The angle between the lines 2x - y + 5 = 0 an | vedclass

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(C) The given lines are $2x - y + 5 = 0$ and $3x + y + 4 = 0$. Comparing with $y = mx + c$,the slopes are $m_1 = 2$ and $m_2 = -3$. The angle $	heta$

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(C) The given lines are $2x - y + 5 = 0$ and $3x + y + 4 = 0$. Comparing with $y = mx + c$,the slopes are $m_1 = 2$ and $m_2 = -3$. The angle $	heta$ between two lines with slopes $m_1$ and $m_2$ is given by $	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^o$.

The angle between the lines $2x - y + 5 = 0$ and $3x + y + 4 = 0$ is ...$^o$.

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