The line L is given by frac{x}{5} + frac{y}{b | vedclass

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(C) Since the line $L$ passes through $(13, 32)$,we have:$\frac{13}{5} + \frac{32}{b} = 1$ $\Rightarrow \frac{32}{b} = 1 - \frac{13}{5} = -\frac{8}{5}

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Since the line $L$ passes through $(13, 32)$,we have:$\frac{13}{5} + \frac{32}{b} = 1$ $\Rightarrow \frac{32}{b} = 1 - \frac{13}{5} = -\frac{8}{5}$ $\Rightarrow b = -20$.Thus,the equation of line $L$ is $\frac{x}{5} - \frac{y}{20} = 1$,which simplifies to $4x - y = 20$.Since line $K$ is parallel to $L$,its equation is of the form $4x - y = k$.Given the equation of $K$ is $\frac{x}{c} + \frac{y}{3} = 1$,we can rewrite it as $y = -\frac{3}{c}x + 3$,or $3x + cy = 3c$. Comparing slopes,the slope of $L$ is $4$,so the slope of $K$ is $-\frac{3}{c} = 4$,which gives $c = -\frac{3}{4}$.Substituting $c$ into the equation $\frac{x}{c} + \frac{y}{3} = 1$,we get $\frac{x}{-3/4} + \frac{y}{3} = 1$ $\Rightarrow -\frac{4x}{3} + \frac{y}{3} = 1$ $\Rightarrow -4x + y = 3$,or $4x - y = -3$.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 $L$ is given by $\frac{x}{5} + \frac{y}{b} = 1$ and 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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