Find the equation of the line passing through | vedclass

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(D) Let the points be $A(-5, 6)$ and $B(-6, 5)$.The slope of the line $AB$ is given by $m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 6}{-6 - (-5)} =

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$x + y - 5 = 0$

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(D) Let the points be $A(-5, 6)$ and $B(-6, 5)$.The slope of the line $AB$ is given by $m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 6}{-6 - (-5)} = \frac{-1}{-1} = 1$.Since the required line is perpendicular to $AB$,its slope $m_2$ must satisfy $m_1 	imes m_2 = -1$.Thus,$1 	imes m_2 = -1$,which gives $m_2 = -1$.The equation of a line with slope $m$ passing through $(x_0, y_0)$ is given by $y - y_0 = m(x - x_0)$.Substituting $m = -1$ and the point $(2, 3)$,we get:$y - 3 = -1(x - 2)$$y - 3 = -x + 2$$x + y - 5 = 0$.

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

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