Question

The area (in s units) of the triangle formed by the lines $x^{2}-3 x y+y^{2}=0$ and $x+y+1=0$, is

• A. $\frac{2}{\sqrt{3}}$
• B. $\frac{\sqrt{3}}{2}$
• C. $5 \sqrt{2}$
• D. $\frac{1}{2 \sqrt{5}}$

Answer 1

The Correct Option is D

Given equations of line are $x^{2}-3 x y+y^{2}=0$ and $x+y+1=0$ Let, $m_{1}$ and $m_{2}$ be the slope of the line $x^{2}-3 x y+y^{2}=2$ $\therefore m_{1}+m_{2} =-\frac{\text { Coefficient of } x y}{\text { Coefficient of } y^{2}} $ $=+\frac{3}{1}=3\,\,\,...(i)$ and $m_{1} m_{2} =\frac{\text { Coefficient of } x^{2}}{\text { Coefficient of } y^{2}} $ $=\frac{1}{1}=1\,\,\,...(ii)$ Now, $m_{1}-m_{2}=\sqrt{\left(m_{1}+m_{2}\right)^{2}-4 m_{1} m_{2}}$ $\Rightarrow =\sqrt{(3)^{2}-4 \times 1} $ $ =\sqrt{9-4}=\sqrt{5} $ $\Rightarrow m_{1}-m_{2} =\sqrt{5}\,\,\,...(iii)$ On solving Eqs. (i) and (iii), we get $m_{1}=\frac{3+\sqrt{5}}{2} $ and $m_{2}=\frac{3-\sqrt{5}}{2}$ $\therefore$ Equation of lines will be $y=\frac{3+\sqrt{5}}{2} x \,\,\,...(iv)$ and $y=\frac{3-\sqrt{5}}{2} x\,\,\,...(v)$ and third line is given $x+y+1=0\,\,\,...(vi)$ $\therefore$ The points of intersection of these lines are $A(0,0), B\left(-\frac{2}{5+\sqrt{5}}, \frac{3+\sqrt{5}}{5+\sqrt{5}}\right), $ and $C\left(-\frac{2}{5-\sqrt{5}}, \frac{3-\sqrt{5}}{5-\sqrt{5}}\right)$ $\therefore$ Area of triangle $=\frac{1}{2}\begin{vmatrix}0 & 0 & 1 \\ -\frac{2}{5+\sqrt{5}} & \frac{3+\sqrt{5}}{5+\sqrt{5}} & 1 \\ -\frac{2}{5-\sqrt{5}} & \frac{3-\sqrt{5}}{5-\sqrt{5}} & 1\end{vmatrix}$ $=\frac{1}{2}\left[0+0+1\left(\frac{-2(3-\sqrt{5})}{5^{2}-(\sqrt{5})^{2}}+\frac{2(3+\sqrt{5})}{5^{2}-(\sqrt{5})^{2}}\right]\right.$ $=\frac{1}{2}\left[\frac{-6+2 \sqrt{5}+6+2 \sqrt{5}}{25-5}\right]$ $ =\frac{1}{2}\left[\frac{4 \sqrt{5}}{20}\right]=\frac{1}{2 \sqrt{5}}$