(Q34)Find the point on the X-axis which is equidistant form (2,-5) and (-2,9).

Answer

Given points A(2,-5),B(-2,9)

Let P(x,0) be the point on X-axis

which is equidistant from A and B i.e. PA=PB

Distance formula = √((x2 - x1)2 + (y2 - y1)2)

PA=√((-x)2 + (--0)2)

=√x2-√4x + +

=√x2-√4x +

PB=√(--x)2 + (-0)2

=√x2 + 4x + +

=√x2 + 4x +

But PA=PB

√x2-√4x + = √x2 + 4x +

Squaring on both sides we get

x2-4x+ = x2+4x+

-4x-4x=-

-8x=

x=-

(x,0)=(-,0) is the point which is equidistant from the given points.