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

\[y - 1 = \frac{0 - 1}{2 - 0}\left( x - 0 \right)\]

\[ \Rightarrow y - 1 = \frac{- 1}{2}\left( x - 0 \right)\]

\[ \Rightarrow 2y - 2 = - x\]

\[ \Rightarrow x + 2y - 2 = 0\]

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

\[y + 2 = \frac{\frac{1}{2} + 2}{1 + 1}\left( x + 1 \right)\]

\[y + 2 = \frac{5}{4}\left( x + 1 \right)\]

\[ \Rightarrow 4y + 8 = 5x + 5\]

\[ \Rightarrow 5x - 4y - 3 = 0\]