Question

find the point on xaxis which is eqidistant from the points (-2,5) and (-2,9)

Answer 1

let the point be Q(x,y) at equidistant from P(-2,5) R(-2,9)
[tex]( x+2) ^{2} +(y-5) ^{2}= (x+2) ^{2} +(y-9) ^{2} \\ x^{2} +4x+4+ y^{2}-10y+25= x^{2} +4x+4+ y^{2}-18y+81 \\ -10y+18y+25-81=0 \\ 8y=56 \\ y=7 \\ x=0 [/tex]

Answer 2

x1 + x2   ,     y1  +  y2
     2                    2

-2+(-2)      ,   5+9
    2                 2

0/2   ,14/2

(0,7)

7 is the  point on the x axis which is equidistant from the points (-2,5) and (-2,9)