# EQUATION OF THE LINE ALONG THE ALTITUDE OR MEDIAN OF TRIANGLE

Equation of the Line Along the Altitude or Median of Triangle :

Here we are going to see how to find the equation of the line along the altitude or median of the triangle.

## Equation of the Line Along the Altitude or Median of Triangle - Examples

Question 14 :

Vertices of triangle ABC are A (2,-4) ,B(3,3) and  C (-1,5). Find the equation of the straight line along the altitude from the vertex B.

Solution :

A line drawn from the vertex B is perpendicular to the line AC.So the product of their slopes will be equal to -1.

Slope of AC = (y₂ - y₁)/(x₂ - x₁)

x₁ = 2, y₁ = -4, x₂ = -1 , y₂ = 5

=  (5 - (-4))/(-1 - 2)

=  (5 + 4)/(-3)

=  -9/3 = -3

slope of line drawn from the vertex B = -1/m

= -1/(-3)  =  1/3

Equation of the line drawn from the vertex B:

(y - y₁) = m (x - x₁)

B (3,3)  slope = 1/3

x₁ = 3, y₁ = 3

(y - 3) = (1/3) (x - 3)

3 (y - 3) = 1 (x - 3)

3 y - 9 = x - 3

x - 3y - 3 + 9 = 0

x - 3y + 6 = 0

Question 15 :

If the vertices of triangle ABC are A (-4,4), B(8,4) and C(8,10). Find the equation of the along the median from the vertex A.

Solution :

Since AD is median,it passes through the midpoint of the side BC.

x₁ = 8, y₁ = 4, x₂ = 8 , y₂ = 10

=  [(x + x)/2 , (y₁ + y₂)/2]

=  [(8 + 8)/2 , (4 + 10)/2]

=  16/2 , 14/2

=  (8,7)

(y - y₁)/(y₂ - y₁) =   (x - x₁)/(x₂ - x₁)

(-4,4) (8,7)

x₁ = -4, y₁ = 4, x₂ = 8 , y₂ = 7

(y -4)/(7-4) = (x-(-4))/(8-(-4))

(y -4)/3 = (x+4)/(8+4)

(y -4)/3 = (x+4)/12

12(y - 4) = 3(x + 4)

12 y - 48 = 3 x + 12

3 x - 12 y + 12 + 48 = 0

3 x - 12y + 60 = 0

Dividing the whole equation by 3

x - 4 y + 20 = 0

After having gone through the stuff given above, we hope that the students would have understood, how to find the equation of the line along the altitude or median of triangle

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