Question

$A(5,4),B(-3,-2)$ and $C(1,-8)$ are the vertices of a $△ABC$. Find the equations of median AD and line parallel to AC passing through the point B.

Answer 1

Here, $A(5,4),B(-3,-2)$ and $C(1,-8)$ and AD is median.
$\therefore$ D is the midpoint of BC
$\therefore D=(\frac{-3+1}{2},\frac{-2(-8)}{2})$
$=(-1,-5)$
and equation of A(5, 4) and $D(-1,-5)$ is
$\frac{y-4}{x-5}=\frac{4-(-5)}{5-(-1)}=\frac{9}{6}=\frac{3}{2}$
$2y-8=3x-15$
$\therefore 3x-2y-7=0$ is the equation of median AD.
Now, Slope of $AC=\frac{4-(-8)}{5-1}$ $=\frac{12}{4}$$=3$
The required line is parallel to AC and passes through $B(-3,-2)$.
So, equation of required line is
$y-(-2)=3[x-(-3\left)\right]$

$y+2=3x+9$

$y=3x+7$

Hence, $3x-y+7=0$ is the required equation of line.