Answers

2016-01-18T13:40:42+05:30
Let the point be Q(x,y) at equidistant from P(-2,5) R(-2,9)
( 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

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answer is -7 and we have to find the value of x-axis
2016-01-18T16:24:35+05:30
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)
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