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.

Answer:

Let ABCD be a parallelogram  the equation of whose sides AB, BC, CD and DA are given as:2021-02-01_8-39-59

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.

Answer:

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

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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 .

Answer:

Given equations:2021-02-01_14-38-22

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}