Question

Find the equations of the medians of $$\triangle \mathrm{ABC}$$ whose vertices are $$\mathrm{A}(-1, 2), \mathrm{B}(2, 1)$$ and $$\mathrm{C}(0, 4)$$. Hence, find the co-ordinates of the centroid of $$\triangle \mathrm{ABC}$$.

Answer 1

Let $$D$$, $$E$$ and $$F$$ be the mid-points of sides $$BC$$, $$CA$$ and $$AB$$ respectively.

Mid-point of $$BC$$: 

$$D = \left(\frac{2 + 0}{2}, \frac{1 + 4}{2}\right)$$

$$= \left(1, \frac{5}{2}\right)$$

Equation of median $$AD$$ passing through $$A(-1, 2)$$ and $$D(1, 2.5)$$:

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

$$= \frac{\frac{1}{2}}{2}$$

$$= \frac{1}{4}$$

$$y - 2 = \frac{1}{4}(x - (-1))$$ 

$$4(y - 2) = x + 1$$ 

$$4y - 8 = x + 1$$ 

$$x - 4y + 9 = 0$$

Mid-point of $$CA$$: 

$$E = \left(\frac{0 + (-1)}{2}, \frac{4 + 2}{2}\right)$$

$$= \left(-\frac{1}{2}, 3\right)$$

Equation of median $$BE$$ passing through $$B(2, 1)$$ and $$E(-0.5, 3)$$: 

$$m_{BE} = \frac{3 - 1}{-\frac{1}{2} - 2}$$

$$= \frac{2}{-\frac{5}{2}}$$

$$= -\frac{4}{5}$$ 

$$y - 1 = -\frac{4}{5}(x - 2)$$ 

$$5(y - 1) = -4(x - 2)$$ 

$$5y - 5 = -4x + 8$$ 

$$4x + 5y - 13 = 0$$

Mid-point of $$AB$$: 

$$F = \left(\frac{-1 + 2}{2}, \frac{2 + 1}{2}\right)$$

$$= \left(\frac{1}{2}, \frac{3}{2}\right)$$

Equation of median $$CF$$ passing through $$C(0, 4)$$ and $$F(0.5, 1.5)$$: 

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

$$= \frac{-\frac{5}{2}}{\frac{1}{2}}$$

$$= -5$$ 

Since $$C(0, 4)$$ lies on the $$y$$-axis, $$c = 4$$: 

$$y = -5x + 4$$ 

$$5x + y - 4 = 0$$

Centroid of $$\triangle ABC$$: 

$$x = \frac{-1 + 2 + 0}{3} = \frac{1}{3}$$ 

$$y = \frac{2 + 1 + 4}{3} = \frac{7}{3}$$

So, the coordinates of the centroid are $$\left(\frac{1}{3}, \frac{7}{3}\right)$$.