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Orthocenter Calculator + Online Solver With Free Steps
The Orthocenter Calculator is a free online calculator that illustrates the intersection of a triangle’s three altitudes.
For all triangles, the orthocenter serves as a crucial point of intersection in the middle. The orthocenter’s position perfectly describes the type of triangle that is being studied.
What Is an Orthocenter Calculator?
An orthocenter Calculator is an online tool used to calculate a centroid or point where the triangle’s altitudes meet.
That’s because a triangle’s altitude is defined as a line that goes through each of its vertices and is perpendicular to the other side, there are three possible altitudes: one from each vertex.
We can state that the orthocenter of the triangle is the place at which all three elevations consistently intersect.
How To Use an Orthocenter Calculator
You can use the Orthocenter Calculator by following these detailed guidelines, and the calculator will automatically show you the results.
Step 1
Fill in the appropriate input box with the three Coordinates (A, B, and C) of a triangle.
Step 2
Click on the “Calculate Orthocenter” button to determine the center for the given coordinates and also the whole step-by-step solution for the Orthocenter Calculator will be displayed.
How Does Orthocenter Calculator Work?
The Orthocenter Calculator works by using two of the altitudes intersecting to calculate the third intersect. The orthocenter of a triangle is the intersection point where all three of the triangle’s altitudes come together, according to mathematics. We are aware that there are various kinds of triangles, including scalene, isosceles, and equilateral triangles.
For each type, the orthocenter will be different. The orthocenter is located on the triangle for a right triangle, outside the triangle for an obtuse triangle, and inside the triangle for an acute triangle.
The orthocenter of any triangle can be calculated in 4 steps, which are listed below.
Step 1: Use the following formula to determine the triangle’s side slopes
Slope of a line $= \frac{y_2−y_1}{x_2−x_1}$
Step 2: Determine the sides’ perpendicular slope using the formula below:
The perpendicular slope of the line $=− \frac{1}{Slope of a line}$
Step 3: Using the following formula, find the equation for any two altitudes and their corresponding coordinates: y−y1=m(x − x1)
Step 4: Resolving equations for altitude (any two altitude equations of Step 3)
Orthocenter Properties and Trivia
Some interesting orthocenter characteristics include:
- Correlates with an equilateral triangle’s circumcenter, incenter, and centroid.
- Correlates with a right triangle’s right-angled vertex.
- For acute triangles, lies within the triangle.
- In obtuse triangles, lies outside the triangle.
Solved Examples
Let’s explore some examples to better understand the Orthocenter Calculator.
Example 1
A triangle ABC has the vertex coordinates: A = (1, 1), B = (3, 5), C = (7, 2). Find its Orthocenter.
Solution
Find the slope:
AB side slope \[ = \frac{(5 – 1) }{(3 – 1)} = 2 \]
Calculate the slope of the perpendicular line:
Perpendicular slope to AB side \[ = – \frac{1}{2} \]
Find the line equation:
\[ y – 2 = – \frac{1}{2} (x – 7) \]
so
y = 5.5 – 0.5 (x)
Repeat for another side, e.g., BC;
BC side slope \[= \frac{ (2 – 5) }{(7 – 3)} = – \frac{3}{4} \]
Perpendicular slope to BC side \[= \frac{4}{3} \]
\[ y – 1 = \frac{4}{3} (x – 1) \] so \[ y = – \frac{1}{3} + \frac{4}{3} (x) \]
Solve the system of linear equations:
y = 5.5 – 0.5 . x
and
y = -1/3 + 4/3 . x
So,
\[5.5 – 0.5 \times x = – \frac{1}{3} + \frac{4}{3} \times x \]
\[ \frac{35}{6} = x \times \frac{11}{6} \]
\[ x = \frac{35}{11} \approx 3.182 \]
Substituting x into either equation will give us:
\[ y = \frac{43}{11} \approx 3.909 \]
Example 2
Find the coordinates of the orthocenter of a triangle whose vertices are (2, -3) (8, -2) and (8, 6).
Solution
The given points are A (2, -3) B (8, -2), C (8, 6)
We now need to work on the AC slope. From there, we must determine the perpendicular line through B’s slope.
Slope of AC \[= \frac{(y2 – y1)}{(x2 – x1)}\]
Slope of AC \[= \frac{(6 – (-3))}{(8 – 2)} \]
Slope of AC \[= \frac{9}{6} \]
Slope of AC \[= \frac{3}{2} \]
Slope of the altitude BE \[= – \frac{1}{slope of AC} \]
Slope of the altitude BE \[ = – \frac{1}{(\frac{3}{2})} \]
Slope of the altitude BE \[ = – \frac{2}{3} \]
Equation of the altitude BE is given as :
\[(y – y1) = m (x – x1) \]
Here B (8, -2) and $m = \frac{2}{3}$
\[ y – (-2) = (-\frac{2}{3})(x – 8) \]
3(y + 2) = -2 (x – 8)
3y + 6 = -2x + 16
2x + 3y -16 + 6 = 0
2x + 3y – 10 = 0
We must now calculate BC’s slope. From there, we must determine the perpendicular line through D’s slope.
Slope of BC \[ = \frac{(y_2 – y_1)}{(x_2 – x_1)} \]
B (8, -2) and C (8, 6)
Slope of BC \[ = \frac{(6 – (-2))}{(8 – 8)} \]
Slope of BC \[ = \frac{8}{0} = \infty \]
Slope of the altitude AD \[= – \frac{1}{slope of AC} \]
\[= -\frac{1}{\infty} \]
= 0
The equation of the altitude AD is as follows:
\[(y – y_1) = m (x – x_1) \]
Here A(2, -3) and $m = 0$
\[ y – (-3) = 0 (x – 2) \]
\[ y + 3 = 0 \]
\[ y = -3 \]
By putting the value of x in the first equation:
\[ 2x + 3(-3) = 10 \]
\[ 2x – 9 = 10 \]
\[ 2x = 10 + 9 \]
\[ 2x = 19 \]
\[ x = \frac{1}{2} \]
\[ x = 9.2 \]
So the orthocenter is (9.2,-3).