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(A) The given line is $2x - 3y = 5$,which can be written as $y = \frac{2}{3}x - \frac{5}{3}$. The slope of this line is $m_1 = \frac{2}{3}$.Since the

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$3x + 2y - 1 = 0$

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(A) The given line is $2x - 3y = 5$,which can be written as $y = \frac{2}{3}x - \frac{5}{3}$. The slope of this line is $m_1 = \frac{2}{3}$.Since the required line is perpendicular to this line,its slope $m_2$ must satisfy $m_1 	imes m_2 = -1$.Therefore,$m_2 = -\frac{1}{m_1} = -\frac{3}{2}$.The equation of a line with slope $m = -\frac{3}{2}$ passing through $(x_1, y_1) = (1, -1)$ is given by $y - y_1 = m(x - x_1)$.Substituting the values: $y - (-1) = -\frac{3}{2}(x - 1)$.$y + 1 = -\frac{3}{2}(x - 1)$.$2(y + 1) = -3(x - 1)$.$2y + 2 = -3x + 3$.$3x + 2y - 1 = 0$.

Find the equation of the line perpendicular to the line $2x - 3y = 5$ and passing through the point $(1, -1)$.

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