Question

Show that the area of the triangle formed by the lines y = m₁ x, y = m₂ x and y = c is equal to \[\frac{c^2}{4}\left( \sqrt{33} + \sqrt{11} \right),\] where m₁, m₂ are the roots of the equation \[x^2 + \left( \sqrt{3} + 2 \right)x + \sqrt{3} - 1 = 0 .\]

Answer 1

The given lines are as follows:
y = m₁ x      ... (1)
y = m₂ x      ... (2)
y = c            ... (3)
Solving (1) and (2), we get (0, 0) as their point of intersection.
Solving (1) and (3), we get 

\[\left( \frac{c}{m_1}, c \right)\]  as their point of intersection.
Similarly, solving (2) and (3), we get 

\[\left( \frac{c}{m_2}, c \right)\]  as their point of intersection.

∴ Area of the triangle formed by these lines = \[\frac{1}{2}\begin{vmatrix}0 & 0 & 1 \\ \frac{c}{m_1} & c & 1 \\ \frac{c}{m_2} & c & 1\end{vmatrix} = \frac{1}{2}\left( \frac{c^2}{m_1} - \frac{c^2}{m_2} \right) = \frac{c^2}{2}\left| \frac{m_2 - m_1}{m_1 m_2} \right|\]

It is given that m₁ and m₂ are the roots of the equation

\[x^2 + \left( \sqrt{3} + 2 \right)x + \sqrt{3} - 1 = 0 .\]

\[\therefore m_1 + m_2 = - \left( \sqrt{3} + 2 \right), m_1 m_2 = \sqrt{3} - 1\]

\[ \Rightarrow m_2 - m_1 = \sqrt{\left( m_1 + m_2 \right)^2 - 4 m_1 m_2}\]

\[ \Rightarrow m_2 - m_1 = \sqrt{\left\{ - \left( \sqrt{3} + 2 \right) \right\}^2 - 4\sqrt{3} + 4}\]

\[ \Rightarrow m_2 - m_1 = \sqrt{7 + 4\sqrt{3} - 4\sqrt{3} + 4} = \sqrt{11}\]

\[\therefore \text { Area  }= \frac{c^2}{2}\left| \frac{\sqrt{11}}{\sqrt{3} - 1} \right| = \frac{c^2}{2}\left| \frac{\left( \sqrt{3} + 1 \right)\sqrt{11}}{\left( \sqrt{3} + 1 \right)\left( \sqrt{3} - 1 \right)} \right|\]

\[ = \frac{c^2}{2}\left| \frac{\left( \sqrt{33} + \sqrt{11} \right)}{2} \right| = \frac{c^2}{4}\left( \sqrt{33} + \sqrt{11} \right)\]