HOW TO FIND THE EQUATION OF ALTITUDE OF A TRIANGLE

Example :

A(-3, 0) B(10, -2) and C(12, 3) are the vertices of ΔABC. Find the equations of the altitudes through A, B and C in general form.

Solution :

Equation of the altitude through A :

slope of BC = (y₂ - y₁)/(x₂ - x₁)

Substitute (x₁, y₁) = B(10, -2) and (x₂, y₂) = C(12, 3).

= (3 + 2)/(12 - 10)

slope of BC = 5/2

slope of AD = -1/slope of BC

= -1/(5/2)

= -2/5

Equation of the altitude AD :  

y = mx + b

Substitute m = 5/2.

y = (-2/5)x + b ----(1)

Substitute (x, y) = A(-3, 0). 

0 = (-2/5)(-3) + b

0 = 6/5 + b

-6/5 = b

(1)----> y = (-2/5)x - 6/5

Multiply each side by 5.

5y = -2x - 6

2x + 5y + 6 = 0

Equation of the altitude through B :

slope of AC = (y₂ - y₁)/(x₂ - x₁)

Substitute (x₁, y₁) = A(-3, .0) and (x₂, y₂) = C(12, 3).

= (3 - 0)/(12 + 3)

= 3/15

slope of AC = 1/5

slope of BE = -1/slope of AC

= -1/(1/5)

= -5

Equation of the altitude BE :  

y = mx + b

Substitute m = -5.

y = -5x + b ----(2)

Substitute (x, y) = B(10, -2). 

-2 = -5(10) + b

-2 = -50 + b

48 = b

(2)----> y = -5x + 48

5x + y - 48 = 0

Equation of the altitude through C :

slope of AB = (y₂ - y₁)/(x₂ - x₁)

Substitute (x₁, y₁) = A(-3, 0) and (x₂, y₂) = B(10, -2).

= (-2 - 0)/(10 + 3)

= -2/13

slope of AB = -2/13

slope of CF = -1/slope of AB

= -1/(-2/13)

= 13/2

Equation of the altitude CF :  

y = mx + b

Substitute m = 13/2.

y = (13/2)x + b ----(3)

Substitute (x, y) = C(12, 3). 

3 = (13/2)(12) + b

3 = 78 + b

-75 = b

(3)----> y = (13/2)x - 75

Multiply each side by 2. 

2y = 13x - 150

-13x + 2y + 150 = 0

13x - 2y - 150 = 0

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