Question

Find the equation of the side BC of the triangle ABC whose vertices are (−1, −2), (0, 1) and (2, 0) respectively. Also, find the equation of the median through (−1, −2).

Answer 1

The vertices of triangle ABC are A (−1, −2), B (0, 1) and C (2, 0).

So, the equation of BC is

$$\begin{gathered} y-1=\frac{0-1}{2-0}\left(x-0\right) \\ \Rightarrow y-1=\frac{-1}{2}\left(x-0\right) \\ \Rightarrow 2y-2=-x \\ \Rightarrow x+2y-2=0 \end{gathered}$$

Let D be the midpoint of BC.

$\therefore D\equiv \left(\frac{0+2}{2},\frac{1+0}{2}\right)\equiv \left(1,\frac{1}{2}\right)$

So, the equation of median AD is

$$\begin{gathered} y+2=\frac{{\displaystyle \frac{1}{2}}+2}{1+1}\left(x+1\right) \\ y+2=\frac{5}{4}\left(x+1\right) \\ \Rightarrow 4y+8=5x+5 \\ \Rightarrow 5x-4y-3=0 \end{gathered}$$