Question

If $a d-b c \neq 0$, then the area (in sq. units) of the parallelogram formed by the lines $a x+b y+2=0, a x+b y+5=0, c x+d y+3=0$ and $c x+d y+7=0$ is

• A. $\frac{1}{|a d-b c|}$
• B. $\frac{5}{|a d-b c|}$
• C. $\frac{7}{|a d-b c|}$
• D. $\frac{12}{|a d-b c|}$

Answer 1

Answer: (D)

We have, $a x+b y+2=0$ and $a x+b y+5=0$ are parallel lines.
$a x+b y+2=0 $
$\Rightarrow y=\frac{-a}{b} x-\frac{2}{b}$
and $a x+b y+5=0$
$ \Rightarrow y=\frac{-a}{b} x \frac{-5}{b}$
So, $m_{1}=\frac{-a}{b}, c_{1}=\frac{-2}{b}, c_{2}=-\frac{5}{b}$
Now, again
$c x+d y+3=0 $
$\Rightarrow y=\frac{-c}{d} x-\frac{3}{d}$
and $c x+d y+7=0 $
$\Rightarrow y=\frac{-c}{d} x-\frac{7}{d}$
So, $m_{2}=\frac{-c}{d}, d_{1}=\frac{-3}{d}, d_{2}=\frac{-7}{d}$
Then, area of parallelogram $=\left|\frac{\left(c_{1}-c_{2}\right)\left(d_{1}-d_{2}\right)}{m_{1}-m_{2}}\right|$
$=\left|\frac{\left(\frac{-2}{b}+\frac{5}{b}\right)\left(\frac{-3}{d}+\frac{7}{d}\right)}{-\frac{a}{b}+\frac{c}{d}}\right|$
$=\left|\frac{\frac{3}{b} \times \frac{4}{d}}{\frac{-(a d-b c)}{b d}}\right|=\left|\frac{-12}{a d-b c}\right|=\frac{12}{|a d-b c|}$