Question

Which point in the x - axis is equidistant from (7,6) and (-3,4) ?

Answer 1

let a point on x axis p(x,0) and q(7,6)and r(-3,4)
therefore
       pq=pr
or,   (x-7)^2+(0-6)^2=(x-*-3)^2+(0-4)^2
or,    x^2-14x+49+36=x^2+6x+9+16
or,  -20x+85=25
or,  -20x=-60
or,  x=3 
therefore, p(3,0)

Answer 2

Let the point on X axis be P(x, 0) which is equidistant from the points A(7,6) and B(–3,4)
That means, PA = PB
Using the distance between two points formula $\sqrt{(x_{2} - x_{1})^2+(y_{2} - y_{1})^2}$

⇒$\sqrt{(7 - x)^2+(6 - 0)^2} = \sqrt{(-3 - x)^2+(4 - 0)^2}$

⇒ $49 + x^2 -14x + 36 = 9 + x^2 +6x + 16$

⇒$85 -14x = 25 +6x$

⇒ 20x = 60

⇒x = 3

The point on X axis which is equidistant from the points (7,6) and (–3,4) is (3,0)