The slopes of the lines,making an angle of 45 | vedclass

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(A) Let the slope of the given line $2x - 3y = 5$ be $m_1$. Rewriting the equation as $3y = 2x - 5$,we get $y = \frac{2}{3}x - \frac{5}{3}$. Thus,$m_1

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$5, \frac{-1}{5}$

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(A) Let the slope of the given line $2x - 3y = 5$ be $m_1$. Rewriting the equation as $3y = 2x - 5$,we get $y = \frac{2}{3}x - \frac{5}{3}$. Thus,$m_1 = \frac{2}{3}$.Let the slope of the required lines be $m$. The angle $	heta$ between the lines is $45^{\circ}$.Using the formula $	an 	heta = \left| \frac{m - m_1}{1 + m \cdot m_1} ight|$,we have:$	an 45^{\circ} = \left| \frac{m - \frac{2}{3}}{1 + m \cdot \frac{2}{3}} ight|$$1 = \left| \frac{3m - 2}{3 + 2m} ight|$This gives two cases:Case $1$: $\frac{3m - 2}{3 + 2m} = 1$ $\Rightarrow 3m - 2 = 3 + 2m$ $\Rightarrow m = 5$.Case $2$: $\frac{3m - 2}{3 + 2m} = -1$ $\Rightarrow 3m - 2 = -3 - 2m$ $\Rightarrow 5m = -1$ $\Rightarrow m = \frac{-1}{5}$.Therefore,the slopes are $5$ and $\frac{-1}{5}$.

The slopes of the lines,making an angle of $45^{\circ}$ with the line $2x - 3y = 5$,are

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