Question 1: Prove that the area of the parallelogram formed by the lines:

$a_1x+b_1y+c_1=0, \ \ a_1x+b_1y+d_1=0, \ \ a_2x+b_2y+c_2=0, \ \ a_2x+b_2y+d_2=0$ is

$\Big|$ $\frac{(d_1-c_1)(d_2-c_2)}{a_1b_2-a_2b_1}$ $\Big|$ sq. units. Deduce the condition for these lines to form a rhombus.

Let $ABCD$ be a parallelogram  the equation of whose sides $AB, BC, CD$ and $DA$ are given as:

$AB: \ \ a_1x+b_1y+c_1=0$

$BC: \ \ a_2x+b_2y+c_2=0$

$CD: \ \ a_1x+b_1y+d_1=0$

$DA: \ \ a_2x+b_2y+d_2=0$

Let $p_1$ and $p_2$ be the distances between the pair of parallel sides of the parallelogram.

In $\triangle ALD$ and $\triangle AMB$ we get,

$\sin \theta =$ $\frac{p_1}{AD}$  and $\sin \theta =$ $\frac{p_2}{AB}$ respectively.

$\Rightarrow AD =$ $\frac{p_1}{\sin \theta}$  and $AB =$ $\frac{p_2}{\sin \theta}$

Now, Area of parallelogram $ABCD = AB \times p_1 =$ $\frac{p_1p_2}{\sin \theta}$       $\Big[ \because AB =$ $\frac{p_2}{\sin \theta}$ $\Big]$

Also, Area of parallelogram $ABCD = AD \times p_2 =$ $\frac{p_1p_2}{\sin \theta}$       $\Big[ \because AB =$ $\frac{p_1}{\sin \theta}$ $\Big]$

Therefore, the area of a parallelogram is $\frac{p_1p_2}{\sin \theta}$,  where $p_1$ and $p_2$ are the distances between pairs of parallel sides and $\theta$ is the angle between the two adjacent sides.

Let $m_1$ and $m_2$ be the slopes of sides $AB$ and $AD$ respectively. Then,

$m_1 = -$ $\frac{a_1}{b_1}$  and $m_2 = -$ $\frac{a_2}{b_2}$

The angle $\theta$ between $AB$ and $AD$ is given by

$\tan \theta =$ $\frac{m_1-m_2}{1+m_1m_2}$

$\Rightarrow \tan \theta =$ $\frac{\frac{-a_1}{b_1}+\frac{a_2}{b_2}}{1+\frac{a_1a_2}{b_1b_2}}$

$\Rightarrow \tan \theta =$ $\frac{a_2b_1-a_1 b_2}{a_1a_2+b_1b_2}$

$\Rightarrow \sin \theta =$ $\frac{a_2b_1-a_1 b_2}{\sqrt{({a_1}^2+{a_2}^2)({b_1}^2+{b_2}^2)}}$

We have $p_1 =$ Distance between parallel sides $AB$ and $AD =$ $\frac{|c_1-d_1|}{\sqrt{{a_1}^2+{b_1}^2}}$

and $p_2 =$ Distance between parallel sides $AB$ and $AD =$ $\frac{|c_2-d_2|}{\sqrt{{a_2}^2+{b_2}^2}}$

Therefore are of parallelogram $ABCD =$ $\frac{p_1p_2}{\sin \theta }$

$= \frac{|c_1-d_1|}{\sqrt{{a_1}^2+{b_1}^2}} \times \frac{|c_2-d_2|}{\sqrt{{a_2}^2+{b_2}^2}} \times \frac{\sqrt{({a_1}^2+{a_2}^2)({b_1}^2+{b_2}^2)}} {a_2b_1-a_1 b_2}$

$= \frac{|c_1-d_1||c_2-d_2|}{| a_2b_1-a_1 b_2 |}$

$= \Big| \frac{(d_1-c_1)(d_2-c_2)}{ a_2b_1-a_1 b_2 } \Big|$

In case of a rhombus, $p_1$ and $p_2$ are equal. Hence the condition is:

$\frac{|c_1-d_1|}{\sqrt{{a_1}^2+{b_1}^2}}$ $=$ $\frac{|c_2-d_2|}{\sqrt{{a_2}^2+{b_2}^2}}$

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Question 2: Prove that the area of the parallelogram formed by the lines $3x -4y+a = 0, \ \ 3x -4y+3a = 0, \ \ 4x-3y-a=0, \ \ 4x-3y-2a=0$ is $\frac{2a^2}{7}$ sq. units.

Given lines are:

$3x -4y+a = 0$     … … … … … i)

$3x -4y+3a = 0$     … … … … … ii)

$4x-3y-a=0$     … … … … … iii)

$4x-3y-2a=0$     … … … … … iv)

Area of parallelogram $=$ $\frac{|c_1-d_1||c_2-d_2|}{| a_2b_1-a_1 b_2 |}$ $=$ $\frac{|(3-3a)||(2a-a)|}{|-9+16 |}$ $=$ $\frac{2a^2}{7}$ sq. units

$\\$

Question 3: Show that the diagonals of  the parallelogram whose sides are $lx+my+n=0, \ \ lx+my+n'=0, \ \ mx+ly+n = 0, \ \ mx+ly +n'=0$ include an angle $\pi/2$.

Given equations:

$lx+my+n=0$     … … … … … i)

$mx+ly+n' = 0$     … … … … … iii)

$lx+my+n'=0$     … … … … … ii)

$mx+ly +n =0$     … … … … … iv)

Solving i) and ii) we get the point of intersection $B= \Big($ $\frac{mn'-ln}{l^2-m^2}$ $,$ $\frac{mn-ln'}{l^2-m^2}$ $\Big)$

Solving ii) and iii) we get the point of intersection $C= \Big($ $\frac{-n'}{m+l}$ $,$ $\frac{-n'}{m+l}$ $\Big)$

Solving iii) and iv) we get the point of intersection $D= \Big($ $\frac{mn-ln'}{l^2-m^2}$ $,$ $\frac{mn'-ln'}{l^2-m^2}$ $\Big)$

Solving iv) and i) we get the point of intersection $E= \Big($ $\frac{-n}{m+l}$ $,$ $\frac{-n}{m+l}$ $\Big)$

Slope of $AC (m_1) = \Bigg($ $\frac{ \frac{-n'}{m+l} - ( \frac{-n}{m+l} ) }{ \frac{-n'}{m+l} - ( \frac{-n}{m+l} ) }$ $\Bigg) = 1$

Slope of $BD (m_2) = \Bigg($ $\frac{ \frac{mn'-ln}{l^2-m^2} - \frac{mn-ln'}{l^2-m^2} }{ \frac{mn-ln'}{l^2-m^2} - \frac{mn'-ln}{l^2-m^2} }$ $\Bigg) = -1$

Since $m_1 \times m_2 = -1$ .

Therefore the diagonals are perpendicular to each other.

Hence the diagonals include an angle $\frac{\pi}{2}$