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.

Answer:

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 .

Answer:

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}