A) \[\frac{27}{4}sq\]. units
B) \[27sq.\]units
C) \[\frac{27}{2}sq.\] units
D) None of these
Correct Answer: A
Solution :
The lines represented by \[{{y}^{2}}-9xy+18{{x}^{2}}=0\] are \[6x-y=0\]and \[3x-y=0\]and third line is \[y=9\]. Therefore, coordinates of vertices of triangle are given by \[A(0,0)\];\[B\text{ }(3,9)\] and \[C\left( \frac{3}{2},9 \right)\]. Hence, area of DABC \[=\frac{1}{2}\left| \begin{matrix} 0 & 0 & 1 \\ 3 & 9 & 1 \\ 3/2 & 9 & 1 \\ \end{matrix} \right|=\frac{27}{4}sq.\,units.\] Aliter: Applying the formula given in the theory part, the required area is \[\frac{{{(-9)}^{2}}\sqrt{{{(9/2)}^{2}}-18}}{18\times 1+9\times 0\times 1+1\times 0}=\frac{81}{18}\sqrt{\frac{81}{4}-18}\] \[=\frac{81}{18}\times \frac{3}{2}=\frac{27}{4}\,sq.\ units.\]
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