The angle between the lines 2x - y - 15 = 0 a | vedclass

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(B) The given lines are $2x - y - 15 = 0$ $(i)$ and $3x + y + 4 = 0$ $(ii)$.Comparing these with the slope-intercept form $y = mx + c$,the slopes are

Vedclass · Accepted Answer

$45$

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(B) The given lines are $2x - y - 15 = 0$ $(i)$ and $3x + y + 4 = 0$ $(ii)$.Comparing these with the slope-intercept form $y = mx + c$,the slopes are $m_1 = 2$ and $m_2 = -3$.If $	heta$ is the angle between the lines,then $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.Substituting the values,$	an 	heta = \left| \frac{2 - (-3)}{1 + (2)(-3)} ight| = \left| \frac{5}{1 - 6} ight| = \left| \frac{5}{-5} ight| = 1$.Since $	an 	heta = 1$,we have $	heta = 45^\circ$.

The angle between the lines $2x - y - 15 = 0$ and $3x + y + 4 = 0$ is .....$^o$.

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