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Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).

2 Answers

+1 vote
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Answer:
We have to find a point on x-axis. Therefore, its y-coordinate will be 0.
Let the point on x-axis be (x,0)image
(- 2)2 + 25 = (x - 2)2 + 81
x2 + 4 - 4x + 25 = x2 + 4 + 4x + 81
8x = 25 -81
8x = -56
x = -7
Therefore, the point is ( - 7, 0).

+1 vote
by (48.0k points)

According to the question we have 

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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