find the point on the x axis which is equidistace from (2-5) (-2 9)
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Step-by-step explanation:
Given points A(2,−5) and B(−2,9)
Let the points be P(x,0).
So, AP=PB and AP
2
=PB
2
⇒(x−2)
2
+(0+5)
2
=(x+2)
2
+(0−9)
2
⇒x
2
+4−4x+25=x
2
+4+4x+81
⇒x
2
+29−4x=x
2
+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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