If the points (2k, k), (k, 2k) and (k, k) wit | vedclass

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Question

$\Delta=\frac{1}{2}\left|\begin{array}{lll}{2 k} & {k} & {1} \ {k} & {2 k} & {1} \ {k} & {k} & {1}\end{array}ight|=\pm 18$

Vedclass · Accepted Answer

$(8, 8)$

Vedclass · Answer

$\Delta=\frac{1}{2}\left|\begin{array}{lll}{2 k} & {k} & {1} \ {k} & {2 k} & {1} \ {k} & {k} & {1}\end{array}ight|=\pm 18$

$\frac{k^{2}}{2}\left|\begin{array}{lll}{2} & {1} & {1} \ {1} & {2} & {1} \ {1} & {1} & {1}\end{array}ight|=\pm 18$

$\frac{\mathrm{k}^{2}}{2}[2(2-1)-1(1-1)+1(1-2)]=\pm 18$

$\frac{\mathrm{k}^{2}}{2}=\pm 18 \Rightarrow \mathrm{k}=6$

Centroid $\left(\frac{12+6+6}{3}, \frac{6+12+6}{3}ight)$

$=(8,8)$

If the points $(2k, k), (k, 2k)$ and $(k, k)$ with $k > 0$ enclose a triangle of area $18$ square unit then centroid of triangle is equal to

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