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The Centroid

P2,-1,Q1,-2,R...

Question

$P (2, - 1), Q (1, - 2), R (3, - 3)$ are the mid points of the sides $B C, C A$ and $A B$ respectively of the triangle $A B C$ . If $G$ is the centroid of triangle $A B C$ and $H$ is the centroid of the triangle $P Q R$ , then $G H$ is equal to

3

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2 \sqrt{3}

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2 \sqrt{2}

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0

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Solution

The correct option is D $0$
$a + e = 2$
$b + f = 6$
$a + c = 6$
$b + d = - 6$
$c + e = 4$
$d + f = - 2$
$\frac{c - e = 4 c + e = 4}{2 c = 8 c = 4} \begin{matrix} a = 2 d = - 2 \end{matrix}$
$e = 0$
$\frac{b - d = - 4 b + f = - 4}{2 b = - 8 b = - 4 f = 0}$
$A = (2, - 4)$
$B = (4, - 4)$
$C = (0, 0)$
cebtroid $\Rightarrow (2, - 2)$
$G H = 0$
$P = (3, - 3)$
$Q = (1, - 2)$
$R = (2, - 1)$
Centroid $= (2, - 2)$

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P(2,−1),Q(1,−2),R(3,−3) are the mid points of the sides BC,CA and AB respectively of the triangle ABC. If G is the centroid of triangle ABC and H is the centroid of the triangle PQR, then GH is equal to

The correct option is D 0a+e=2b+f=6a+c=6b+d=−6c+e=4d+f=−2c−e=4c+e=42c=8c=4a=2d=−2e=0b−d=−4b+f=−42b=−8b=−4f=0A=(2,−4)B=(4, −4)C=(0,0)cebtroid ⇒(2,−2)GH=0P=(3,−3)Q=(1,−2)R=(2,−1)Centroid =(2,−2)

$P (2, - 1), Q (1, - 2), R (3, - 3)$ are the mid points of the sides $B C, C A$ and $A B$ respectively of the triangle $A B C$ . If $G$ is the centroid of triangle $A B C$ and $H$ is the centroid of the triangle $P Q R$ , then $G H$ is equal to