Medians0%

The Straight Line · Topic 8 of 9

Medians

Video lesson · from 48:231 worked example

One lesson video covers all of The Straight Line, so it opens at 48:23 for this topic — not from the beginning.

Theory

Median

A line which joins a vertex to the midpoint of the opposite side.

ABCM

To find the equation:

  1. Find the midpoint of the opposite side.
  2. Find the gradient between the vertex and the midpoint.
  3. Use y−b=m(x−a)y - b = m(x - a) with either the vertex or the midpoint.

The three medians of a triangle meet at the centroid.

Centroid

⚠️ Common Examiner Traps

  • A median goes to a midpoint: so find the midpoint of the opposite side first. No perpendicular gradient is involved — that is an altitude.
  • Use the vertex and the midpoint: the gradient is between those two points, then apply y−b=m(x−a)y - b = m(x - a).
  • Match vertex to opposite side: the median from AA goes to the midpoint of BCBC.
  • Do not confuse the three cevians: median goes to a midpoint, altitude is perpendicular to the opposite side, perpendicular bisector does both but starts from the midpoint.

Worked examples

Example 1

Triangle ABC has vertices A(4,−9)A(4, -9), B(10,2)B(10, 2) and C(4,−4)C(4, -4). Find the equation of the median from A.

The median from A meets the midpoint of BC.

1. Midpoint of BC:

M=(10+42,2−42)=(7,−1)\begin{aligned} M &= \left(\frac{10+4}{2}, \frac{2-4}{2}\right) \\ &= (7, -1) \end{aligned}

2. Gradient from A to M:

mAM=−1−(−9)7−4=83\begin{aligned} m_{AM} &= \frac{-1 - (-9)}{7 - 4} \\ &= \frac{8}{3} \end{aligned}

3. Equation (using point A):

y−(−9)=83(x−4)y - (-9) = \frac{8}{3}(x - 4)
3(y+9)=8(x−4)3(y + 9) = 8(x - 4)
3y+27=8x−323y + 27 = 8x - 32
3y=8x−59 or 8x−3y−59=03y = 8x - 59 \text{ or } 8x - 3y - 59 = 0