The equation of a line passing through (3, -4 | vedclass

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(D) The given line is $3x + 4y = 5$,which can be written as $4y = -3x + 5$ or $y = -\frac{3}{4}x + \frac{5}{4}$.The slope of this line is $m_1 = -\fra

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$y + 4 = \frac{4}{3}(x - 3)$

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(D) The given line is $3x + 4y = 5$,which can be written as $4y = -3x + 5$ or $y = -\frac{3}{4}x + \frac{5}{4}$.The slope of this line is $m_1 = -\frac{3}{4}$.Since the required line is perpendicular to this line,its slope $m_2$ must satisfy $m_1 	imes m_2 = -1$.Thus,$m_2 = -\frac{1}{m_1} = -\frac{1}{-3/4} = \frac{4}{3}$.The equation of a line passing through $(x_1, y_1) = (3, -4)$ with slope $m = \frac{4}{3}$ is given by the point-slope form: $y - y_1 = m(x - x_1)$.Substituting the values,we get $y - (-4) = \frac{4}{3}(x - 3)$,which simplifies to $y + 4 = \frac{4}{3}(x - 3)$.

The equation of a line passing through $(3, -4)$ and perpendicular to the line $3x + 4y = 5$ is

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