HOW TO FIND THE SLOPES OF MEDIAN AND ALTITUDE OF A TRIANGLE

Example 1 :

A triangle has vertices at (6, 7), (2, -9) and (-4, 1). Find the slopes of its medians.

Solution :

Let A(6, 7), B(2, -9) and C(-4, 1) be the vertices of triangle.

Midpoint of the side BC = D

Midpoint of the side AC = E

Midpoint of the side AB = F

Medians of triangle are AD, BE and CF.

Slope of Median AD :

To find the slope median AD, find the point D which is the midpoint of B and D.

Midpoint formula = ((x₁ + x₂)/2, (y₁ + y₂)/2)

Midpoint of BC = ((2 - 4)/2, (-9 + 1)/2)

= (-2/2, -8/2)

= D(-1, -4)

Using the points A and D, find the slope of the median AD.

A(6, 7), D(-1, -4)

Slope of median AD = (y₂ - y₁)/(x₂ - x₁)

= (-4 - 7)/(-1 - 6)

= -11/(-7)

= 11/7

Slope of Medina BE :

A(6, 7), C(-4, 1)

Midpoint of the side AC = ((6 -4)/2, (7 + 1)/2)

=  (2/2, 8/2)

= E(1, 4)

Using the points B and E, find the slope of the median BE.

B(2, -9), E(1, 4)

Slope of median BE = (y₂ - y₁)/(x₂ - x₁)

= (4 + 9)/(1 - 2)

= 13/(-1)

= -13

Slope of Medina CF :

A(6, 7), B(2, -9)

Midpoint of the side AC = (6 + 2)/2, (7 - 9)/2)

= (8/2, -2/2)

= F(4, -1)

Using the points C and F, find the slope of the median CF.

C(-4, 1), F(4, -1)

Slope of median CF = (y₂ - y₁)/(x₂ - x₁)

= (-1 - 1)/(4 + 4)

= -2/8

= -1/4

Example 2 :

A triangle has vertices at (-5, 7), (-4, -5) and (4, 5). Find the slopes of its altitudes.

Solution :

Let A(-5, 7), B(-4, -5) and C(4, 5) be the vertices of triangle.

Altitude drawn through the vertices of the triangle will be perpendicular to the other sides.

Slope of altitude AD :

Slope of BC = (y₂ - y₁)/(x₂ - x₁)

= (5 + 5) / (4 + 4)

= 10/8

= 5/4

Slope of AD = -1/slope of BC

= -1/(5/4)

= -4/5

Slope of altitude BE :

Slope of AC = (y₂ - y₁)/(x₂ - x₁)

= (5 - 7)/(4 + 5)

= -2/9

Slope of BE = -1/slope of AC

= -1/(-2/9)

= 9/2

Slope of altitude CF :

Slope of AB = (y₂ - y₁)/(x₂ - x₁)

= (-5 - 7)/(-4 + 5)

= -12/1

= -12

Slope of CF = -1/slope of AB

= -1/(-12)

= 1/12

Example 3 :

A triangle has vertices A(-4, 2), B(-2, -6), and C(6, -2).

a) Determine the length of the median from vertex A.

b) Determine an equation in the form y = mx + b for the median from vertex A.

Solution :

Median drawn from vertex A will divide bisect the opposite side BC.

Midpoint of BC = (x₁ + x₂)/2, (y₁ + y₂)/2

= (-2 + 6)/2, (-6 - 2)/2

= 4/2, -8/2

= (2, -4)

a) Length of median = √(x₂ - x₁)² + (y₂ - y₁)²

A(-4, 2) and (2, -4)

= √(2+4)² + (-4-2)²

= √6² + (-6)²

= √(36+36)

= √72

= 6√2

b) 

Slope of median :

A(-4, 2) and (2, -4)

m = (y₂ - y₁)/(x₂ -x₁)

= (-4 - 2) / (2 + 4)

= -6/6

= -1

Equation of median :

(y - y₁) = m(x - x₁)

Applying the point A(-4, 2) 

(y - 2) = -1(x - (-4))

y - 2 = -1(x + 4)

y - 2 = -x - 4

y = -x - 4 + 2

y = -x - 2

Example 4 :

Determine an equation for the right bisector of the line segment with endpoints D(-3, 5) and M(7, -9).

Solution :

Midpoint of the line joining the points D(-3, 5) and M(7, -9)

Midpoint = (x₁ + x₂)/2, (y₁ + y₂)/2

= (-3 + 7)/2, (5 + (-9))/2

= 4/2, -4/2

= (2, -2)

Slope of the line joining the points D(-3, 5) and M(7, -9)

= (-9 - 5) / (7 - (-3))

= -14/(7 + 3)

= 14/10

= 7/5

The right bisector and the line joining the given points will be perpendicular. So, the product of their slopes will be equal to -1.

Equation of right bisector :

(y - y₁) = m(x - x₁)

y - (-2) = (7/5)(x - 2)

5(y + 2) = 7(x - 2)

5y + 10 = 7x - 14

7x - 5y - 14 - 10 = 0

7x - 5y - 24 = 0

Example 5 :

Given triangle XYZ with X (-1, 6), Y (-4, 0) and Z (3, 2), draw median from vertex X. Calculate the slope of the median.

Solution :

Median drawn from the vertex X will divide YZ into two equal parts.

Midpoint of YZ = (-4 + 3)/2, (0 + 2)/2

= (-1/2, 2/2)

= (-1/2, 1)

Slope of median :

X(-1, 6) and (-1/2, 1)

= (1 - 6)/(-1/2 + 1)

= -5/(1/2)

= -10

So, the slope of the median is -10.

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