Question

find the point on the X-axis which is equidistant from (2,-5) and (-2,9).
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Answer 1

Answer:

Let A and B are the points that are on the graph.

Let P be the point that is equidistant from A and B.

AP=BP

Step-by-step explanation:

A(2,-5) (X1,y1)

B(-2,9) (X2 Y2)

P(x,y)

By using distance formula

PA=√(x-x1)²+(y-y1)²

=√(x-2)²+(y-(-5))²

=√x²+2²-2(2)(x)+(y+5)² { by applying (a+b)²}

=√x²+4-4x+y²+(5)²+2(y)(5) {by applying (a-b)²}

=√x²+4-4x+y²+25+10y

PB=√(x-x2)²+(y-y2)²

=√(X-(-2))²+(y-9)²

=√(x+2)²+y²+9²+2(9)(y) {by applying (a-b)²}

=√(x+2)²+y²+81+18y

=√x²+4+4x+y²+81+18y

√x²-4x+4+y²+10y+25=√x²+4x+4+y²+18y+81

(PA=PB)

Squaring on both sides

x²-4x+4+y²+10y+25=x²+4x+4+y²+18y+81

x²-4x+y²+10y+29=x²+4x+y²+18y+85

x²-4x+y²+10y-x²-4x-y²-18y=85-29