Two vertices of ∆ABC are A(−1, 4) and B(5, 2) and its centroid is G(0, −3)

Two vertices of ∆ABC are A(−1, 4) and B(5, 2) and its centroid is G(0, −3). Then, the coordinates of C are

(a) (4, 3)
(b) (4, 15)
(c) (−4, −15)
(d) (−15, −4)



(c) (−4, −15)

Two vertices of $\triangle A B C$ are $A(-1,4)$ and $B(5,2)$.

Let the third vertex be C(a, b).
Then, the coordinates of its centroid are

$G\left(\frac{-1+5+a}{3}, \frac{4+2+b}{3}\right)$

i.e. $G\left(\frac{4+a}{3}, \frac{6+b}{3}\right)$

But it is given that the centroid is $G(0,-3)$.


$\frac{4+a}{3}=0$ and $\frac{6+b}{3}=-3$

$\Rightarrow 4+a=0$ and $6+b=-9$

$\Rightarrow a=-4$ and $b=-15$

Hence, the third vertex of $\Delta A B C$ is $C(-4,-15)$.



Leave a comment

Please enter comment.
Please enter your name.