HOW TO FIND THE EQUATION OF ALTITUDE OF A TRIANGLE

Example 1 :

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

Example 2 :

Write the equation of the perpendicular bisector that goes through the line segment with end points of (9, 5) and (-7, 13).

Solution :

Perpendicular bisector will divide the opposite side into equal parts and must create the angle measure 90 degree.

Midpoint of the line segment = (x₁ + x₂)/2, (y₁ + y₂)/2

= (9 - 7)/2, (5 + 13)/2

= 2/2, 18/2

= (1, 9)

Slope of the perpendicular bisector (or) slope of altitude

= -1/Slope of the line joining the given points 

= (y₂ - y₁)/(x₂ - x₁)

= (13 - 5)/(-7 - 9)

= 8/-16

= -1/2

Slope of altitude = -1/(-1/2)

= 2

Equation of altitude :

(y - y1) = m(x - x1)

y - 9 = 2(x - 1)

y = 2x - 2 + 9

y = 2x + 7

Example 3 :

Write the equation of the perpendicular bisector that goes through the line segment with end points of (10, -10) and B (2, 2).

Solution :

Midpoint of the line segment = (x₁ + x₂)/2, (y₁ + y₂)/2

= (10 + 2)/2, (-10 + 2)/2

= 12/2, -8/2

= (6, -4)

Slope of the perpendicular bisector (or) slope of altitude

= -1/Slope of the line joining the given points 

= (y₂ - y₁)/(x₂ - x₁)

= (2 - (-10))/(2 - 10)

= (2 + 10)/-8

= -12/8

= -3/2

Slope of altitude = -1/(-3/2)

= 2/3

Equation of altitude :

(y - y1) = m(x - x1)

y + 4 = (2/3)(x - 6)

3y + 12 = 2x - 12

2x - 3y - 12 + 12 = 0

2x - 3y = 0

Example 4 :

Write the equation of the perpendicular bisector that goes through the line segment with end points of (-10, -8) and B (-14 , 8).

Solution :

Midpoint of the line segment = (x₁ + x₂)/2, (y₁ + y₂)/2

= (-10 - 14)/2, (-8 + 8)/2

= -24/2, 0/2

= (-12, 0)

Slope of the perpendicular bisector (or) slope of altitude

= -1/Slope of the line joining the given points 

= (y₂ - y₁)/(x₂ - x₁)

= (8 + 8)/(-14 - 10)

= 16 / (-24)

= -2/3

Slope of altitude = -1/(-2/3)

= -3/2

Equation of altitude :

(y - y1) = m(x - x1)

y - 0 = (-3/2)(x + 12)

2y = -3(x + 12)

2y = -3x + 36

y = (-3/2)x + 36/2

y = (-3/2)x + 18

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