Question

Find a point on the $x$-axis, which is equidistant from the points $(7,6)$ and $(3,4)$.

Answer 1

Let $(a,0)$ be the point on $x$-axis that is equidistant from the points $(7,6),(3,4)$.
$\Rightarrow \sqrt{(7-a{)}^2+(6-0{)}^2}$$=\sqrt{(3-a{)}^2+(4-0{)}^2}$
$\Rightarrow \sqrt{49+a^2-14a+36}$$=\sqrt{9+a^2-6a+16}$
$\Rightarrow \sqrt{a^2-14a+85}$$=\sqrt{a^2-6a+25}$
On squaring both sides, we obtain
$a^2-14a+85=a^2-6a+25$
$\Rightarrow -14a+6a=25-85$
$\Rightarrow 8a=60$
$\Rightarrow a=\frac{60}{8}=\frac{15}{2}$
Thus, the required point on $x$-axis is $(\frac{15}{2},0)$