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 2a²/5 sq. units.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

The sides of parallelogram are

2x - 3y + a = 0

3x - 2y - a = 0

2x - 3y + 3a = 0

and

3x - 2y - 2a = 0

Let assume that ABCD is the required parallelogram such that

Equation of AB : 2x - 3y + a = 0

Equation of BC : 3x - 2y - a = 0

Equation of CD : 2x - 3y + 3a = 0

and

Equation of AD : 3x - 2y - 2a = 0

We know that,

Area of parallelogram bounded by the lines

$\rm \: y = m_1x + c_1$

$\rm \: y = m_2x + c_3$

$\rm \: y = m_1x + c_2$

$\rm \: y = m_2x + c_4$

is given by

$\boxed{\sf{ \:Area \: = \: \bigg | \frac{(c_1 - c_2)(c_3 - c_4)}{m_1 - m_2} \bigg| \: \: }}$

So, let first find all these values.

Now, Consider Equation of AB and CD as both are parallel lines having same slope

$\rm \: 2x - 3y + a = 0$

$\rm \: 3y = 2x + a$

$\rm\implies \:y = \dfrac{2}{3}x + \dfrac{a}{3}$

So, on comparing with slope intercept form of the line, we get

$\rm \: m_1 = \dfrac{2}{3}$

and

$\rm \: c_1 = \dfrac{a}{3}$

Now, Consider Equation of CD

$\rm \: 2x - 3y + 3a = 0$

$\rm \: 3y = 2x + 3a$

$\rm\implies \:y = \dfrac{2}{3}x + a$

So, on comparing with slope intercept form of the line, we get

$\rm \: m_1 = \dfrac{2}{3}$

and

$\rm \: c_2 = a$

Now, Consider Equation of line BC and AD as they are parallel lines.

$\rm \: 3x - 2y - a = 0$

$\rm \: 2y = 3x -a$

$\rm\implies \:y = \dfrac{3}{2}x - \dfrac{a}{2}$

So, on comparing with slope intercept form of the line, we get

$\rm \: m_2 = \dfrac{3}{2}$

and

$\rm \: c_3 = - \dfrac{a}{2}$

Now, Consider Equation of AD

$\rm \: 3x - 2y - 2a = 0$

$\rm \: 2y = 3x -2a$

$\rm\implies \:y = \dfrac{3}{2}x - a$

So, on comparing with slope intercept form of the line, we get

$\rm \: m_2 = \dfrac{3}{2}$

and

$\rm \: c_4 = a$

Now, we know that

Area of parallelogram is given by

$\boxed{\sf{ \:Area \: = \: \bigg | \frac{(c_1 - c_3)(c_2 - c_4)}{m_1 - m_2} \bigg| \: \: }}$

So, on substituting the values, we get

$\rm \: Area = \bigg |\dfrac{\bigg(\dfrac{a}{3} -a\bigg)\bigg( - \dfrac{a}{2} + a \: \bigg) }{\dfrac{2}{3} - \dfrac{3}{2} } \bigg|$

$\rm \: Area = \bigg |\dfrac{\bigg( - \dfrac{2a}{3}\bigg)\bigg(\dfrac{a}{2} \ \bigg) }{\dfrac{4 - 9}{6} } \bigg|$

$\rm \: Area = \bigg |\dfrac{\bigg( - \dfrac{2 {a}^{2} }{6}\bigg) \ \bigg) }{\dfrac{ - 5}{6} } \bigg|$

$\rm\implies \:Area \: = \: \dfrac{ {2a}^{2} }{5} \: square \: units$

Hence, Proved