A) \[\frac{7}{12}\]
B) \[\frac{5}{12}\]
C) \[\frac{1}{12}\]
D) \[\frac{1}{6}\]
Correct Answer: C
Solution :
Given,\[{{x}^{2}}-2xy-xy+2{{y}^{2}}=0\] \[\Rightarrow \] \[(x-2y)(x-y)=0\] \[\Rightarrow \] \[x=2y,\,\,x=y\] ? (i) Also \[x+y+1=0\] ... (ii) On solving Eqs. (i) and (ii), we get \[A\left( -\frac{2}{3},\,\,-\frac{1}{3} \right),\,\,B\left( -\frac{1}{2},\,\,-\frac{1}{2} \right),\,\,C(0,\,\,0)\] \[\therefore \]Area of\[\Delta ABC=\frac{1}{2}\left| \begin{matrix} -\frac{2}{3} & -\frac{1}{3} & 1 \\ -\frac{1}{2} & -\frac{1}{2} & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right|\] \[=\frac{1}{2}\left[ \frac{1}{3}-\frac{1}{6} \right]=\frac{1}{2}\left[ \frac{1}{6} \right]=\frac{1}{12}\]
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