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Distance between Parallel Lines

The line L1...

Question

The line $L_{1}$ given by $\frac{x}{5} + \frac{y}{b} = 1$ passes through the point $M (13, 32)$ . The line $L_{2}$ is parallel to $L_{1}$ and has the equation $\frac{x}{c} + \frac{y}{3} = 1$ . Then the distance between $L_{1}$ and $L_{2}$ is

A

\sqrt{17}

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B

\frac{17}{\sqrt{15}}

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C

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

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D

\frac{23}{\sqrt{15}}

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Solution

The correct option is B $\frac{23}{\sqrt{17}}$
Consider the equation of line $L_{1}$ .

$\frac{x}{5} + \frac{y}{b} = 1$

Here, $L_{1}$ passes through $M (13, 32)$ .

So,

$\frac{13}{5} + \frac{32}{b} = 1$

$13 b + 160 = 5 b$

$b = - \frac{160}{8} = - 20$

Therefore, equation of line $L_{1}$ is,

$- 20 x + 5 y = - 100$ …… (1)

Consider the equation of line $L_{2}$ .

$\frac{x}{c} + \frac{y}{3} = 1$

$3 x + c y = 3 c$ ……. (2)

It is given that $L_{1}$ and $L_{2}$ (2) are parallel. So,

$\frac{3}{- 20} = \frac{c}{5}$

$c = - \frac{3}{4}$

Therefore, equation of line $L_{2}$ is,

$3 x - \frac{3 y}{4} = 3 \times (- \frac{3}{4})$

$12 x - 3 y = - 9$

$- 20 x + 5 y = - 9 \times (- \frac{5}{3})$

$- 20 x + 5 y = 15$ …… (3)

Therefore, distance between $L_{1}$ and $L_{2}$ is,

$D = \frac{| - 100 - 15 |}{\sqrt{400 + 25}} = \frac{115}{5 \sqrt{17}} = \frac{23}{\sqrt{17}}$

Hence, this is the required result.

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Similar questions

L_{1} : \frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 3}{1}

L_{2} : \frac{x}{1} = \frac{y}{- 1} = \frac{z - 5}{3}

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L_{2}

and passing through the point of intersection of

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be the lines whose equation are

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respectively A and B are two points on

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