find the point equidistant fr. (-6,-1) & (-1,2) & at a distance 5 fr. (-2,7)?

((-6-1)/2, (2-1)/2)

= (-7/2,1/2).

The gradient of line AB is:

(2 – (-1)) / (-1 -(-6))

= 3 / 5.

The gradient of the perpendicular bisector of AB is therefore:

-5/3.

The equation of the perpendicular bisector is:

y – 1/2 = (-5/3)(x – (-7/2))

y – 1/2 = -5x/3 – 35/6

y = -5x/3 – 16/3 ……….(1)

The distance of a point (x,y) from (-2,7) is:

sqrt((x – (-2))^2 + (y – 7)^2)

= sqrt(x + 2)^2 + (y – 7))^2.

The requirement for the distance to be 5 is:

(x + 2)^2 + (y – 7)^2 = 25 ………(2)

The point must also satisfy (1).

Substitute for y from (1) in (2):

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

(x + 2)^2 + ( -5x/3 – 37/3 )^2 = 25

9(x + 2)^2 + ( -5x – 37 )^2 = 225

9x^2 + 36x + 18 + 25x^2 + 1369 + 370x = 225

34x^2 + 406x + 1162 = 0

17x^2 + 203x + 581 = 0

x = ( -203 +/- sqrt(203^2 – 4*17*581 ) / 34

= ( – 203 +/- sqrt(1701) ) / 34

Substituting these two values of x in turn in (1) gives the corresponding y co-ordinates:

y = -5(( – 203 +/- sqrt(1701) ) / 34 ) / 3 – 16/3

= -(5/102)( – 203 +/- sqrt(1701) ) – 16/3

= -(1/102)( – 1559 +/- 5sqrt(1701) ).