EQUATION OF MEDIAN OF A TRIANGLE

In a triangle, median is a line segment joining a vertex to the midpoint of the corresponding opposite side. There are three medians for a triangle. 

In ΔABC shown below, D is the midpoint of side BC and AD is the median through the vertex A. 

We can find the equation of median as explained below. 

Step 1 : 

Using midpoint formula, find the midpoint of BC, which is D.

Step 2 : 

Find the slope of median AD using the points A and D.

Step 3 : 

Using point-slope form equation y - y1 = m(x - x1), find the equation of the median AD. 

Note : 

When point-slope form is used to find the equation of median, usually the point at the vertex is taken. 

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

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