Question

The vertices of a triangle are A(3, 4), B(2, 0), and C(−1, 6). Find the equation of the line containing the midpoints of sides AB and BC

Answer 1

Let P, Q be the midpoints of sides AB and BC respectively.

Then P ≡ ${\displaystyle (\frac{3+2}{2},\frac{4+0}{2})=(\frac{5}{2},2)}$

and Q ≡ ${\displaystyle (\frac{2-1}{2},\frac{0+6}{2})=(\frac{1}{2},3)}$

∴ equation of the required line, i.e., line PQ is

${\displaystyle \frac{y-2}{x-\frac{5}{2}}=\frac{3-2}{\frac{1}{2}-\frac{5}{2}}}$

∴ ${\displaystyle \frac{2y-4}{2x-5}=\frac{1}{-2}}$

∴ – 4y + 8 = 2x – 5

∴ 2x + 4y –  13 = 0.