(-2.7) is 5 units from (-6,-1) and (-1,2)

all points 5 units from (-2,7) are on a circle

(x+2)^2 + (y-7)^2 = 25

find where the perpendicular line and the circle intersect

x^2 + 4x +4 + (-5x/3 -16/3 -7)^2 =25

x^2 +4x +(5x/3 +37/3)^2 = 21

9x^2 +36x + 25x^2 +370x + 1369= 189

34x^2 +406x +1180 =0

17x^2 +203x + 590 = 0

x=-203/34 + or - (1/34)sqr(203^2 -4(17)(590))

x=-203/34 +33/34 = -170/34 = -5

y=-5(-5)/3 -16/3 = 9/3 = 3

(-5,3) is one point 5 units from (-2,7) and equidistant from (-6,-1) and (-1,2)

5^2 =(-5+2)^2 + (7-3)^2 = 3^2 +4^2= 25

1^2 +4^2 = 4^2+1^2 =17

(-5,3) is sqr17 from (-6,1) and (-1,2)

the line intersects the circle in two points, so there is a 2nd point also 5 units from (-2,7) and equidistant to the 2 points.

It has a smaller x coordinate and larger y coordinate.

It may help to plot the points, even with a rough sketch.