The line L_1 given by frac{x}{p} + frac{y}{2} | vedclass

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(A) Step $1$: Find $p$ for $L_1$. Since $L_1$ passes through $(5, 0)$, $\frac{5}{p} + \frac{0}{2} = 1 \implies p = 5$. So, $L_1: \frac{x}{5} + \frac{y

Vedclass · Accepted Answer

$\frac{10}{\sqrt{29}}$

Vedclass · Answer

(A) Step $1$: Find $p$ for $L_1$. Since $L_1$ passes through $(5, 0)$, $\frac{5}{p} + \frac{0}{2} = 1 \implies p = 5$. So, $L_1: \frac{x}{5} + \frac{y}{2} = 1 \implies 2x + 5y - 10 = 0$.Step $2$: Since $L_2$ is parallel to $L_1$, its equation is $2x + 5y + k = 0$. Given $L_2: \frac{x}{10} + \frac{y}{q} = 1$, we rewrite it as $qx + 10y - 10q = 0$. Comparing coefficients, $\frac{q}{2} = \frac{10}{5} \implies q = 4$. Thus $L_2: 4x + 10y - 40 = 0 \implies 2x + 5y - 20 = 0$.Step $3$: The distance between parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.Step $4$: Here $A=2, B=5, C_1=-10, C_2=-20$. So, $d = \frac{|-10 - (-20)|}{\sqrt{2^2 + 5^2}} = \frac{10}{\sqrt{4 + 25}} = \frac{10}{\sqrt{29}}$.

The line $L_1$ given by $\frac{x}{p} + \frac{y}{2} = 1$ passes through the point $(5, 0)$. The line $L_2$ given by $\frac{x}{10} + \frac{y}{q} = 1$ is parallel to $L_1$. Find the distance between the lines $L_1$ and $L_2$.

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