Question

Find the point on the x-axis which is a equidistant from $(5,4)$ and $(-2,3)$.

Answer 1

Let the required point be $(x,0)$. Then,

$(x-5{)}^2+(0-4{)}^2=(x+2{)}^2+(0-3{)}^2$

$(x-5{)}^2+16=(x+2{)}^2+9$

$x^2+25-10x+16=x^2+4+4x+9$

$14x=25+16-13$

$14x=28$

$x=2$

Hence, the required point is $(2,0)$.