Question

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

• A. $\frac{2a^2}{13}$ square units
• B. $\frac{2a^2}{5}$ square units
• C. $\frac{2a}{5}$ square units
• D. $\frac{4a^2}{5}$ square units

Answer 1

The correct option is B $\frac{2a^2}{5}$ square units

Given lines

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

On Comparing both equation with $y=mx+c$

$m_1=\frac{2}{3},c_1=\frac{a}{3},c_2=a$ and

$3x-2y-a=0$ and $3x-2y-2a=0$

On Comparing both equation with $y=mx+d$

$m_2=\frac{3}{2},d_1=\frac{-a}{2},d_2=-a$

Area $=\left|\frac{(c_1-c_2)(d_1-d_2)}{m_1-m_2}\right|$

Area $=\left|\frac{(\frac{a}{3}-a)(\frac{-a}{2}+a)}{\frac{2}{3}-\frac{3}{2}}\right|$

Area $=\left|\frac{\frac{-2a}{3}\left(\frac{a}{2}\right)}{\frac{-5}{6}}\right|$

Area $=\left|\frac{\frac{-2a^2}{6}}{\frac{-5}{6}}\right|$

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