Find the point on the x- …

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Yogita Ingle 4 years, 10 months ago
Given points A(2,−5) and B(−2,9)
Let the points be P(x,0).
So, AP=PB and AP2=PB2
⇒(x−2)2+(0+5)2=(x+2)2+(0−9)2
⇒x2+4−4x+25=x2+4+4x+81
⇒x2+29−4x=x2+85+4x
⇒−4x−4x=85−29
⇒−8x=56
⇒x=−7
Hence, point on the x-axis which is equidistant from (2,−5) and (−2,9) is (−7,0).
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