Question

If the point P(x, 3) is equidistant from the point A(7, −1) and B(6, 8), then find the value of x and find the distance AP. [CBSE 2014]

Answer 1

It is given that P(x, 3) is equidistant from the point A(7, −1) and B(6, 8).

∴ AP = BP

$\Rightarrow \sqrt{{\left(x-7\right)}^2+{\left[3-\left(-1\right)\right]}^2}=\sqrt{{\left(x-6\right)}^2+{\left(8-3\right)}^2}$ (Distance formula)

Squaring on both sides, we get

$$\begin{gathered} {\left(x-7\right)}^2+16={\left(x-6\right)}^2+25 \\ \Rightarrow x^2-14x+49+16=x^2-12x+36+25 \\ \Rightarrow -14x+12x=61-65 \\ \Rightarrow -2x=-4 \\ \Rightarrow x=2 \end{gathered}$$

Thus, the value of x is 2.

$\therefore \mathrm{AP}=\sqrt{{\left(2-7\right)}^2+{\left[3-\left(-1\right)\right]}^2}=\sqrt{{\left(-5\right)}^2+4^2}=\sqrt{25+16}=\sqrt{41}$ units