Question

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

Answer 1

Given: Vertices of a $\Delta ABC$ are A (5, 4), B (-3, -2) and C (1, -8)
Let $A(5,4)=(x_1,y_1),B(-3,-2)=(x_y,y_2),C(1,-8)=(x_3,y_3)$ and D is the midpoint of BC = D(x, y)
$\therefore$ Co-ordinate of D by midpoint formula.
$\frac{x_2+x_3}{2}=\frac{y_2+y_3}{2}$
$\frac{-3+1}{2}=\frac{-2-8}{2}$
$\frac{-2}{2}=\frac{-10}{2}$
$\therefore -1,-5$
Co-ordinate of D (-1, -5)
Equation of median AD in two point form
$\frac{x-x_1}{x_1-x_2}=\frac{y-y_1}{y_1-y_2}$
$\frac{x-5}{5-(-1)}=\frac{y-4}{4-(-5)}$
$\frac{x-5}{5+1}=\frac{y-4}{4+5}$
$9(x-5)=6(y-4)$
$9x-45=6y-24$
$9x-6y-21=0$
$3x-2y-7=0$ (Dividing all by 3)
$\therefore$ Equation of median $AD=3x-2y-7=0$
If a line is || to AC then slope of that line has to be equal to slope of AC
Let $A(5,4)=(x_1,y_1),C(1,-8)=(x_2,y_2)$
Slope of line $AC=\frac{y_2-y_1}{x_2-x_1}=\frac{-8-4}{1-5}=\frac{-12}{-4}=3$
$\therefore$ Slope of line $AC=3$
$\therefore$ Slope of line passing through $B(-3,-2)$ and parallel to $AC$ is $3$
$\therefore$ Slope $\left(m\right)=3$
By point - slope form
$y-y_1=m(x-x_1)$
$y-(-2)=3(x-(-3\left)\right)$
$y+2=3(x+3)$
$y+2=3x+9$
$-3x+y+2-9=0$
$-3x+y-7=0$
$\therefore$ Equation of median $AD$ is $3x-2y-7=0$
The equation of line parallel to $AC$ and passing through point $B$ is $3x-y+7=0.$