Question

If the point $P(x,3)$ is equidistant from the points $A(7,-1)$ and $B(6,8)$, find the value of $x$ and find the distance $AP$.

Answer 1

Given $\Rightarrow$ $(PA{)}^2=(PB{)}^2$

So by distance formula, we have,

$\begin{array}{cc}(x-7{)}^2+(3+1{)}^2& =(x-6{)}^2+(3-8{)}^2\\ x^2+49-14x+16& =x^2+36-12x+25\\ 49+16-36-25& =2x\\ 65-61& =2x\\ 2x& =4\\ x& =2\end{array}$

Distance:

$AP=\sqrt{(2-7{)}^2+(-1-3{)}^2}$ (by distance formula)

$=\sqrt{(-5{)}^2+(-4{)}^2}$

$=\sqrt{25+16}$

$=\sqrt{41}$