Orthocenter comes under the concept of Geometry. The term ortho means right and it is considered to be the intersection point of all three altitudes drawn from the vertices of a triangle. The Orthocenter is the intersecting point of all the altitudes of a triangle.
Orthocenter Formula
The orthocenter is the point in a triangle where altitudes are met. Altitudes are perpendicular lines from one side of the triangle to the opposite vertex. In the below diagram, Orthocenter is denoted by the letter "O" and the altitudes are AE, BF, CD, and the vertices are A, B, C.
There is no particular formula to calculate the orthocenter of a triangle. But it can figure out by the below steps:
- Find the Slope of the sides of the triangle.
Slope of a side = (y2 - y1)/(x2 - x1)
Where (x1, y1) and (x2, y2) are endpoints of the side.
- To find the altitude slope which is perpendicular to the sides of the triangle can be calculated by,
Perpendicular Slope of a line (Altitude) = -1/(Slope of a side)
- To calculate the equation for altitude, the point-slope formula is,
(y - y1) = m × (x - x1)
So by finding any two slopes and equation of altitudes, the orthocenter can be calculated by solving those equations.
Sample Problems
Question 1: Find the slope of the line between two points A(1, 10), B(5, 5).
Solution:
Given A(1, 10), B(5, 5)
Slope Of line AB = (y2 - y1)/(x2 - x1)
= (5 - 10)/(5 - 1)
= -5/4
Slope of line AB = -5/4
Question 2: Find the slope of the line between two points A(5, 4), B(-1, 2).
Solution:
Given A(5, 4), B(-1, 2)
Slope Of line AB = (y2 - y1)/(x2 - x1)
= (2 - 4)/(-1 - 5)
= -2/-6
= 1/3
Slope of line AB = 1/3
Question 3: Find the Perpendicular slope of AB if the slope of AB is -5/4?
Solution:
Given Slope of AB = -5/4
Perpendicular Slope of AB = -1/Slope of AB
= -1/(-5/4)
= 4/5
Perpendicular slope of AB is 4/5.
Question 4: Find the orthocenter of the triangle when the vertices are A(1, 0), B(3, 1), C(2, 3).
Solution:
Given Vertices:
A(1, 0), B(3, 1), C(2, 3)
Slope of AB = (y2 - y1)/(x2 - x1)
= (1 - 0)/(3 - 1)
= 1/2
Slope of CD = Perpendicular slope of AB
= -1/Slope of AB
= -1/(1/2)
= -2
The equation of cd is
(y - y1) = m × (x - x1)
(y - 3) = -2 × (x - 2)
y - 3 = -2x + 4
2x + y - 3 - 4 = 0
2x + y - 7 = 0 ⇢ (1)
Slope of AC = (y2 - y1)/(x2 - x1)
= (3 - 0)/(2 - 1)
= 3
Slope of BF = Perpendicular slope of AC
= -1/Slope of AC
= -1/3
The equation of BF is,
(y - y1) = m × (x - x1)
(y - 1) = (-1/3) × (x - 3)
3(y - 1) = -1(x - 3)
3y - 3 = -x + 3
x + 3y - 6 = 0 ⇢ (2)
Solve Equation (1) & (2)
(1) × 1 => 2x + y - 7 = 0
(2) × 2 => 2x + 6y - 12 = 0
-5y + 5 = 0
-5y = -5
y = 1
Substitute y value in equation - (1)
2x + y - 7 = 0
2x + 1 - 7 = 0
2x = 6
x = 3
Orthocenter is O(3, 1)
Question 5: Find the orthocenter of the triangle when the vertices are A(0, 0), B(2, -1), C(1, 3).
Solution:
Given Vertices:
A(0, 0), B(2, -1), C(1, 3)
Slope of AB = (y2 - y1)/(x2 - x1)
= (-1 - 0)/(2 - 0)
= -1/2
Slope of CD = Perpendicular slope of AB
= -1/Slope of AB
= -1/(-1/2)
= 2
The equation of CD is,
(y - y1) = m × (x - x1)
(y - 3) = 2 × (x - 1)
y - 3 = 2x - 2
2x - y + 1 = 0 ⇢ (1)
Slope of AC = (y2 - y1)/(x2 - x1)
= (3 - 0)/(1 - 0)
= 3
Slope of BF = Perpendicular slope of AC
= -1/Slope of AC
= -1/3
The equation of BF is,
(y - y1) = m × (x - x1)
(y - (-1)) = (-1/3) × (x - 2)
3(y + 1) = -1(x - 2)
3y + 3 = -x + 2
x + 3y + 1 = 0 ⇢ (2)
Solve Equation (1) & (2)
(1) × 1 => 2x - y + 1 = 0
(2) × 2 => 2x + 6y + 2 = 0
-7y - 1 = 0
-7y = 1
y = -1/7
Substitute y value in equation-(1)
2x - y + 1 = 0
2x - (-1/7) + 1 = 0
2x + (1/7) + 1 = 0
2x = -1 - (1/7)
2x = (-7 - 1)/7
x = -8/14
x = -4/7
The orthocenter for the given data is O(-4/7, -1/7)
Question 6: Find the orthocenter of the triangle when the vertices are A(3, 2), B(0, 3), C(-2, 1)
Solution:
Given Vertices:
A(3, 2), B(0, 3), C(-2, 1)
Slope of AB = (y2 - y1)/(x2 - x1)
= (3 - 2)/(0 - 3)
= -1/3
Slope of CD = Perpendicular slope of AB
= -1/Slope of AB
= -1/(-1/3)
= 3
The equation of CD is,
(y - y1) = m × (x - x1)
(y - 1) = 3 × (x + 2)
y - 1 = 3x + 6
3x - y + 7 = 0 ⇢ (1)
Slope of AC = (y2 - y1)/(x2 - x1)
= (1 - 2)/(-2 - 3)
= -1/-5
= 1/5
Slope of BF = Perpendicular slope of AC
= -1/Slope of AC
= -1/(1/5)
= -5
The equation of BF is
(y - y1) = m × (x - x1)
(y - 3) = (-5) × (x - 0)
y - 3 = -5x
5x + y - 3 = 0 ⇢ (2)
Solve Equation (1) & (2)
(1) × 1 => 3x - y + 7 = 0
(2) × 1 => 5x + y - 3 = 0
8x + 4 = 0
8x = -4
x = -4/8
x = -1/2
Substitute x value in equation-(1)
3x - y + 7 = 0
3(-1/2) - y + 7 = 0
(-3/2) - y + 7 = 0
y = 7 - (3/2)
y = (14 - 3)/2
y = 11/2
The orthocenter for the given data is O(-1/2, 11/2)
Question 7: Find the equation of a line if the slope and a data point are -1/2, (3, 3) respectively.
Solution:
Given (x1, y1) = (3, 3)
Slope (m) = -1/2
Equation of Line is,
(y - y1) = m × (x - x1)
(y - 3) = (-1/2) × (x - 3)
2(y - 3) = -1(x - 3)
2y - 6 = -x + 3
x + 2y - 9 = 0
The equation of line for the given slope and data point is x + 2y - 9 = 0.
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