What is the orthocenter of a triangle with corners at (6 ,3 ), (2 ,5 ), and (7 ,9 )?

Mar 17, 2016

The Orthocenter is the point where the perpendicular line to AB and BC meet, and is $O \left(\frac{62}{13} , \frac{59}{13}\right) \implies \frac{1}{13} \left(62 , 59\right) , \mathmr{and} O \left(4.77 , 4.54\right)$ Explanation:

First find the equation of line of the side of the triangle,
$\overline{A B} , \overline{B C} \text{ and } \overline{C A}$

Equation of line sets for: $\textcolor{red}{\overline{A B}}$:

${y}_{\overline{A B}} = {m}_{\overline{A B}} x + {b}_{\overline{A B}}$ with $m = \frac{3 - 5}{6 - 2} = - \frac{1}{2}$
Use B(2,5) which is on ${y}_{\overline{A B}} = {m}_{\overline{A B}} x + {b}_{\overline{A B}}$
Letting $y = 5$ and $x = 2$ solve for ${b}_{\overline{A B}}$
5=-1/2*2+b_(bar(AB)); b_(bar(AB)) = 6 thus
${y}_{\overline{A B}} = - \frac{1}{2} x + 6$
A perpendicular equation of line will have a slope
m_(bar(AB))=-1/m_(bar(AB)); -1/2=-1/m_(bar(AB));m_(Pr_bar(AB))= 2
Thus the perpendicular equation of linr to AB is:
${y}_{P {r}_{\overline{A B}}} = 2 x + {b}_{P {r}_{\overline{A B}}}$
use $A \left(7 , 9\right)$ to determine ${b}_{P {r}_{\overline{A B}}}$
9 = 2*7+ b_(Pr_bar(AB)); b_(Pr_bar(AB))=-5
So the pair of equation sets, for $\textcolor{red}{\overline{A B}}$ are:
${\left[\begin{matrix}y \\ \textcolor{b l u e}{{y}_{p r}}\end{matrix}\right]}_{\overline{A B}} = \left[\begin{matrix}- \frac{1}{2} & 6 \\ \textcolor{b l u e}{2} & \textcolor{b l u e}{- 5}\end{matrix}\right] \cdot \left[\begin{matrix}x \\ 1\end{matrix}\right]$

Equation of line sets for: $\textcolor{red}{\overline{B C}}$:

${y}_{\overline{A = B C}} = {m}_{\overline{B C}} x + {b}_{\overline{B C}}$ with $m = \frac{9 - 5}{7 - 2} = \frac{4}{5}$
5=4/5*2+b; b=17/5 the perpendicular has slope and y-intercept:
m_(pr) = -1/m; m_(pr)=-5/4
Using point$A \left(6 , 3\right)$ to determine ${b}_{p r}$:
3=-5/4*6+b_(pr); b_(pr)=21/2
${\left[\begin{matrix}y \\ \textcolor{m a \ge n t a}{{y}_{p r}}\end{matrix}\right]}_{\overline{B C}} = \left[\begin{matrix}\frac{4}{5} & \frac{17}{5} \\ \textcolor{m a \ge n t a}{- \frac{5}{4}} & \textcolor{m a \ge n t a}{\frac{21}{2}}\end{matrix}\right] \cdot \left[\begin{matrix}x \\ 1\end{matrix}\right]$

The Orthocenter is the point where the perpendicular line to AB and BC meet, i.e.: color(blue)(y_(Pr_bar(AB))) = color(magenta)(y_(Pr_bar(BC))
color(blue)(2x-5)=color(magenta)(-5/4x +21/2); x=62/13#
Using x=62/13 solve for y:
$y = 2 x - 5 {|}_{x = \left(\frac{62}{13}\right)}$, $y = 2 \cdot \frac{62}{13} - 5 = \frac{59}{13}$ 