If the angle between two lines is frac{pi}{4} | vedclass

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(N/A) We know that the acute angle $	heta$ between two lines with slopes $m_{1}$ and $m_{2}$ is given by $	an 	heta = \left| \frac{m_{2} - m_{1}}{1

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(N/A) We know that the acute angle $	heta$ between two lines with slopes $m_{1}$ and $m_{2}$ is given by $	an 	heta = \left| \frac{m_{2} - m_{1}}{1 + m_{1}m_{2}} ight| \dots (1)$Let $m_{1} = \frac{1}{2},$ $m_{2} = m,$ and $	heta = \frac{\pi}{4}.$Now,substituting these values in $(1),$ we get$	an \frac{\pi}{4} = \left| \frac{m - \frac{1}{2}}{1 + \frac{1}{2}m} ight| \implies 1 = \left| \frac{m - \frac{1}{2}}{1 + \frac{1}{2}m} ight|.$This gives two cases:Case $1: \frac{m - \frac{1}{2}}{1 + \frac{1}{2}m} = 1 \implies m - \frac{1}{2} = 1 + \frac{1}{2}m \implies \frac{1}{2}m = \frac{3}{2} \implies m = 3.$Case $2: \frac{m - \frac{1}{2}}{1 + \frac{1}{2}m} = -1 \implies m - \frac{1}{2} = -1 - \frac{1}{2}m \implies \frac{3}{2}m = -\frac{1}{2} \implies m = -\frac{1}{3}.$Hence,the slope of the other line is $3$ or $-\frac{1}{3}.$

If the angle between two lines is $\frac{\pi}{4}$ and the slope of one of the lines is $\frac{1}{2},$ find the slope of the other line.

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