Question

Show that the area of the parallelogram formed by the lines
$2x-3y+a=0$, $3x-2y-a=0,2x-3y+3a=0$ and $3x-2y-2a=0$ is $\frac{2a^2}{5}$ sq. units.

Answer 1

R.E.F image

$CD=\sqrt{(\frac{99}{5}-\frac{129}{5}{)}^2+(\frac{119}{5}-\frac{139}{5}{)}^2}$

$=\frac{\sqrt{139}}{5}$

$AE=\frac{\frac{169}{5}-\frac{219}{5}+39}{\sqrt{13}}$ (perpendicular distance formula)

$=2a/\sqrt{3}$

Area of parallelogram $=b\times h$

$=\frac{29}{\sqrt{13}}\times \frac{\sqrt{13}9}{5}$

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