What is the equation of the line passing thro | vedclass

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(A) The given line is $2x + 3y + 4 = 0$,which can be written as $3y = -2x - 4$,or $y = -\frac{2}{3}x - \frac{4}{3}$.The slope of this line is $m_1 = -

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$2(y - 1) = 3(x + 1)$

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(A) The given line is $2x + 3y + 4 = 0$,which can be written as $3y = -2x - 4$,or $y = -\frac{2}{3}x - \frac{4}{3}$.The slope of this line is $m_1 = -\frac{2}{3}$.Since the required line is perpendicular to the given line,its slope $m_2$ must satisfy $m_1 	imes m_2 = -1$.Therefore,$m_2 = -\frac{1}{m_1} = -\frac{1}{-2/3} = \frac{3}{2}$.The equation of a line passing through $(x_1, y_1) = (-1, 1)$ with slope $m_2 = \frac{3}{2}$ is given by $y - y_1 = m_2(x - x_1)$.Substituting the values,we get $y - 1 = \frac{3}{2}(x - (-1))$,which simplifies to $y - 1 = \frac{3}{2}(x + 1)$.Multiplying both sides by $2$,we get $2(y - 1) = 3(x + 1)$.

What is the equation of the line passing through $(-1, 1)$ and perpendicular to the line $2x + 3y + 4 = 0$?

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