A point P moves on the line 2x - 3y + 4 = 0. | vedclass

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(C) Let the coordinates of point $P$ be $(\alpha, \beta)$. Since $P$ lies on the line $2x - 3y + 4 = 0$,we have $2\alpha - 3\beta + 4 = 0$. Let $(h, k

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with slope $\frac{2}{3}$

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(C) Let the coordinates of point $P$ be $(\alpha, \beta)$. Since $P$ lies on the line $2x - 3y + 4 = 0$,we have $2\alpha - 3\beta + 4 = 0$. Let $(h, k)$ be the centroid of $\Delta PQR$. The coordinates of the centroid are given by: $h = \frac{\alpha + 1 + 3}{3}$ $\Rightarrow 3h = \alpha + 4$ $\Rightarrow \alpha = 3h - 4$ $k = \frac{\beta + 4 - 2}{3}$ $\Rightarrow 3k = \beta + 2$ $\Rightarrow \beta = 3k - 2$ Substituting these values into the equation of the line: $2(3h - 4) - 3(3k - 2) + 4 = 0$ $6h - 8 - 9k + 6 + 4 = 0$ $6h - 9k + 2 = 0$ The locus of the centroid $(h, k)$ is $6x - 9y + 2 = 0$. Rewriting in slope-intercept form: $9y = 6x + 2$ $\Rightarrow y = \frac{6}{9}x + \frac{2}{9}$ $\Rightarrow y = \frac{2}{3}x + \frac{2}{9}$. The slope of this line is $\frac{2}{3}$.

$A$ point $P$ moves on the line $2x - 3y + 4 = 0$. If $Q(1, 4)$ and $R(3, -2)$ are fixed points,then the locus of the centroid of $\Delta PQR$ is a line

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