The line L given by frac{x}{5}+frac{y}{b}=1 p | vedclass

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(D) Line $L$ passes through $(13, 32)$.$\frac{13}{5} + \frac{32}{b} = 1$$\frac{32}{b} = 1 - \frac{13}{5} = -\frac{8}{5}$$b = -20$So,the equation of $L

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$\frac{23}{\sqrt{17}}$

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(D) Line $L$ passes through $(13, 32)$.$\frac{13}{5} + \frac{32}{b} = 1$$\frac{32}{b} = 1 - \frac{13}{5} = -\frac{8}{5}$$b = -20$So,the equation of $L$ is $\frac{x}{5} - \frac{y}{20} = 1$,which simplifies to $4x - y - 20 = 0$.The slope of $L$ is $m_1 = 4$.Since line $K$ is parallel to $L$,its slope $m_2$ must be $4$.The equation of $K$ is $\frac{x}{c} + \frac{y}{3} = 1$,which can be written as $y = -\frac{3}{c}x + 3$.Thus,$-\frac{3}{c} = -4 \Rightarrow c = \frac{3}{4}$.The equation of $K$ is $\frac{x}{3/4} + \frac{y}{3} = 1$ $\Rightarrow \frac{4x}{3} + \frac{y}{3} = 1$ $\Rightarrow 4x + y - 3 = 0$.Wait,re-evaluating the slope: The equation $\frac{x}{c} + \frac{y}{3} = 1$ gives $y = -\frac{3}{c}x + 3$. For this to be parallel to $4x - y - 20 = 0$ (i.e.,$y = 4x - 20$),we need $-\frac{3}{c} = 4$,so $c = -\frac{3}{4}$.The equation of $K$ becomes $\frac{x}{-3/4} + \frac{y}{3} = 1$ $\Rightarrow -\frac{4x}{3} + \frac{y}{3} = 1$ $\Rightarrow -4x + y = 3$ $\Rightarrow 4x - y + 3 = 0$.The distance between $4x - y - 20 = 0$ and $4x - y + 3 = 0$ is $\frac{|-20 - 3|}{\sqrt{4^2 + (-1)^2}} = \frac{23}{\sqrt{17}}$.

The line $L$ given by $\frac{x}{5}+\frac{y}{b}=1$ passes through the point $(13,32)$. The line $K$ is parallel to line $L$ and has the equation $\frac{x}{c}+\frac{y}{3}=1$. Then the distance between $L$ and $K$ is $\qquad$ units.

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