Question

find the point on the x axis which is equidistace from (2-5) (-2 9)​

Answer 1

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).