Wilmer said:
Right triangle ABC.
BC = a = sqrt(17)
AC = b = sqrt(68)
AB = c = sqrt(85)
A's coordinates: 0,12
B's coordinates: 6,5
What's EASIEST way to get C's coordinates?
Let's make a drawing:
\begin{tikzpicture}[>=stealth']
%preamble \usetikzlibrary{arrows}
\def\a{sqrt(17)}
\def\b{sqrt(68)}
\def\c{sqrt(85)}
\coordinate (A) at (0,12);
\coordinate (B) at (6,5);
\coordinate (C1) at (2,4);
\coordinate (C2) at ({38/5},{44/5});
\coordinate (D) at ({\b*\b/\c/\c * 6}, {12+\b*\b/\c/\c * (5-12)});
\coordinate (E) at ({\b*\b/\c/\c * 6 + \a*\b/\c/\c * (12-5)}, {12+\b*\b/\c/\c * (5-12) + \a*\b/\c/\c * 6});
\draw (A) -- (B) -- (C1) -- cycle;
\draw (A) -- (B) -- (C2) -- cycle;
\draw[->, thick] (A) -- node[right, xshift=6] {$b\cos\alpha\cdot\frac{\overrightarrow{AB}}{c}$} (D);
\draw[->, thick] (D) -- node[above, xshift=-18] {$b\sin\alpha\cdot\frac{\overrightarrow{AB}^\perp}{c}$} (E);
\path (A) node[above left] {A} -- (B) node[below right] {B} -- (C2) node[above right] {C} -- (C1) node[below left] {C'};
\path (A) -- node[below left] {c} (B) -- node
{a} (C2) -- node[above] {b} (A) -- (D) node[below left] {D};
\path (A) node[below right, xshift=10, yshift=-6] {$\alpha$};
\end{tikzpicture}
It follows that:
$$\overrightarrow C = \overrightarrow A + b\cos\alpha\cdot\frac{\overrightarrow{AB}}{c} \pm b\sin\alpha\cdot \frac{\overrightarrow{AB^\perp}}{c}
= \overrightarrow A + b\cdot \frac bc \cdot\frac{\overrightarrow{AB}}{c} \pm b\cdot \frac ac\cdot \frac{\overrightarrow{AB^\perp}}{c} \\
= \binom{0}{12} + \frac {68}{85} \cdot\binom{6-0}{5-12} \pm \frac {\sqrt{17\cdot 68}}{85}\cdot\binom{12-5}{6-0}
$$
