Question

The sum of the tangents of the interior angles of a triangle formed by the lines $L_1:2x+3y+3=0;L_2:y-x-2=0$ and $L_3:2x-3y-3=0$, is

• A. $-\frac{12}{5}$
• B. $\frac{1}{5}$
• C. $\frac{12}{5}$
• D. $-\frac{1}{5}$

Answer 1

The correct option is A $-\frac{12}{5}$

Let the triangle be $△ABC$
Given,
$L_1:2x+3y+3=0\Rightarrow m_1=-\frac{2}{3}$
$L_2:y-x-2=0\Rightarrow m_2=1$
$L_3:2x-3y-3=0\Rightarrow m_3=\frac{2}{3}$
$m_1=-ve\to \text{obtuse angle}$
$m_2=+ve\to \text{acute angle}$
$m_3=+ve\to \text{acute angle}$
But $m_2>m_3,$ so $m_2$ will be second largest angle and $m_3$ will be the least angle of the triangle.
So, arranging slopes in decreasing order with respect to angles, we get
$m_1>m_2>m_3$

Now,
$\tan A=\frac{m_1-m_2}{1+m_1{m}_2}=\frac{-\frac{2}{3}-1}{1-\frac{2}{3}}=-5\tan B=\frac{m_2-m_3}{1+m_2{m}_3}=\frac{1-\frac{2}{3}}{1+\frac{2}{3}}=\frac{1}{5}\tan C=\frac{m_3-m_1}{1+m_3{m}_1}=\frac{\frac{2}{3}+\frac{2}{3}}{1-\frac{4}{9}}=\frac{12}{5}\Rightarrow \tan A+\tan B+\tan C=-5+\frac{1}{5}+\frac{12}{5}\therefore \tan A+\tan B+\tan C=-\frac{12}{5}$