Question

If $ad-bc\ne 0$, then the area (in sq. units) of the parallelogram formed by the lines $ax+by+2=0,ax+by+5=0,cx+dy+3=0$ and $cx+dy+7=0$ is

• A. $\frac{1}{|ad-bc|}$
• B. $\frac{5}{|ad-bc|}$
• C. $\frac{7}{|ad-bc|}$
• D. $\frac{12}{|ad-bc|}$

Answer 1

Answer: (D)

We have, $ax+by+2=0$ and $ax+by+5=0$ are parallel lines.
$ax+by+2=0$
$\Rightarrow y=\frac{-a}{b}x-\frac{2}{b}$
and $ax+by+5=0$
$\Rightarrow y=\frac{-a}{b}x\frac{-5}{b}$
So, $m_1=\frac{-a}{b},c_1=\frac{-2}{b},c_2=-\frac{5}{b}$
Now, again
$cx+dy+3=0$
$\Rightarrow y=\frac{-c}{d}x-\frac{3}{d}$
and $cx+dy+7=0$
$\Rightarrow y=\frac{-c}{d}x-\frac{7}{d}$
So, $m_2=\frac{-c}{d},d_1=\frac{-3}{d},d_2=\frac{-7}{d}$
Then, area of parallelogram $=\left|\frac{\left(c_1-c_2\right)\left(d_1-d_2\right)}{m_1-m_2}\right|$
$=\left|\frac{\left(\frac{-2}{b}+\frac{5}{b}\right)\left(\frac{-3}{d}+\frac{7}{d}\right)}{-\frac{a}{b}+\frac{c}{d}}\right|$
$=\left|\frac{\frac{3}{b}\times \frac{4}{d}}{\frac{-(ad-bc)}{bd}}\right|=\left|\frac{-12}{ad-bc}\right|=\frac{12}{|ad-bc|}$