Find the third vertex of ∆ABC if two of its vertices are B(−3, 1) and C(0, −2)

Two vertices of ∆ABC are B(−3,1) and C(0, −2). Let the third vertex be A(a, b).
Then, the coordinates of its centroid are

$\left(\frac{-3+0+a}{3}, \frac{1-2+b}{3}\right)$

i. e. $\left(\frac{-3+a}{3}, \frac{-1+b}{3}\right)$

But it is given that the centroid is at the origin, that is G(0, 0). Therefore,

$0=\frac{-3+a}{3}, 0=\frac{-1+b}{3}$

$\Rightarrow 0=-3+a, 0=-1+b$

$\Rightarrow 3=a, 1=b$

$\Rightarrow a=3, b=1$

Therefore, the third vertex of ∆ABC is A(3, 1).