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 \((\frac {15}{2}, 0)\).