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How to find the Orthocenter of a Triangle?

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  • Last Updated : 20 May, 2022

A simple polygon having three sides and three vertices are called a triangle. The intersection point of the three altitudes of a triangle is called the “orthocenter of a triangle,” and it is generally represented by the letter “H.” An altitude of a triangle is a line segment drawn from each vertex to the opposite side that is perpendicular to its opposite side. As a triangle has three vertices and three sides, it has three altitudes, and the point of intersection of these three sides is called the orthocenter. 

 

For every triangle, the position of the orthocenter varies; i.e; for an equilateral triangle, the orthocenter, circumcenter, incenter, and centroid are the same, but in the case of the other triangles,, the position will be different. 

  • In the case of an acute-angled triangle, the orthocenter lies inside the triangle.
  • In the case of an obtuse-angled triangle, the orthocenter lies outside the triangle.
  • In the case of a right-angled triangle, the orthocenter lies on the vertex of the right angle.

 

Determining the orthocenter of a triangle

Let’s consider a triangle ABC to determine the orthocenter of a triangle. AD, BE, and CF are the perpendiculars drawn from the vertices A (x1, y1), B (x2, y2), and C (x3, y3) to their respective opposite sides BC, AC, and AB and “H” is the point of their intersection.

Step 1: Calculate the slopes of the sides of the triangle ABC using the slope formula;

m = (y2 – y1)/(x2 – x1)

Let the slope of the AB be mAB.

mAB = (y2 – y1)/(x2 – x1)

Let the slope of the BC be mBC

So, mBC = (y3 – y2)/(x3 – x2)

Step 2: Using the slopes of the sides of a triangle, find the slopes of altitudes.

We know that the altitude is perpendicular to the side. 

Product of slopes of two perpendicular slopes lines = m1 × m2 = -1 

So, the slope of the altitude = -1/slope of the side = -1/m

Now, the slopes of the respective altitudes CF and AD are,

mCF = -1/mAB

mAD = -1/mBC

Step 4: With the help of the point-slope form equation, find the equations of the altitudes using the slopes and the coordinates of the opposite vertices.

The equation of CF is (y – y3) = mCF(x – x3)

The equation of AD is (y – y1) = mAD(x – x1)

Step 4: Solve the equations of any two altitudes and the values of x and y obtained by solving both equations are the coordinates of the orthocenter of the triangle.

Sample Problems

Problem 1: Determine the coordinates of the orthocenter of a triangle whose vertices are A (3, 1), B (-5, 2), and C (0, 4).

Solution:

Given,

The vertices of a triangle are A (x1, y1) = (3,1), B (x2, y2) = (-5,2) and C (x3, y3) = (0,4)

Now, the slope of the side AB = (y2 – y1)/(x2 – x1)

⇒ mAB = (2 – 1)/(-5 -3) = -(1/8)

The slope of the line perpendicular to AB i.e., slope of CF = -(1/slope of AB) = 8

So, the equation of the line CF with point C (0,4) and slope = 8 is y – y3 = m(x – x3) [point-slope form]

⇒ y – 4 = 8 (x – 0) 

⇒ y – 4 = 8x

⇒ 8x – y = -4 ⇢ (1)

Slope of the side BC = (y3 – y2)/(x3 – x2)

⇒ mBC = (4 – 2)/(0 – (-5)) = 2/5

Now, the slope of the line perpendicular to BC i.e., the slope of AD = -(1/slope of BC) = -(5/2)

So, the equation of the line AD with point A (3,1) and slope = -(5/2) is y – y1 = m(x – x1) [point slope form]

⇒ y – 1 = -(5/2) (x – 3)

⇒ 2(y – 1) = -5(x – 3)

⇒ 2y – 2 = -5x + 15

⇒ 5x + 2y = 17 ⇢ (2)

Now, multiply the equation (1) with “2” on both sides and add both equations (1) and (2).

16x – 2y = -8

5x + 2y = 17

21x = 9 ⇒ x = 3/7

Now, substitute the value of x = 3/7 in equation (1)

⇒ 8(3/7) – y = -4

⇒ y = 24/7 + 4 = 52/7

Hence, the coordinates of the orthocenter (H) are (3/7, 52/7).

Problem 2:  Determine the coordinates of the orthocenter of a triangle whose vertices are A (5, -3), B (7, 0), and C (4, 9).

Solution:

Given,

The vertices of a triangle are A (x1, y1) = (5, -3), B (x2, y2) = (7, 0) and C (x3, y3) = (4, 9).

Now, the slope of the side AB = (y2 – y1)/(x2 – x1)

⇒ mAB = (0 – (-3))/(7 – 5) = 3/2

The slope of the line perpendicular to AB i.e., slope of CF =  -(1/slope of AB) = -(2/3)

So, the equation of the line CF with point C (4, 9) and slope = -(2/3)  is y – y3 = m(x – x3) [point-slope form]

⇒ y – 9 = -(2/3) (x – 4)

⇒ 3(y – 9) = -2(x – 4)

⇒ 3y – 27 = -2x + 8

⇒ 2x + 3y = 35 ⇢ (1)

Slope of the side BC = (y3 – y2)/(x3 – x2)

⇒ mBC = (9 – 0)/(4 – 7) = -(9/3) = -3

Now, the slope of the line perpendicular to BC i.e., the slope of AD = -(1/slope of BC) = 1/3

So, the equation of the line AD with point A (5, -3) and slope = 1/3 is y – y1 = m(x – x1) [point slope form]

⇒ y – (-3) = (1/3) (x – 5)

⇒ 3(y + 3) = x – 5

⇒ 3y + 9 = x – 5

⇒ x – 3y = 14 ⇢ (2)

Now, add equations (1) and (2)

2x + 3y = 35

x – 3y = 14

3x = 49  ⇒ x =49/3

Now, substitute the value of x = 49/3 in equation (2)

⇒ 49/3 – 3y = 14 ⇒ 3y = -1

⇒ y =7/9

Hence, the coordinates of the orthocenter (H) are (49/3, 7/9).

Problem 3: Find the orthocenter of a triangle whose vertices are A (2, -7), B (6, 3), and C (-8, 0).

Solution:

Given,

The vertices of a triangle are A (x1, y1) = (5, -3), B (x2, y2) = (7, 0) and C (x3, y3) = (4, 9)

Now, the slope of the side AB = (y2 – y1)/(x2 – x1)

⇒ mAB = (3 – (-7))/(6 – 2) = 10/4 = 5/2

The slope of the line perpendicular to AB i.e., slope of CF = -(1/slope of AB) = -(2/5)

So, the equation of the line CF with point C (-8, 0) and slope = -(2/5)  is y – y3 = m(x – x3) [point-slope form]

⇒ y – 0 = -(2/5) (x – (-8))

⇒ 5y = -2(x + 8)

⇒ 5y = -2x -16

⇒ 2x + 5y = -16 ⇢ (1)

Slope of the side AC = (y3 – y1)/(x3 – x1)

⇒ mAC = (0 – (-7))/(-8 – 2) = -(7/10)

Now, the slope of the line perpendicular to AC i.e., the slope of BE = -(1/slope of AC) = 10/7

So, the equation of the line BE with point B (6, 3) and slope = 10/7 is y – y2 = m(x – x2) [point slope form]

⇒ y – 3 = (10/7) (x – 6)

⇒ 7(y – 3) = 10(x – 6)

⇒ 7y – 21 = 10x – 60

⇒ 10x – 7y = 39 ⇢ (2)

Multiply equation (1) with “5” on both sides and subtract both equations.

10x + 25y = -80

10x – 7y = 39

(-)   (+)     (-)

——————

32y = -119 ⇒ y = – 119/32

Now, substitute the value of y = -119/32 in equation (1)

2x + 5(-119/32) = -16

⇒ 2x – 595/32 = -16 ⇒ 2x = 595/32 – 16

⇒ 2x = 83/32 ⇒ x = 83/64

Hence, the coordinates of the orthocenter (H) are (83/64, -119/32).

Problem 4: Find the orthocenter of a triangle whose vertices are A (6, 2), B (1, 1), and C (-4, 7).

Given,

The vertices of a triangle are A (x1, y1) = (6, 2), B (x2, y2) = (1, 1) and C (x3, y3) = (-4, 7).

Now, the slope of the side AC = (y3 – y1)/(x3– x1)

⇒ mAC = (7 – 2)/(-4 – 6) = -(5/10) = -1/2

The slope of the line perpendicular to AC i.e., slope of BE = -(1/slope of AC) = 2

So, the equation of the line BE with point B (1,1) and slope = 2  is y – y2 = m(x – x2) [point-slope form]

⇒ y – 1 = 2(x – 1)

⇒ y – 1 = 2x – 2

⇒ 2x – y = 1 ⇢ (1)

Slope of the side BC = (y3 – y2)/(x3 – x2)

⇒ mBC =(7 – 1)/(-4 – 1) = -(6/5)

Now, the slope of the line perpendicular to BC i.e., the slope of AD = -(1/slope of BC) = 5/6

So, the equation of the line BE with point A (6, 2) and slope = 10/7 is y – y1 = m(x – x1) [point slope form]

⇒ y – 2 = (5/6) (x – 6)

⇒ 6(y – 2) = 5(x – 6)

⇒ 6y – 12 = 5x – 30

⇒ 5x – 6y = 18 ⇢ (2)

Now, multiply equation (1) with “6” on both sides and subtract both equations.

12x – 6y = 6

5x – 6y = 18

(-) (+)     (-)

—————— 

7x = -12 ⇒ x = -12/7

Now, substitute the value of x = -12/7 in equation (1)

2(-12/7) – y = 1

⇒ y = -24/7 – 1 ⇒ y = -31/7

Hence, the coordinates of the orthocenter (H) are (-12/7, -31/7).

Problem 5: Determine the coordinates of the orthocenter of a triangle whose vertices are A (0,-5), B (3,-2), and C (-6, 0).

Given,

The vertices of a triangle are A (x1, y1) = (3,1), B (x2, y2) = (-5,2) and C (x3, y3) = (0,4)

Slope of the side BC = (y3 – y2)/(x3 – x2)

⇒ mBC =(0 – (-2))/(-6 – 3) = -(2/9)

Now, the slope of the line perpendicular to BC i.e., the slope of AD = -(1/slope of BC) = (9/2)

So, the equation of the line AD with point A (3,1) and slope = (9/2) is y – y1 = m(x – x1) [point slope form]

⇒ y – (-5) = (9/2) (x – 0)

⇒ 2(y + 5) = 9x

⇒ 2y + 10 = 9x

⇒ 9x – 2y = 10 ⇢ (1)

Now, the slope of the side AB = (y2 – y1)/(x2 – x1)

⇒ mAB = (-2 – (-5))/(3 – 0) = 3/3 = 1

The slope of the line perpendicular to AB i.e., slope of CF =  -(1/slope of AB) = -1

So, the equation of the line CF with point C (-6, 0) and slope = -1 is y – y3 = m(x – x3) [point-slope form]

⇒ y – 0 = (-1)(x – (-6))

⇒ y = -(x + 6)

⇒ y = -x – 6

⇒ x + y = -6 ⇢ (2)

Now, multiply equation (2) with “2” on both sides and add both equations.

9x – 2y = 10

2x + 2y = -12

11x = -2 ⇒ x = -2/11

Now, substitute the value of x = -2/11 in equation (2)

⇒ -2/11 + y = -6

⇒ y = -6 + 2/11 ⇒ y = -64/11

Now, by solving the equations of the lines AD and CF we get the coordinates of the orthocenter (H) are (-2/11, -64/11).


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