Question

The area of the parallelogram formed by the lines $3y-2x=a,2y-3x+a=0$,
$2x-3y+3a=0$ and $3x-2y=2a$ is

• A. $2a^2/5$
• B. $2a/5$
• C. $a^2/5$
• D. $2a^2/7$

Answer 1

The correct option is A $2a^2/5$

$L_1=3y-2x=a$

$L_2=2x-3y+3a=0$

$L_3=2y-3x+a=0$ Ref. image

$L_4=3x-2y=2a$

$d_1={\perp }^r$ distance between to 11 lines

$=\frac{3a-a}{\sqrt{3^2+2^2}}=\frac{2a}{\sqrt{13}}$

$d_2=\frac{2a-a}{\sqrt{3^2+2^2}}=\frac{a}{\sqrt{13}}$

angle between $AD$ and $AB$,

$\tan \theta =\left|\frac{\frac{3}{2}-\frac{2}{3}}{1+\frac{3}{2}\times \frac{2}{3}}\right|$

$=\frac{5}{12}$

$\therefore \sin \theta =\frac{5}{13}$

Now,

ar. of $11gm=\frac{1}{2}\frac{d_1{d}_2}{\sin \theta }$

$=\frac{2a}{\sqrt{13}}\times \frac{a}{\sqrt{13}}\times \frac{13}{5}$

$=\frac{2a^2}{5}$

$\therefore A$ is the correct answer.