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(A) The slope $m_1$ of the line joining the points $(2, 5)$ and $(-3, 6)$ is given by $m_1 = \frac{6 - 5}{-3 - 2} = \frac{1}{-5} = -\frac{1}{5}$.Two l

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$5x - y + 20 = 0$

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(A) The slope $m_1$ of the line joining the points $(2, 5)$ and $(-3, 6)$ is given by $m_1 = \frac{6 - 5}{-3 - 2} = \frac{1}{-5} = -\frac{1}{5}$.Two lines are perpendicular if the product of their slopes is $-1$. Let the slope of the required line be $m_2$.Then $m_1 	imes m_2 = -1 \implies -\frac{1}{5} 	imes m_2 = -1 \implies m_2 = 5$.The equation of the line passing through $(-3, 5)$ with slope $m_2 = 5$ is given by the point-slope form: $y - y_1 = m_2(x - x_1)$.Substituting the values: $y - 5 = 5(x - (-3))$.$y - 5 = 5(x + 3)$.$y - 5 = 5x + 15$.$5x - y + 20 = 0$.

Find the equation of the line passing through $(-3, 5)$ and perpendicular to the line passing through the points $(2, 5)$ and $(-3, 6)$.

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