Question

If $m_{1}$ and $m_{2}$ are the roots of the equation $x^{2}+\left(\sqrt{3} + 2\right)x+\sqrt{3}-1=0$ , then the area (in sq. units) of the triangle bounded by the lines $y=m_{1}x,y=m_{2}x$ and $y=2$ is

• A. $\sqrt{33}+\sqrt{11}$
• B. $\sqrt{33}-\sqrt{11}$
• C. $2\sqrt{33}$
• D. $2\sqrt{11}$

Answer 1

Answer: (A)

Vertices of the triangle formed by the lines $y=m_{1}x,y=m_{2}x\&y=2$ are
$\left(0,0\right),\left(\frac{2}{m_{1}} , 2\right)$ & $\left(\frac{2}{m_{2}} , 2\right)$
Area of the triangle $=\begin{vmatrix} \frac{1}{2} \begin{vmatrix} 0 & 0 & 1 \\ \frac{2}{m_{1}} & 2 & 1 \\ \frac{2}{m_{2}} & 2 & 1 \end{vmatrix} \end{vmatrix}$
$=2\left|\frac{m_{2} - m_{1}}{m_{1} m_{2}}\right|=2\left|\frac{\frac{\sqrt{D}}{\left|a\right|}}{\frac{c}{a}}\right|=2\left|\frac{\sqrt{D}}{c}\right|=2\left|\frac{\sqrt{b^{2} - 4 a c}}{c}\right|$
$=2\frac{\sqrt{7 + 4 \sqrt{3} - 4 \left(\sqrt{3} - 1\right)}}{\sqrt{3} - 1}=\frac{2 \sqrt{11}}{\sqrt{3} - 1}$
$=\sqrt{33}+\sqrt{11}$ sq. units