The angle between the straight lines 3x + 4y | vedclass

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(A) Given,the equations of the straight lines are:$l_1: 3x + 4y + 9 = 0$ ...$(i)$$l_2: x - 7y - 22 = 0$ ...(ii)The slope of line $l_1$ is $m_1 = -\fra

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$\frac{\pi}{4}$

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(A) Given,the equations of the straight lines are:$l_1: 3x + 4y + 9 = 0$ ...$(i)$$l_2: x - 7y - 22 = 0$ ...(ii)The slope of line $l_1$ is $m_1 = -\frac{3}{4}$.The slope of line $l_2$ is $m_2 = -\frac{1}{-7} = \frac{1}{7}$.Let $	heta$ be the angle between the lines $l_1$ and $l_2$. The formula for the angle between two lines is:$	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$Substituting the values of $m_1$ and $m_2$:$	an 	heta = \left| \frac{-\frac{3}{4} - \frac{1}{7}}{1 + (-\frac{3}{4})(\frac{1}{7})} ight|$$	an 	heta = \left| \frac{-\frac{21+4}{28}}{1 - \frac{3}{28}} ight| = \left| \frac{-\frac{25}{28}}{\frac{25}{28}} ight|$$	an 	heta = |-1| = 1$Since $	an 	heta = 1$,we have $	heta = \frac{\pi}{4}$.

The angle between the straight lines $3x + 4y + 9 = 0$ and $x - 7y - 22 = 0$ is

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