Question

ABC is a triangle and \(\mathrm{G}(4, 3)\) is its centroid. If \(\mathrm{A}(1, 3), \mathrm{B}(4, b)\) and \(\mathrm{C}(a, 1)\) be the vertices, find the values of a and b and hence find the length of side BC.

Answer 1

Using the centroid formula:

$$4 = \frac{1 + 4 + a}{3}$$

$$\Rightarrow 12 = 5 + a$$

$$\Rightarrow a = 7$$

$$3 = \frac{3 + b + 1}{3}$$

$$\Rightarrow 9 = b + 4$$

$$\Rightarrow b = 5$$

So $$a = 7$$ and $$b = 5$$, i.e. $$B(4, 5)$$ and $$C(7, 1)$$.

Now the length of $$BC$$:

$$BC = \sqrt{(7 - 4)^2 + (1 - 5)^2}$$

$$= \sqrt{3^2 + (-4)^2}$$

$$= \sqrt{9 + 16}$$

$$= \sqrt{25}$$

$$= 5\text{ units}$$