Find the point (x,y) on the x-axis that is equidistant from the points (−9,−1) and (8,0)

Question

Please help, my answer was wrong

Answer 1

(x-9)^2 + 1 = (x+8)^2

x^2 -18x + 82 = x^2 +16x + 64

34x = 18

x = 18/34 = 9/17

the equidistant point is (-9/17, 0) on the x axis

it's a distance of 8 9/17 = 145/17 = about 8.53 from (8,0)

it's a distance of sqr[(9- 9/17)^2 +1^2] from (-9,-1)

= sqr[(8 8/17)^2 +1]

= sqr((144/17)^2 +1)

= sqr(21025/17^2)

= 145/17

= 8 9/17 = about 8.53 from (-9,-1)

If you plot the two points, the point near (-1/2, 0) is very close to equidistant from the two points, about 8 1/2 from each point.

-9/17 is very close to -1/2

the y coordinate has to = 0, to be on the x axis