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The equation of the lines through the point (2,3) and making an intercept of length 2 units between the lines $\mathrm{y+2 x=3 \: and \: y+2 x=5}$ are

Option: 1
$\mathrm{x-2=0,3 x+4 y=18}$

Option: 2
$\mathrm{x+3=0,3 x+4 y=12}$

Option: 3
$\mathrm{y-2=0,4 x-3 y=6}$

Option: 4
none of these

Answers (1)

best_answer

Let line through $\mathrm{A(2,3)}$ be

$\mathrm{\frac{x-2}{\cos \theta}=\frac{y-3}{\sin \theta} =A B, B C }$

$\mathrm{ =\lambda, \lambda+2}$

$\therefore \mathrm{B}$ is $\mathrm{(2+\lambda \cos \theta, 3+\lambda \sin \theta)}$ which lies on $\mathrm{y+2 x=3}$ ..........(1)

The condition is $\mathrm{3+\lambda \sin \theta+2(2+\lambda \cos \theta)=3}$

Similarly, the condition for $\mathrm{C(2+(\lambda+2) \cos \theta, 3+(\lambda+2) \sin \theta)}$ is

$\mathrm{3+(\lambda+2) \sin \theta+2\{2+(\lambda+2) \cos \theta\}=5}$ ..........(2)

On eliminate $\lambda$ from (1) and (2), we obtain

$\mathrm{2 \sin \theta+4 \cos \theta=2 }$

$\mathrm{\Rightarrow 2 \cos \theta=1-\sin \theta \text { or } 4\left(1-\sin ^2 \theta\right)=(1-\sin \theta)^2 }$

$\mathrm{\Rightarrow \sin \theta=1 \text { or } 4(1+\sin \theta)=1-\sin \theta }$

$\mathrm{\Rightarrow \theta=\frac{\pi}{2} \text { and } \sin \theta=-\frac{3}{5} \Rightarrow \tan \theta=\infty,-\frac{3}{4} }$

The lines are $\mathrm{x-2=0 \: and \: 3 x+4 y=18}$ .

Hence option 1 is correct.

Posted by

Gaurav

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