Question

The centroid of a triangle formed by $(7,p),(q,-6),(9,10)$ is $(6,3)$, then find $p+q$.

Answer 1

$(\frac{7+q+9}{3},\frac{p-6+10}{3})=(6,3)$

$\Rightarrow (p,q)=(5,2)$

The vertices of a triangle are $(7,p),(q,-6),(9,10)$.

Formula for centroid of a triangle:

$(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

Centroid of given triangle is

$(\frac{7+q+9}{3},\frac{p-6+10}{3})$

It is given that centroid of given triangle is (6,3).

$(\frac{7+q+9}{3},\frac{p-6+10}{3})=(6,3)$

On comparing both sides we get

The value of q is 2.

The value of p is 5.

Therefore, the value of $p+q$ is 7