Find the points 'A' on the x-axis with distance from (8,8) to A is twice as far the distance from (-1,2) to A

Answer 1

A=(x,0),
d(A, (8,8))= 2 d(A,(-1,2)

then
sqr( (8-x)^2+8^2 )= 2 sqr ( (-1-x)^2 + 2^2 )
squaring both sides we get:
(8-x)^2+8^2 = 4 ( (-1-x)^2 + 2^2 )
64 -16x + x^2 + 64 = 4 ( 1+2x +x^2 + 4)
128 -16x + x^2 = 4 +8x +4x^2 + 16
-3x^2 -24x +108=0
dividing by -3, we get:
x^2 + 8x - 36 =0
then
x=(-8+14.42)/2=3.21
x=(-8-14.42)/2=-11.21

Answer 2

Since the point is on the x axis it is of the form (h,0)
Use the distance formula to solve for h.
sqrt[(h-8)^2+8^2]=2sqrt[(h+1)^2+2^2]