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  4. The angle between the lines 2x + 11y - 7 = 0 and x + 3y + 5 = 0 is equal to

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Q. The angle between the lines $2x + 11y - 7 = 0$ and $x + 3y + 5 = 0$ is equal to

Straight Lines

Solution:

Let $\theta$ be the angle between the two lines
$\therefore tan\,\theta=\left|\frac{m_{1}-m_{2}}{1+m_{1}m_{2}}\right|$
where $m_{1}$ and $m_{2}$ are slopes of given lines.
$2x + 11 y - 7 = 0$ and $x + 3y + 5 = 0$
$\therefore m_{1}=-\frac{2}{11}, m_{2}=\frac{-1}{3}$

$\therefore tan\,\theta=\left|\frac{-\frac{1}{3}+\frac{2}{11}}{1+\left(\frac{-2}{11} \times \frac{-1}{3}\right)}\right|=\left|\frac{-11+6}{33+2}\right|$

$=\left|\frac{-5}{35}\right|=\frac{1}{7}$

$\Rightarrow \theta=tan^{-1}\left(\frac{1}{7}\right)$