• # question_answer Area of a triangle formed by the lines x2 + 4xy + y2 = 0 and x + y = 1 is A)  $\sqrt{3}$                                        B)  2 C)  1                                             D)  $\frac{\sqrt{3}}{2}$

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.