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

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(D) Given line $L: \frac{x}{5} + \frac{y}{b} = 1$ passes through $(13, 32)$.
Substituting the point in $L$: $\frac{13}{5} + \frac{32}{b} = 1 \implies

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given line $L: \frac{x}{5} + \frac{y}{b} = 1$ passes through $(13, 32)$.
Substituting the point in $L$: $\frac{13}{5} + \frac{32}{b} = 1 \implies \frac{32}{b} = 1 - \frac{13}{5} = -\frac{8}{5}$.
Thus,$b = \frac{32 	imes 5}{-8} = -20$.
The equation of $L$ is $\frac{x}{5} - \frac{y}{20} = 1$,which simplifies to $4x - y - 20 = 0$.
Since line $K: \frac{x}{c} + \frac{y}{3} = 1$ is parallel to $L$,the slopes must be equal. The slope of $L$ is $4$. The slope of $K$ is $-\frac{1/c}{1/3} = -\frac{3}{c}$.
Equating slopes: $4 = -\frac{3}{c} \implies c = -\frac{3}{4}$.
Substituting $c$ in $K$: $\frac{x}{-3/4} + \frac{y}{3} = 1 \implies -\frac{4x}{3} + \frac{y}{3} = 1 \implies 4x - y + 3 = 0$.
The distance $d$ between parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
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 its equation is $\frac{x}{c} + \frac{y}{3} = 1$. Find the distance between $L$ and $K$.

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