Question

The area of the parallelogram formed by the lines $3x-4y+1=0,3x-4y+3=0,4x-3y-1=0$ and $4x-3y-2=0,$ is

• A. $\frac{1}{6}sq.units$
• B. $\frac{2}{7}sq.units$
• C. $\frac{3}{8}sq.units$
• D. none of the above

Answer 1

Answer: (B)

$\frac{2}{7}sq.units$

Given equations of lines are
$3x-4y+1=0,3x-4y+3=0,4x-3y-1=0$ and $4x-3y-2=0,$

Here $m_1=\frac{3}{4},m_2=\frac{4}{3}$
and $c_1=\frac{1}{4},c_2=\frac{3}{4},d_1=-\frac{1}{3},d_2=-\frac{2}{3}$
Area of parallelogram $=\left|\frac{(c_1-c_2)(d_1-d_2)}{m_1-m_2}\right|$
$=\frac{2}{7}$ sq.units