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 AC passing through point B.

Answer 1

Given: Points A(5, 4), B($-$3, $-$2) and C(1, $-$8) are the vertices of $△$ABC and AD is the median of the triangle, i.e., D is the midpoint of side BC.


∴ x-coordinate of D = $\frac{-3+1}{2}=\frac{-2}{2}=-1$

Also, y-coordinate of D = $\frac{-8-2}{2}=\frac{-10}{2}=-5$

The coordinates of point D is ($-$1,$-$5).
Now, slope of the median AD is given by

$$\begin{gathered} m=\frac{y_1-y_2}{x_1-x_2} \\ =\frac{4-(-5)}{5-(-1)} \\ =\frac{9}{6} \\ =\frac{3}{2} \end{gathered}$$

So, equation of the median AD is given below:
y $-$ y₁ = m (x $-$ x₁)
⇒ y $-$ 4 = $\frac{3}{2}$(x $-$ 5)
⇒ 2y $-$ 8 = 3x $-$15
⇒ 3x $-$ 2y $-$ 7 = 0

Slope of the line AC is given below:

$$\begin{gathered} m=\frac{y_1-y_2}{x_1-x_2} \\ =\frac{4-(-8)}{5-1} \\ =\frac{12}{4} \\ =3 \end{gathered}$$

We know that parallel lines have equal slopes.
So, the equation of the line parallel to AC and passing through B is given by

y $-$ y₁ = m (x $-$ x₁)
⇒ y $-$ ($-$2) = 3{x $-$ ($-$3)}
⇒ y + 2 = 3(x + 3)
⇒ y + 2 = 3x + 9
⇒ 3x $-$ y + 7 = 0