Question

find point on x-axis which is equidistant from p(2,-5) andQ(-2,9)

Answer 1

tP(2,-5) = (x1,y1)
Q(-2,9) =(x2,y2)
$( x = \frac{x1 + x2}{2} )(y = \frac{y1 + y2}{2} )$
$x = \frac{2 + - 2} {2}$
$x = \frac{0}{2}$
$y = \frac{ - 5 + 9}{2}$ y=\frac{4}{2} [/tex]
y=2

Answer 2

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