In triangle ABC, the co-ordinates of vertices A, B and C are (4, 7), (-2, 3) and (0, 1) respectively. Find the equation of median through vertex A.

Also, find the equation of the line through vertex B and parallel to AC.

Let AD be the median through A. So, D will be the mid-point of BC.

Co-ordinates of D = $\Big(\dfrac{-2 + 0}{2}, \dfrac{3 + 1}{2}\Big) = \Big(\dfrac{-2}{2}, \dfrac{4}{2}\Big) = (-1, 2)$.

By formula,

Slope = $\dfrac{y$2 - y1}{x2 - x1}

$\text{Slope of AD} = \dfrac{2 - 7}{-1 - 4} \\[1em] = \dfrac{-5}{-5} = 1.$

By point-slope form,

Equation : y - y₁ = m(x - x₁)

Substituting values we get,

⇒ y - 7 = 1(x - 4)

⇒ y - 7 = x - 4

⇒ x - y - 4 + 7 = 0

⇒ x - y + 3 = 0.

$\text{Slope of AC} = \dfrac{1 - 7}{0 - 4} \\[1em] = \dfrac{-6}{-4} = \dfrac{3}{2}.$

Since, parallel lines have equal slope, equation of line passing through B and parallel to AC is

⇒ y - 3 = $\dfrac{3}{2}$[x - (-2)]

⇒ 2(y - 3) = 3(x + 2)

⇒ 2y - 6 = 3x + 6

⇒ 3x - 2y + 12 = 0.

Hence, equation of median through A is x - y + 3 = 0 and equation of line passing through B and parallel to AC is 3x - 2y + 12 = 0.