JEE Main 2019 — Straight Lines Question with Solution

Let, coordinate of centroid is $(h, k)$ and coordinate of point $P$ is $(\alpha , \beta )$

We know that the centroid of a triangle with vertices $\left(x_1, y_1\right), \left(x_2, y_2\right)$ and $\left(x_3, y_3\right)$ is $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right).$

Hence, in $∆PQR,$ we have $\frac{\alpha +1+3}{3}=h$ and $\frac{\beta +4-2}{3}=k$

$\Rightarrow \alpha =3h-4,$ $\beta =3k-2$

Since $(\alpha , \beta )$ lies on the line $2x-3y+4=0$

Hence, $2\left(3h-4\right)-3\left(3k-2\right)+4=0$

$\Rightarrow 6h-9k+2=0$

For finding the locus replace $\left(h, k\right)$ by $\left(x, y\right)$

Hence, the locus of centroid is $6x-9y+2=0$ which has slope $=-\frac{6}{\left(-9\right)}=\frac{2}{3}$ since the slope of a line $ax+by+c=0$ is $-\frac{a}{b}.$