Question

A (5, 4), B (–3,–2) and C (1,–8) are the vertices of a triangle ABC. Find the equation of median AD and line parallel to AB passing through point C.

Answer 1

It is given that A(5, 4), B(–3, –2) and C(1, –8) are the vertices of a triangle ABC. 
Let the coordinates of point D be (a, b).
Also, D is the mid-point of side BC

So, 

\[a = \frac{- 3 + 1}{2}, b = \frac{- 2 + \left( - 8 \right)}{2}\]

\[ \Rightarrow a = \frac{- 2}{2}, b = \frac{- 10}{2}\]

\[ \Rightarrow a = - 1, b = - 5\]

Thus, the coordinates of point D are (−1, −5).

The equation of median AD is

\[y - y_1 = \left( \frac{y_2 - y_1}{x_2 - x_1} \right)\left( x - x_1 \right)\]

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

\[ \Rightarrow y - 4 = \left( \frac{- 9}{- 6} \right)\left( x - 5 \right)\]

\[ \Rightarrow y - 4 = \frac{3}{2}\left( x - 5 \right)\]

\[ \Rightarrow 2y = 3x - 7\]

A line parallel to AB will have the same slope as AB. 

Slope of AB = \[\frac{y_2 - y_1}{x_2 - x_1} = \frac{- 2 - 4}{- 3 - 5} = \frac{- 6}{- 8} = \frac{3}{4}\]

So, the equation of line parallel to AB and passing through the point C(1, –8) is

\[y - y_1 = m\left( x - x_1 \right)\]

\[ \Rightarrow y - \left( - 8 \right) = \frac{3}{4}\left( x - 1 \right)\]

\[ \Rightarrow 3x - 4y - 35 = 0\]