Question

Find the equations of the medians of a \(\triangle\) ABC whose vertices are A(2, 5), B(-4, 9) and C(-2, -1).

Answer 1

The vertices of \(\triangle\) ABC are A(2, 5), B(-4, 9) and C(-2, -1).

Let D, E, F be the midpoints of BC, CA and AB respectively.

Then, these points are

D \(\left (\frac{-4-2}{2}, \frac{9-1}{2} \right ), \left (\frac{-2+2}{2}, \frac{-1+5}{2} \right )\) and F \(\left (\frac{2-4}{2}, \frac{5+9}{2} \right )\)

i.e., D(-3 , 4), E(0, 2) and F(-1, 7).

Thus, AD, BE and CF are the medians of \(\triangle \) ABC.

Equation of median AD is \(\frac{y-5}{x-2} = \frac{4-5}{-3-2} \Rightarrow \frac{y-5}{x-2} = \frac{1}{5}\)

\(\Rightarrow 5(y-5)=x-2\)

\(\Rightarrow x - 5y + 23 = 0\)

Equation of median BE is \(\frac{y-9}{x+4} = \frac{2-9}{0+4} \Rightarrow \frac{y-9}{x+4} = \frac{-7}{4}\)

\(\Rightarrow 4(y-9) + 7(x + 4) = 0\)

\(\Rightarrow 7x + 4y - 8 = 0\)

Equation of median CF is \(\frac{y+1}{x+2} = \frac{7+1}{-1+2} \Rightarrow (y+1)=8(x+2)\)

\(\Rightarrow 8x - y + 15 = 0\)