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(A, B) Let the slope of the given line $2x + 3y = 5$ be $m_1 = -\frac{2}{3}$.Let the slope of the required line be $m$.The angle between the two lines

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$5x + y - 13 = 0$

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(A, B) Let the slope of the given line $2x + 3y = 5$ be $m_1 = -\frac{2}{3}$.Let the slope of the required line be $m$.The angle between the two lines is $	heta = \frac{\pi}{4}$.Using the formula $	an 	heta = |\frac{m - m_1}{1 + m \cdot m_1}|$,we have:$	an \frac{\pi}{4} = |\frac{m - (-2/3)}{1 + m(-2/3)}|$$1 = |\frac{3m + 2}{3 - 2m}|$This gives two cases:Case $1$: $\frac{3m + 2}{3 - 2m} = 1 \implies 3m + 2 = 3 - 2m \implies 5m = 1 \implies m = \frac{1}{5}$.The equation of the line is $y - 3 = \frac{1}{5}(x - 2) \implies 5y - 15 = x - 2 \implies x - 5y + 13 = 0$.Case $2$: $\frac{3m + 2}{3 - 2m} = -1 \implies 3m + 2 = -3 + 2m \implies m = -5$.The equation of the line is $y - 3 = -5(x - 2) \implies y - 3 = -5x + 10 \implies 5x + y - 13 = 0$.Thus,the possible equations are $x - 5y + 13 = 0$ or $5x + y - 13 = 0$.

Find the equation of the line passing through the point $(2, 3)$ and making an angle of $\frac{\pi}{4}$ with the line $2x + 3y = 5$.

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