Question

Area of the parallelogram formed by the lines $y=mx,y=mx+1,y=nx,y=nx+1$ is :

• A. $\frac{m+n}{(m-n{)}^2}$
• B. $\frac{2}{(m-n{)}^2}$
• C. $\frac{1}{(m+n)}$
• D. $\frac{1}{|m-n|}$

Answer 1

The correct option is D $\frac{1}{|m-n|}$

$y=mx+1$ and $y=nx+1$
$y=mx$ and $y=nx$
This two lines This two lines are also parallel
are parallel
distance between $y=nx+1$ and $y=nx$ is a
So, $d=\left|\frac{1}{\sqrt{1+n^2}}\right|$
and distance between $y=mx+1$ and $y=mx$ is b
So, $b=\left|\frac{1}{\sqrt{1+m^2}}\right|$
$\therefore$ area of parallelogram ABCD$=d\times \frac{\sqrt{1+n^2}}{|(n-m)|}=d\times AB$
$=\frac{1}{\sqrt{1+n^2}}\frac{\sqrt{1+n^2}}{|n-m|}$
$=\frac{1}{|n-m|}$