EQUATION OF MEDIAN OF A TRIANGLE

Example 1 :

Find the equations of three medians of ΔABC with vertices A(1, -3), B(-2, 5) and C(-3, 4).

Solution :

Let D, E and F be the midpoints of the sides AB, BC, and AC respectively.

Equation of Median AD :

Midpoint of BC :

= ((x₁ + x₂)/2, (x₁ + x₂)/2)

Substitute (x₁, y₁) = B(-2, 5) and (x₂, y₂) = C(-3, 4). 

= D((-2 - 3)/2, (5 + 4)/2)

= D(-5/2, 9/2)

Slope of AD :

= (y₂ - y₁)/(x₂ - x₁)

Substitute (x₁, y₁) = A(1, -3) and (x₂, y₂) = D(-5/2, 9/2). 

= (9/2 + 3)/(-5/2 - 1) 

= (15/2)/(-7/2)

= -15/7

Equation of median AD :

y - y₁ = m(x - x₁)

Substitute (x₁, y₁) = A(1, -3) and m = -15/7.

y + 3 = (-15/7)(x - 1)

7(y + 3) = -15(x - 1)

7y + 21 = -15x + 15

15x + 7y + 6 = 0

Equation of Median BE :

Midpoint of AC :

= ((x₁ + x₂)/2, (x₁ + x₂)/2)

Substitute (x₁, y₁) = A(1, -3) and (x₂, y₂) = C(-3, 4). 

= E((1 - 3)/2, (-3 + 4)/2)

= E(-1, 1/2)

Slope of BE :

= (y₂ - y₁)/(x₂ - x₁)

Substitute (x₁, y₁) = B(-2, 5) and (x₂, y₂) = E(-1, 1/2). 

= (1/2 - 5)/(-1 + 2) 

= (-9/2)/1

= -9/2

Equation of median BE :

y - y₁ = m(x - x₁)

Substitute (x₁, y₁) = B(-2, 5) and m = -9/2.

y - 5 = (-9/2)(x + 2)

2(y - 5) = -9(x + 2)

2y - 10 = -9x - 18

9x + 2y + 8 = 0

Equation of Median CF :

Midpoint of AB :

= ((x₁ + x₂)/2, (x₁ + x₂)/2)

Substitute (x₁, y₁) = A(1, -3) and (x₂, y₂) = B(-2, 5). 

= F((1 - 2)/2, (-3 + 5)/2)

= E(-1/2, 1)

Slope of CF :

= (y₂ - y₁)/(x₂ - x₁)

Substitute (x₁, y₁) = C(-3, 4) and (x₂, y₂) = F(-1/2, 1). 

= (1 - 4)/(-1/2 + 3) 

= -3/(5/2)

= -6/5

Equation of median CF :

y - y₁ = m(x - x₁)

Substitute (x₁, y₁) = C(-3, 4) and m = -6/5.

y - 4 = (-6/5)(x + 3)

5(y - 4) = -6(x + 3)

5y - 20 = -6x - 18

6x + 5y - 2 = 0

Example 2 :

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

Solution :

Let D be the midpoint of the side BC. 

Midpoint of BC :

= ((x₁ + x₂)/2, (x₁ + x₂)/2)

Substitute (x₁, y₁) = B(8, 4) and (x₂, y₂) = C(8, 10). 

= D((8 + 8)/2, (4 + 10)/2)

= D(8, 7)

Slope of AD :

= (y₂ - y₁)/(x₂ - x₁)

Substitute (x₁, y₁) = A(-4, 4) and (x₂, y₂) = D(8, 7). 

= (7 - 4)/(8 + 4) 

= 3/12

= 1/4

Equation of the line along the median from the vertex A :

y - y₁ = m(x - x₁)

Substitute (x₁, y₁) = A(-4, 4) and m = 1/4.

y - 4 = (1/4)(x + 4)

4(y - 4) = 1(x + 4)

4y - 16 = x + 4

x - 4y + 20 = 0

Example 3 :

The vertices of ∆ PQR are P(2, 1), Q(−2, 3) and R (4, 5). Find the equation of the median through the vertex R. 

Solution :

Median which passes through vertex R will divide the opposite side into two equal parts.

Midpoint = (2 + (-2)) / 2, (1 + 3)/2

=  (0/2, 4/2)

= S (0, 2)

Equation of the line joining the points R (4, 5) and S (0, 2)

Slope of RS :

= (2 - 5) / (0 - 4)

= -3/(-4)

= 3/4

Slope of the line RS = 3/4

Equation of median line :

y = mx + b

2 = (3/4) (0) + b

b = 2

y = (3/4) x + 2

Example 4 :

For triangle A(-6, 6) B(4, -4) C(2, 10) find the equation of the median from vertex B (start by making a rough sketch of the median on the diagram) :

The median drawn from the point B will bisect the side AC. The midpoint of AC is D.

Midpoint of AC = (-6 + 2)/2, (6 + 10)/2

= -4/2, 16/2

= D (-2, 8)

Slope of BD :

B(4, -4) and D(-2, 8)

Slope = (8 - (-4)) / (-2 - 4)

= (8 + 4) / (-6)

= 12/(-16)

= -2

Equation of median BD :

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

y + 4 = -2(x - 4)

y = -2x + 8 - 4

y = -2x + 4

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