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Straight Lines and Pair of Straight Lines question

2010 · Shift 0 · Q43

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Straight Lines and Pair of Straight Lines
2010 · Shift 0 · Q43

Straight Lines and Pair of Straight Lines question

2010 · Shift 0 · Q43

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1

The line

L

given by

{x \over 5} + {y \over b} = 1

passes through the point

\left( {13,32} \right)

. The line K is parrallel to

L

and has the equation

{x \over c} + {y \over 3} = 1.

Then the distance between

L

and

K

is :

A
$\sqrt {17}$
B
${{17} \over {\sqrt {15} }}$
C
${{23} \over {\sqrt {17} }}$
D
${{23} \over {\sqrt {15} }}$

View written solutionFree

Correct answer: C

Find the equation of line $L$

Given $\frac{x}{5}+\frac{y}{b}=1$ and it passes through $(13,32)$ .

Substitute $(x,y)=(13,32)$ : $\frac{13}{5}+\frac{32}{b}=1$ $\frac{32}{b}=1-\frac{13}{5}=\frac{5-13}{5}=-\frac{8}{5}$ $b=\frac{32}{-8/5}=-20$

So line $L$ is $\frac{x}{5}+\frac{y}{-20}=1$ $\frac{x}{5}-\frac{y}{20}=1$

Multiply by $20$ : $4x-y=20$
Find line $K$ parallel to $L$

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

Multiply by $3c$ : $3x+cy=3c$

Since $K$ is parallel to $L$ , their slopes must be equal.

From $L: 4x-y=20$ , we get $y=4x-20$ so slope of $L$ is $4$ .

From $K$ : $\frac{x}{c}+\frac{y}{3}=1$ $y=3-\frac{3}{c}x$ so slope of $K$ is $-\frac{3}{c}$ .

For parallel lines: $-\frac{3}{c}=4$ $c=-\frac{3}{4}$

Substitute into $K$ : $\frac{x}{-3/4}+\frac{y}{3}=1$ $-\frac{4x}{3}+\frac{y}{3}=1$ Multiply by $3$ : $-4x+y=3$ or $4x-y=-3$
Distance between the parallel lines

The two lines are: $L: 4x-y-20=0$ $K: 4x-y+3=0$

Distance between parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$ is $d=\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$

Here, $a=4,\quad b=-1,\quad c_1=-20,\quad c_2=3$

Thus $d=\frac{|(-20)-3|}{\sqrt{4^2+(-1)^2}}=\frac{23}{\sqrt{17}}$
Match with options

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

This is Option C.

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