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(A) The equation of the given line is $\frac{x}{4} + \frac{y}{6} = 1$. Multiplying by $12$,we get $3x + 2y = 12$,or $3x + 2y - 12 = 0$. The slope of t

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$2x - 3y + 18 = 0$

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(A) The equation of the given line is $\frac{x}{4} + \frac{y}{6} = 1$. Multiplying by $12$,we get $3x + 2y = 12$,or $3x + 2y - 12 = 0$. The slope of this line is $m_1 = -\frac{3}{2}$. The slope of a line perpendicular to this line is $m_2 = -\frac{1}{m_1} = -\frac{1}{(-3/2)} = \frac{2}{3}$. The given line meets the $y-$ axis where $x = 0$. Substituting $x = 0$ in $\frac{x}{4} + \frac{y}{6} = 1$,we get $\frac{y}{6} = 1$,so $y = 6$. The point of intersection is $(0, 6)$. The equation of the line with slope $m_2 = \frac{2}{3}$ passing through $(0, 6)$ is given by $y - y_1 = m_2(x - x_1)$. $y - 6 = \frac{2}{3}(x - 0)$. $3(y - 6) = 2x$. $3y - 18 = 2x$. $2x - 3y + 18 = 0$.

Find the equation of a line drawn perpendicular to the line $\frac{x}{4}+\frac{y}{6}=1$ through the point where it meets the $y-$ axis.

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