Question 1: Prove that the area of the parallelogram formed by the lines: $\displaystyle a_1x+b_1y+c_1=0, \ \ a_1x+b_1y+d_1=0, \ \ a_2x+b_2y+c_2=0, \ \ a_2x+b_2y+d_2=0$ is $\displaystyle \Big| \frac{(d_1-c_1)(d_2-c_2)}{a_1b_2-a_2b_1} \Big| \text{sq. units. }$

Deduce the condition for these lines to form a rhombus.

Let $\displaystyle ABCD$ be a parallelogram the equation of whose sides $\displaystyle AB, BC, CD \text{ and } DA$ are given as:  $\displaystyle AB: \ \ a_1x+b_1y+c_1=0$ $\displaystyle BC: \ \ a_2x+b_2y+c_2=0$ $\displaystyle CD: \ \ a_1x+b_1y+d_1=0$ $\displaystyle DA: \ \ a_2x+b_2y+d_2=0$

Let $\displaystyle p_1 \text{ and } p_2$ be the distances between the pair of parallel sides of the parallelogram.

In $\displaystyle \triangle ALD \text{ and } \triangle AMB$ we get, $\displaystyle \sin \theta = \frac{p_1}{AD} \text{ and } \sin \theta = \frac{p_2}{AB} \text{ respectively. }$ $\displaystyle \Rightarrow AD = \frac{p_1}{\sin \theta} \text{ and } AB = \frac{p_2}{\sin \theta}$ $\displaystyle \text{Now, Area of parallelogram } ABCD = AB \times p_1 = \frac{p_1p_2}{\sin \theta} \Big[ \because AB = \frac{p_2}{\sin \theta} \Big]$ $\displaystyle \text{Also, Area of parallelogram } ABCD = AD \times p_2 = \frac{p_1p_2}{\sin \theta} \Big[ \because AB = \frac{p_1}{\sin \theta} \Big]$

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

Let $\displaystyle m_1 \text{ and } m_2$ be the slopes of sides $\displaystyle AB \text{ and } AD$ respectively. Then, $\displaystyle m_1 = - \frac{a_1}{b_1} \text{ and } m_2 = - \frac{a_2}{b_2}$

The angle $\displaystyle \theta$ between $\displaystyle AB \text{ and } AD$ is given by $\displaystyle \tan \theta = \frac{m_1-m_2}{1+m_1m_2}$ $\displaystyle \Rightarrow \tan \theta = \frac{\frac{-a_1}{b_1}+\frac{a_2}{b_2}}{1+\frac{a_1a_2}{b_1b_2}}$ $\displaystyle \Rightarrow \tan \theta = \frac{a_2b_1-a_1 b_2}{a_1a_2+b_1b_2}$ $\displaystyle \Rightarrow \sin \theta = \frac{a_2b_1-a_1 b_2}{\sqrt{({a_1}^2+{a_2}^2)({b_1}^2+{b_2}^2)}}$

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

and $\displaystyle p_2 = \displaystyle \text{Distance between parallel sides } AB \text{ and } AD = \frac{|c_2-d_2|}{\sqrt{{a_2}^2+{b_2}^2}}$ $\displaystyle \text{Therefore are of parallelogram } ABCD = \frac{p_1p_2}{\sin \theta }$ $\displaystyle = \frac{|c_1-d_1|}{\sqrt{{a_1}^2+{b_1}^2}} \times \frac{|c_2-d_2|}{\sqrt{{a_2}^2+{b_2}^2}} \times \frac{\sqrt{({a_1}^2+{a_2}^2)({b_1}^2+{b_2}^2)}} {a_2b_1-a_1 b_2}$ $\displaystyle = \frac{|c_1-d_1||c_2-d_2|}{| a_2b_1-a_1 b_2 |}$ $\displaystyle = \Big| \frac{(d_1-c_1)(d_2-c_2)}{ a_2b_1-a_1 b_2 } \Big|$

In case of a rhombus, $\displaystyle p_1 \text{ and } p_2$ are equal. Hence the condition is: $\displaystyle \frac{|c_1-d_1|}{\sqrt{{a_1}^2+{b_1}^2}} = \frac{|c_2-d_2|}{\sqrt{{a_2}^2+{b_2}^2}}$ $\displaystyle \\$

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

Answer: Given lines are: $\displaystyle 3x -4y+a = 0$ … … … … … i) $\displaystyle 3x -4y+3a = 0$ … … … … … ii) $\displaystyle 4x-3y-a=0$ … … … … … iii) $\displaystyle 4x-3y-2a=0$ … … … … … iv) $\displaystyle \text{Area of parallelogram } = \frac{|c_1-d_1||c_2-d_2|}{| a_2b_1-a_1 b_2 |} = \frac{|(3-3a)||(2a-a)|}{|-9+16 |} = \frac{2a^2}{7}$ sq. units $\displaystyle \\$

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

Given equations: $\displaystyle lx+my+n=0$ … … … … … i) $\displaystyle mx+ly+n' = 0$ … … … … … iii) $\displaystyle lx+my+n'=0$ … … … … … ii) $\displaystyle mx+ly +n =0$ … … … … … iv) $\displaystyle \text{Solving i) and ii) we get the point of intersection } B= \Big( \frac{mn'-ln}{l^2-m^2} , \frac{mn-ln'}{l^2-m^2} \Big)$ $\displaystyle \text{Solving ii) and iii) we get the point of intersection } C= \Big( \frac{-n'}{m+l} , \frac{-n'}{m+l} \Big)$ $\displaystyle \text{Solving iii) and iv) we get the point of intersection } D= \Big( \frac{mn-ln'}{l^2-m^2} , \frac{mn'-ln'}{l^2-m^2} \Big)$ $\displaystyle \text{Solving iv) and i) we get the point of intersection } E= \Big( \frac{-n}{m+l} , \frac{-n}{m+l} \Big)$ $\displaystyle \text{Slope of } AC (m_1) = \Bigg( \frac{ \frac{-n'}{m+l} - ( \frac{-n}{m+l} ) }{ \frac{-n'}{m+l} - ( \frac{-n}{m+l} ) } \Bigg) = 1$ $\displaystyle \text{Slope of } BD (m_2) = \Bigg( \frac{ \frac{mn'-ln}{l^2-m^2} - \frac{mn-ln'}{l^2-m^2} }{ \frac{mn-ln'}{l^2-m^2} - \frac{mn'-ln}{l^2-m^2} } \Bigg) = -1$

Since $\displaystyle m_1 \times m_2 = -1$ .

Therefore the diagonals are perpendicular to each other. $\displaystyle \text{Hence the diagonals include an angle } \frac{\pi}{2}$