If A(4, – 3), B(3, – 2) and C(2, 8) are vertices of a triangle, th

1.	If A(4, – 3), B(3, – 2) and C(2, 8) are vertices of a triangle, then the distance of it's centroid from the y-axis is
Answer» $Let \: A(4,-3) = (x_{1} , y_{1}) , \\ B(3,-2) = (x_{2} , y_{2}) , \: C(2,8) = (x_{3} , y_{3}) \\ are \: Vertices \\of \: \triangle ABC$ $Centroid (G) = \Big( \frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\Big)$ $= \Big( \frac{4+3+2}{3} , \frac{-3-2+8}{3}\Big) \\= \Big( \frac{9}{3} , \frac{3}{3}\Big) \\= ( 3 , 1 )$ $Now, Distance \:of \: Centroid \:from \:the \\y-axis = x - coordinate \:of \: G \\= 3$ THEREFORE., $\red {Distance \:of \: Centroid \:from \:the }$ $\red {y-axis} \green { = 3 }$ •••♪

If A(4, – 3), B(3, – 2) and C(2, 8) are vertices of a triangle, then the distance of it's centroid from the y-axis is

Answer»

$Let \: A(4,-3) = (x_{1} , y_{1}) , \\ B(3,-2) = (x_{2} , y_{2}) , \: C(2,8) = (x_{3} , y_{3}) \\ are \: Vertices \\of \: \triangle ABC$

$Centroid (G) = \Big( \frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\Big)$

$= \Big( \frac{4+3+2}{3} , \frac{-3-2+8}{3}\Big) \\= \Big( \frac{9}{3} , \frac{3}{3}\Big) \\= ( 3 , 1 )$

$Now, Distance \:of \: Centroid \:from \:the \\y-axis = x - coordinate \:of \: G \\= 3$

THEREFORE.,

$\red {Distance \:of \: Centroid \:from \:the }$

$\red {y-axis} \green { = 3 }$

•••♪

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