Question

Find a point on the X-axis which is equidistant from the point (7,6),(-3,4) which formula

Answer 1

Answer:

Point P(3,0)

Step-by-step explanation:

Let the points be A(7,6),B(-3,4)&P(x,0)

It is given that

AP=BP

By using the distance formula , we get

(7-x)^2+6^2=(-3-x)^2+4^2

i.e. 20 x =85-25

           x=60/20

       x=3

Therefore the Point P(3,0) is equidistant from the point (7,6),(-3,4) and lies on the X axis

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