A) \[\sqrt{3}\]
B) 2
C) 1
D) \[\frac{\sqrt{3}}{2}\]
Correct Answer: D
Solution :
Idea Find the intersection points from given curve such as \[A({{x}_{1}},{{y}_{1}}),B({{x}_{2}},{{y}_{2}})\]and \[C({{x}_{3}},{{y}_{3}})\] Then, area \[(\Delta ABC)=\frac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|\] Let the two lines of x2 + 4xy + y2 = 0 be \[y={{m}_{1}}x\]and \[y={{m}_{2}}x\] \[\therefore \]\[{{x}^{2}}+4xy+{{y}^{2}}=(y-{{m}_{1}}x)(y-{{m}_{2}}x)\] \[\Rightarrow \]\[{{x}^{2}}+4xy={{y}^{2}}={{m}_{1}}{{m}_{2}}{{x}^{2}}-({{m}_{1}}+{{m}_{2}})xy+{{y}^{2}}\] \[\Rightarrow \]\[{{m}_{1}}{{m}_{2}}=1\]and\[{{m}_{1}}+{{m}_{2}}=-4\] (On comparing the coefficients of x2, y2 and xy) also \[{{({{m}_{1}}-{{m}_{2}})}^{2}}={{({{m}_{1}}+{{m}_{2}})}^{2}}-4\,{{m}_{1}}{{m}_{2}}\] \[=16-4=12\] \[\Rightarrow \] \[{{m}_{1}}-{{m}_{2}}=\pm 2\sqrt{3}\] \[{{m}_{1}}+{{m}_{2}}=-4\] On solving, we get \[{{m}_{1}}=\sqrt{3}-2,{{m}_{2}}=-(\sqrt{3}+2)\] Or \[{{m}_{1}}=(\sqrt{3}-2),{{m}_{2}}=-(\sqrt{3}+2)\] \[\therefore \]Lines are\[y=(\sqrt{3}-2)x\] ?(i) \[y=-(\sqrt{3}+2)x\] ?(ii) and \[x+y=1\] ?(iii) Solving Eqs. (i) and (ii), we get x = 0 = y Solving Eqs. (ii) and (iii), we get \[x=\frac{1-\sqrt{3}}{2},y=\frac{\sqrt{3}+1}{2}\] Solving Eqs. (i) and (iii), we get \[x=\frac{\sqrt{3}+1}{2},y=\frac{1-\sqrt{3}}{2}\] Therefore, required area is \[\frac{1}{2}\left| \begin{matrix} \frac{1-\sqrt{3}}{2} & \frac{\sqrt{3}+1}{2} & 1 \\ \frac{\sqrt{3}+1}{2} & \frac{1-\sqrt{3}}{2} & 1 \\ 0 & 0 & 0 \\ \end{matrix} \right|\] \[=\frac{1}{2}{{\left[ {{\left( \frac{1-\sqrt{3}}{2} \right)}^{2}}-\left( \frac{\sqrt{3}+1}{2} \right) \right]}^{2}}=\frac{\sqrt{3}}{2}\]sq units TEST Edge Geometrical application of two dimensional geometry based questions are asked. To solve such type of question, students- are advised to understand the concept of 2D.
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