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 the x axis that is equidistant from the points (7, 6) and (3, 4).
Accordingly, $\sqrt {(7-a)^+(6-0)^2}= \sqrt {(3-a)^2+(4-0)^2}$
⇒ $\sqrt { 49+a^2-14a+36 }= \sqrt {9+a^2-6a+16}$
⇒ $\sqrt {a^2-14a+85 }= \sqrt {a^2-6a+25}$
On squaring both sides,
$a^2 – 14a + 85$= $a^2 – 6a + 25$
⇒ $–14a + 6a = 25 – 85$
⇒ $–8a = –60$
⇒ $a=\frac {60}{8 }$
⇒ $a = \frac {15}{2}$

Thus, the required point on the x-axis is $(\frac {15}{2}, 0)$.