# Orthocenter Formula

## Orthocenter Formula - Orthocenter of a Triangle Formulas

The three altitudes of any triangle are concurrent line segments (they intersect in a single point) and this point is known as the orthocenter of the triangle. The steps to find the coordinates of the orthocenter of a triangle are relatively simple, given that we know the coordinates of the vertices of the triangle .
Let us consider the following triangle ABC, the coordinates of whose vertices are known. First, we will find the slopes of any two sides of the triangle (say AC and BC).
${m_{AC}} = \frac{{\left( {{y_3} - {y_1}} \right)}}{{\left( {{x_3} - {x_1}} \right)}}\quad \quad {m_{BC}} = \frac{{\left( {{y_3} - {y_2}} \right)}}{{\left( {{x_3} - {x_2}} \right)}}$
Next, we can find the slopes of the corresponding altitudes. Remember that if two lines are perpendicular to each other, they satisfy the following equation.
$\begin{gathered} {m_{AC}} \times {m_{BE}} = - 1\quad \quad {m_{BC}} \times {m_{AD}} = - 1 \hfill \\ {m_{BE}} = \frac{{ - 1}}{{{m_{AC}}}}\quad \quad \,{m_{AD}} = \frac{{ - 1}}{{{m_{BC}}}} \hfill \\ \end{gathered}$
Next, we will use the slope-point form of the equation of a straight line to find the equations of the lines that are coincident with the altitudes BE and AD.
$BE:\frac{{\left( {y - {y_2}} \right)}}{{\left( {x - {x_2}} \right)}} = {m_{BE}}\quad \quad AD:\frac{{\left( {y - {y_1}} \right)}}{{\left( {x - {x_1}} \right)}} = {m_{AD}}$
Next, we can solve the equations of BE and AD simultaneously to find their solution, which gives us the coordinates of the orthocenter H.
Let’s see an example now.

Question: Find the coordinates of the orthocenter of a triangle ABC whose vertices are A(1 ,7), B(−6, 0) and C(3, 4).
Solution:
$\begin{gathered} {m_{AC}} = \frac{{\left( {{y_3} - {y_1}} \right)}}{{\left( {{x_3} - {x_1}} \right)}} = \frac{{\left( {4 - 7} \right)}}{{\left( {3 - 1} \right)}} = \frac{{ - 3}}{2} \Rightarrow {m_{BE}} = \frac{{ - 1}}{{{m_{AC}}}} = \frac{2}{3} \hfill \\ {m_{BC}} = \frac{{\left( {{y_3} - {y_2}} \right)}}{{\left( {{x_3} - {x_2}} \right)}} = \frac{{\left( {4 - 0} \right)}}{{\left( {3 - \left( { - 6} \right)} \right)}} = \frac{4}{9} \Rightarrow {m_{AD}} = \frac{{ - 1}}{{{m_{BC}}}} = \frac{{ - 9}}{4} \hfill \\ BE:\frac{{\left( {y - {y_2}} \right)}}{{\left( {x - {x_2}} \right)}} = {m_{BE}} \Rightarrow \frac{{\left( {y - 0} \right)}}{{\left( {x - \left( { - 6} \right)} \right)}} = \frac{2}{3} \Rightarrow 2x - 3y + 12 = 0 \hfill \\ AD:\frac{{\left( {y - {y_1}} \right)}}{{\left( {x - {x_1}} \right)}} = {m_{AD}} \Rightarrow \frac{{\left( {y - 7} \right)}}{{\left( {x - 1} \right)}} = \frac{{ - 9}}{4} \Rightarrow 9x + 4y - 37 = 0 \hfill \\ \end{gathered}$
Solving the equations for BE and AD , we get the coordinates of the orthocenter H as follows.
$H\left( {\frac{9}{5},\frac{{26}}{5}} \right)$
Why don’t you try to solve a problem to see if you are getting the hang of the methodology?

Question:
Find the coordinates of the orthocenter of a triangle ABC whose vertices are A (−1, −4), B (2, −3) and C (5, 2).
Options:
(a) (10, 9)
(b) (10, −9)
(c) (9, 10)
(d) (9, −10)
$\begin{gathered} {m_{AC}} = \frac{{\left( {{y_3} - {y_1}} \right)}}{{\left( {{x_3} - {x_1}} \right)}} = \frac{{\left( {2 - \left( { - 4} \right)} \right)}}{{\left( {5 - \left( { - 1} \right)} \right)}} = 1 \Rightarrow {m_{BE}} = \frac{{ - 1}}{{{m_{AC}}}} = - 1 \hfill \\ {m_{BC}} = \frac{{\left( {{y_3} - {y_2}} \right)}}{{\left( {{x_3} - {x_2}} \right)}} = \frac{{\left( {2 - \left( { - 3} \right)} \right)}}{{\left( {5 - 2} \right)}} = \frac{5}{3} \Rightarrow {m_{AD}} = \frac{{ - 1}}{{{m_{BC}}}} = \frac{{ - 3}}{5} \hfill \\ BE:\frac{{\left( {y - {y_2}} \right)}}{{\left( {x - {x_2}} \right)}} = {m_{BE}} \Rightarrow \frac{{\left( {y - \left( { - 3} \right)} \right)}}{{\left( {x - 2} \right)}} = - 1 \Rightarrow x + y + 1 = 0 \hfill \\ AD:\frac{{\left( {y - {y_1}} \right)}}{{\left( {x - {x_1}} \right)}} = {m_{AD}} \Rightarrow \frac{{\left( {y - \left( { - 4} \right)} \right)}}{{\left( {x - \left( { - 1} \right)} \right)}} = \frac{{ - 3}}{5} \Rightarrow 3x + 5y + 23 = 0 \hfill \\ \end{gathered}$
$H\left( {9, - 10} \right)$