If the point (x,3) is equidistant from (3,-2) and (7,4), find the value of x.

If the point (x,3) is equidistant from (3,-2) and (7,4), find the value of x.

$\begin{array}{|rcll|} \hline \vec{X} &=& \binom{x}{3} \\ \vec{P_1} &=& \binom{3}{-2} \\ \vec{P_2} &=& \binom{7}{4} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline |\vec{X}-\vec{P_1}| &=& |\vec{P_2}-\vec{X}| \\ |\binom{x}{3}-\binom{3}{-2}| &=& |\binom{7}{4}-\binom{x}{3}| \\ |\binom{x-3}{3-(-2)}| &=& |\binom{7-x}{4-3}| \\ |\binom{x-3}{5}| &=& |\binom{7-x}{1}| \\ (x-3)^2+5^2 &=& (7-x)^2 + 1^2 \\ (x-3)^2+25 &=& (7-x)^2 + 1 \\ x^2 -6x + 9 +25 &=& 49 - 14 x + x^2 + 1 \\ x^2 -6x + 34 &=& 50 - 14 x + x^2 \\ -6x + 34 &=& 50 - 14 x \qquad & | \qquad +14x -34\\ 8x &=& 16 \qquad & | \qquad :8 \\ x &=& 2 \\ \hline \end{array}$

The equidistant is $\sqrt{1+25}=\sqrt{26}=5.09901951359$