Orthocenter Formula
The word "ortho" stands for "right." The orthocenter formula represents the center of all the right angles. It is drawn from the vertices to the opposite sides i.e., the altitudes. This concept holds great importance in understanding the various properties of a triangle, in reference to its other dimensions. The line passing through the vertex and perpendicular to the opposite side is called the altitude of a triangle. Thus, each vertex of a triangle creates altitudes. Let's look into the orthocenter formula.
What is Orthocenter Formula?
The point where the altitudes of a triangle intersect is called the othocenter. The orthocenter formula helps in locating the coordinates of the orthocenter of a triangle. Let's look into the orthocenter formula.
Let us consider a triangle PQR, as shown in the figure below.
PA, QB, RC are the perpendicular lines drawn from the three vertices P(\( x_{1}, y_{1})\), Q(\( x_{2}, y_{2})\), and R(\( x_{3}, y_{3})\) respectively of the \( \bigtriangleup \)PQR. H\( ( x, y) \) is the intersection point of the three altitudes of the triangle.
Step1. Calculate the slope of the sides of the triangle using the formula:
m(slope) = \( \frac{y_{2} \  \ y_{1}}{x_{2} \  \ x_{1}} \)
Let slope of PR be given by mPR.
Hence,
mPR = \( \frac{ y_3 \  \ y_1 }{ x_3 \  \ x_1 }\)
Similarly,
mQR = \( \frac{ y_3 \  \ y_2 }{ x_3 \  \ x_2 }\)
Step2. The slope of the altitudes of the \( \bigtriangleup \)PQR will be perpendicular to the slope of the sides of the triangle.
We know,
\( \begin{align*} \text {Perpendicular slope of line} \ &= \ \frac{1}{ \text {slope of the line}} \\ &= \frac{1}{ \text m} \end{align*}\)
The slope of the respective altitudes:
Slope of PA,
mPA = \( \frac{1}{ \text {mQR}}\)
Slope of QB,
mQB = \( \frac{ 1}{ \text {mPR}}\)
We will use the slopepoint form equation as a straight line to calculate the equations of the lines coinciding with PA and QB.
The generalized equation thus formed by using arbitrary points \( x\) and \( y\) is:
\( \begin{align*} \text {mPA} &= \frac{( y \  \ y_1 )}{( x \  \ x_1 )} \\ \text {mQB} &= \frac{( y \  \ y_2)}{( x \  \ x_2 )} \end{align*} \)
Thus, solving the two equations for any given values the orthocenter of a triangle can be calculated.
Let's solve some examples to understand the orthocenter formula.
Solved Examples Using Orthocenter Formula

Example 1: Calculate the orthocenter of a triangle when their vertices are A(1, 3), B(2, 5), C(3, 4) using orthocenter formula.
Solution
Given, the vertices of the triangle,
\( \begin{align*} \text A &= (1, 3) \\ \text B &= (2, 5) \\ \text C &= (3, 4) \end{align*} \)
Let the coordinates of orthocenter be H(x, y).
Using orthocenter formula,
Slope of AB,
\( \begin{align*} \text {mAB} &= \frac{( y_{2} \  \ y_{1})}{( x_{2} \  \ x_{1})} \\ &= \frac{5  3}{2  1} \\ &= \ 2 \end{align*} \)
Slope of CF,
\( \begin{align*} \text {mCF} &= \text {Perpendicular slope of AB} \\ &= \frac{1}{ \text {mAB}} \\ &= \frac{1}{2} \end{align*} \)
The equation of CF is given as,
\( \begin{align*} ( y – \text y_1) &= \text m ( x – \text x_1) \\ y + 4 &= \frac{1}{2} ( x – 3) \\ 2 y + 8 &=  x + 3 \\ x + 2 y &= 5 \ \ \ \ ( \text {equation} 1) \end{align*} \)
Slope of BC,
\( \begin{align*} \text {mBC} &= \frac{(y_{2} \  \ y_{1})}{(x_{2} \  \ x_{1})} \\ &= \frac{4  5}{3  2} \\ &= 9 \end{align*} \)
Slope of AD,
\( \begin{align*} \text {mAD} &= \text {Perpendicular slope of BC} \\ &= \frac{1}{ \text {mBC}} \\ &= \frac{1}{9} \\ &= \frac{1}{9} \end{align*} \)
The equation of AD is given as,
\( \begin{align*} ( y – \text y_1) &= \text m ( x – x_1) \\ y  3 &= \frac{1}{9} ( x – 1) \\ 9 y  27 &= x  1 \\ x  9 y &= 26 \ \ \ \ ( \text {equation} 2)\end{align*} \)
Subtracting equation (2) and (1),
\[ y = \dfrac{21}{11}\]
Thus, putting this value in equation (1),
\[ x = \dfrac {97}{11}\]
Answer: Therefore, orthocenter = (−97/11,21/11)

Example 2: Point H is the orthocenter of \(\triangle{ABC}\).If \(\angle{B} \ = \ 58^\circ , \ \angle{C}=60^\circ , \ \text {and} \angle{A}=62^\circ\). What is the measure of \(\angle{CHD}\)?
Solution:
In the given figure,
\( \begin{align*} \angle{ \text{CDH}} &= \angle{ \text{BFC}} = 90^\circ \\ \angle{ \text{DCH}} &= \angle{ \text {BCF}} \end{align*} \)
Hence, by AAA property,
\( \triangle{ \text {DCH}}\) and \( \triangle{ \text {FCB}} \) are similar triangles.
Thus,
\( \angle{ \text {CHD}} = \angle{ \text {CBF}} = 58^\circ \)
Answer: Therefore, ∠CHD = 58°
visual curriculum