Find the coordinates of the point on x-axis which is equidistant from the points (–2, 5) and (2, –3).
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Let the point on the x−axis be P(a,0)
Let the other two point be A(−2,5) and B(2,−3)
So by distance formula we have,
Distance between two points =
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
Then,
PA=PB
PA
2
=PB
2
⇒(a+2)
2
+(0−5)
2
=(a−2)
2
+(0+3)
2
⇒a
2
+4+4a+25=a
2
+4−4a+9
⇒4a+29=−4a+13
⇒ 8a=−16⇒a=−2
∴ The required point isP(−2,0)
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