A point P moves on the line 2x - 3y + 4 = 0 . If Q(1, 4) and R(3, -...

Let, coordinate of centroid is $( h, k ) and coordinate of point$P$is$( \alpha, \beta )$</p><p>We know that the centroid of a triangle with vertices \left({x}_{1}, {y}_{1}\right), \left({x}_{2}, {y}_{2}\right) and \left({x}_{3}, {y}_{3}\right) is \left(\dfrac{{x}_{1} + {x}_{2} + {x}_{3}}{3}, \dfrac{{y}_{1} + {y}_{2} + {y}_{3}}{3}\right) .</p><p>Hence,$ in ∆ PQR,$we have \dfrac{\alpha + 1 + 3}{3} =$h and $\dfrac{\beta + 4 - 2}{3} =k$

$⇒ \alpha = 3h - 4, \beta = 3k - 2$

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

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

$⇒ 6h - 9k + 2 = 0$

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 $= - \dfrac{6}{\left(- 9\right)}=\dfrac{2}{3} since$ the slope of a lineax + by + c = 0$is$- $\dfrac{a}{b} .</p>$