Question

$A(5,4),B(-3,-2)$ and $C(1,-8)$ are the vertices of a triangle ABC. Find the equation of median AD.

Answer 1

Given $A(5,4),B(-3,-2),C(1,-8)$ are the vertices of $\Delta ABC$.

Segment $AD$ is the median.

$\therefore D$ is the midpoint of $BC$.

Hence by midpoint formula, $D\equiv (\frac{-3+1}{2},\frac{-2-8}{2})$

$\Rightarrow D\equiv (\frac{-2}{2},\frac{-10}{2})$

$\therefore D\equiv (-1,-5)$

The equation of the median $AD$ in two point form is $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}$

Here $(5,4)\equiv (x_1,y_1),(-1,-5)\equiv (x_2,y_2)$

$\therefore \frac{x-5}{-1-5}=\frac{y-4}{-5-4}$

$\Rightarrow \frac{x-5}{-6}=\frac{y-4}{-9}$

$\Rightarrow \frac{x-5}{2}=\frac{y-4}{3}$

$\Rightarrow 3(x-5)=2(y-4)$

$\Rightarrow 3x-15=2y-8$

$\Rightarrow 3x-2y-15+8=0$

$\Rightarrow 3x-2y-7=0$

Hence the equation of the median $AD$ is $3x-2y-7=0$