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3 and 4 .Determinants and Matrices
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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
A
$(8, 8)$
B
$(4, 4)$
C
$(-4, -4)$
D
$(4 \sqrt 2, 4 \sqrt 2)$
Solution
$\Delta=\frac{1}{2}\left|\begin{array}{lll}{2 k} & {k} & {1} \\ {k} & {2 k} & {1} \\ {k} & {k} & {1}\end{array}\right|=\pm 18$
$\frac{k^{2}}{2}\left|\begin{array}{lll}{2} & {1} & {1} \\ {1} & {2} & {1} \\ {1} & {1} & {1}\end{array}\right|=\pm 18$
$\frac{\mathrm{k}^{2}}{2}[2(2-1)-1(1-1)+1(1-2)]=\pm 18$
$\frac{\mathrm{k}^{2}}{2}=\pm 18 \Rightarrow \mathrm{k}=6$
Centroid $\left(\frac{12+6+6}{3}, \frac{6+12+6}{3}\right)$
$=(8,8)$
Standard 12
Mathematics
