Question

Find the point on the x-axis equidistant from(2,-5)and(-2,5)
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Answer 1

Answer:

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).

Answer 2

Answer:

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).