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Question

Let $C\left( {{x_1},{y_1}} \right)$

Centroid, $E = \left( {\frac{{{x_1} - 5}}{3},\frac{{{x_1} - 3}}{3}} \right)$

Since centroid lies o

Vedclass · Accepted Answer

$3x +4y+3=0$

Vedclass · Answer

Let $C\left( {{x_1},{y_1}} ight)$

Centroid, $E = \left( {\frac{{{x_1} - 5}}{3},\frac{{{x_1} - 3}}{3}} ight)$

Since centroid lies on the line 

$3x + 4y + 2 = 0$

$	herefore 3\left( {\frac{{{x_1} - 5}}{3}} ight) + 4\left( {\frac{{{x_1} - 3}}{3}} ight) + 2 = 0$

$ \Rightarrow 3{x_1} + 4{y_1} + 3 = 0$

Hence vertex $\left( {{x_1},{y_1}} ight)$ lies on the line 

$3x + 4y + 3 = 0$

Let $A (-3, 2)$ and $B (-2, 1)$ be the vertices of a triangle $ABC$. If the centroid of this triangle lies on the line $3x + 4y + 2 = 0$, then the vertex $C$ lies on the line

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