Medians And Centroid Of A Triangle

This is a continuation of my previous post. The problems presented here are about the medians of a triangle and their point of concurrency.

The following theorems are applied in the given problems below:

• In a right triangle,the length of the median to the hypotenuse is one half the length of the hypotenuse (see Problem 1).
• In an isosceles triangle, the median to the base is perpendicular to the base itself (see Problem 2).
• The medians of a triangle have a point of concurrency (see Problem 2).

Problem 1

Given a right triangle ABC with vertices at A(-2, 5), B(2, -5) and C(-5, 2) and its median that intersects the hypotenuse, find the length of hypotenuse AB if median CY = √29.

Discussion

Point Y is the midpoint of hypotenuse AB.

Line segment CY is a median, a line drawn from a vertex to the midpoint of the opposite side.

Solution

Distance Between Two Points

• distance_AC = √ 18
• distance_AB = √ 116
• distance_BC = √ 98
• distance_CY = distance_AY = distance_BY = √ 29

Equations of the Line Segments

• line segment AB: 5x + 2y = 0; x-intercept = 0; y-intercept = 0
• line segment BC: x + y + 3 = 0; x-intercept = -3; y-intercept = -3
• line segment AC: x - y + 7 = 0; x-intercept = -7; y-intercept = 7
• line segment CY: 2x + 5y = 0; x-intercept = 0; y-intercept = 0

Slopes of the Line Segments

• m_AB = - 5/2
• m_BC = - 1
• m_AC = 1
• m_CY = - 2/5

Problem 2

Given isosceles triangle ABC with its medians AE, BF and CD, find their point of concurrency. The vertices of the triangle are A(-6, 5), B(4, 1) and C(-2, -5).

Discussion

In the above figure, the medians of isosceles triangle ABC intersect at point X which is called centroid and is the center of gravity of the triangle.

This point is the trisection point of each median.

Taking median AE, line segment AX = 2/3 AE = 2 EX and line segment EX = 1/3 AE. This property holds true for the other two medians as well.

Solution

Point of Concurrency of the Medians: X (- 4/3, 1/3)

Distance Between Two Points

• distance_AB = distance_AC = √ 116
• distance_BC = √ 72
• distance_AE = √ 98
• distance_BF = distance_CD = √ 65
• distance_AX = 2/3 distance = √ 392/9
• distance_EX = 1/3 distance_AE = √ 98/9
• distance_BX = 2/3 distance_BF = √ 260/9
• distance_FX = 1/3 distance_BF = √ 65/9
• distance_CX = 2/3 distance_CD = √ 260/9
• distance_DX = 1/3 distance_CD = √ 65/9

Midpoints of the Sides

• midpoint of side AB: D(-1, 3)
• midpoint of side BC: E(1, -2)
• midpoint of side AC: F(-4, 0)

Equations of the Line Segments

• line segment AB: 2x + 5y - 13 = 0; x-intercept = 13/2; y-intercept = 13/5
• line segment BC: x - y - 3 = 0; x-intercept = 3; y-intercept = - 3
• line segment AC: 5x + 2y + 20 = 0; x-intercept = - 4; y-intercept = - 10
• line segment AE: x + y + 1 = 0; x-intercept = - 1; y-intercept = - 1
• line segment BF: x - 8y + 4 = 0; x-intercept = - 4; y-intercept = 1/2
• line segment CD: 8x - y + 11 = 0; x-intercept = - 11/8; y-intercept = 11

Slopes of the Line Segments

• m_AB = - 2/5
• m_BC = 1
• m_AC = - 5/2
• m_AE = -1