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$$\Rightarrow \frac{32}{b} = 1 - \frac{13}{5} = -\frac{8}{5}$$\Rightarrow b = -20$So

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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$$\Rightarrow \frac{32}{b} = 1 - \frac{13}{5} = -\frac{8}{5}$$\Rightarrow b = -20$So,the equation of $L$ is $\frac{x}{5}-\frac{y}{20}=1$,which simplifies to $4x-y=20$.The slope of $L$ is $m_1=4$.The slope of line $K$ given by $\frac{x}{c}+\frac{y}{3}=1$ is $m_2=-\frac{3}{c}$.Since $L$ and $K$ are parallel,$m_1=m_2$.$4 = -\frac{3}{c} \Rightarrow c = -\frac{3}{4}$.The equation of line $K$ is $-\frac{4x}{3}+\frac{y}{3}=1$,which simplifies to $4x-y=-3$.The distance between two parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ is given by $d = \frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$.Here,$A=4, B=-1, C_1=-20, C_2=3$.$d = \frac{|-20-3|}{\sqrt{4^2+(-1)^2}} = \frac{|-23|}{\sqrt{16+1}} = \frac{23}{\sqrt{17}}$.

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

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