Find the equations of the medians of ΔABC whose vertices are A(–1, 2), B(2, 1) and C(0, 4). Hence, find the co-ordinates of the centroid of ΔABC.

By using Midpoint formula,

(x, y) = $\Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

D (Midpoint of BC): B(2, 1)and C(0, 4)

D = $\Big(\dfrac{2 + 0}{2}, \dfrac{1 + 4}{2}\Big) = \Big(1, \dfrac{5}{2})$

E (Midpoint of AC): A(-1, 2) and C(0, 4)

E = $\Big(\dfrac{-1 + 0}{2}, \dfrac{2 + 4}{2}\Big) = \Big(-\dfrac{1}{2}, 3\Big)$

F (Midpoint of AB): A(-1, 2) and B(2, 1)

F = $\Big(\dfrac{-1 + 2}{2}, \dfrac{2 + 1}{2}\Big) = \Big(\dfrac{1}{2}, \dfrac{3}{2}\Big)$

Median AD through A(-1, 2) and $D\Big(1, \dfrac{5}{2}\Big)$

By slope formula:

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

Substitute values we get:

$m_{AD} = \dfrac{\dfrac{5}{2} - 2}{1 - (-1)} = \dfrac{\dfrac{1}{2}}{2} = \dfrac{1}{4}$

The equation of the line will be given by two-point form i.e.,

y - y₁ = m(x - x₁)

Substituting values in above equation we get,

⇒ y - 2 = $\dfrac{1}{4}$ (x + 4)

⇒ 4(y - 2) = (x + 1)

⇒ 4y - 8 = x + 1

⇒ x - 4y + 9 = 0

Median BE through B(2, 1) and $E\Big(-\dfrac{1}{2}, 3\Big)$

$m_{BE} = \dfrac{3 - 1}{-\dfrac{1}{2} - 2} = \dfrac{2}{-\dfrac{5}{2}} = -\dfrac{4}{5}$

The equation of the line will be given by two-point form i.e.,

⇒ y - 1 = $-\dfrac{4}{5}$ (x - 2)

⇒ 5(y - 1) = 4(x - 2)

⇒ 5y - 5 = 4x - 8

⇒ 4x + 5y - 13 = 0

Median CF through C(0, 4) and $F\Big(\dfrac{1}{2}, \dfrac{3}{2}\Big)$

$m_{CF} = \dfrac{\dfrac{3}{2} - 4}{\dfrac{1}{2} - 0} = \dfrac{-\dfrac{5}{2}}{\dfrac{1}{2}} = -5$

Since C(0, 4) is the y-intercept, we use y = mx + c:

⇒ y = -5x + 4

⇒ 5x + y - 4 = 0

By using centroid formula,

$G = \Big(\dfrac{x$1 + x2 + x3}{3}, \dfrac{y1 + y2 + y3}{3}\Big)

Using A(-1, 2), B(2, 1), and C(0, 4):

$\Rightarrow G = \Big(\dfrac{-1 + 2 + 0}{3}, \dfrac{2 + 1 + 4}{3}\Big) \\[1em] = \Big(\dfrac{-1 + 2 + 0}{3}, \dfrac{2 + 1 + 4}{3}\Big) \\[1em] = \Big(\dfrac{1}{3}, \dfrac{7}{3}\Big).$

Hence, equations of medians are x - 4y + 9 = 0, 4x + 5y - 13 = 0 and 5x + y - 4 = 0, coordinates of centroid $\Big(\dfrac{1}{3}, \dfrac{7}{3}\Big)$.