Class 11: The Straight Line

$\displaystyle \textbf{Question 1: }\text{Prove that the area of the parallelogram formed by the lines}$
$\displaystyle a_1x+b_1y+c_1=0,\ a_1x+b_1y+d_1=0,\ a_2x+b_2y+c_2=0,\ a_2x+b_2y+d_2=0$
$\displaystyle \text{ is }\left|\frac{(d_1-c_1)(d_2-c_2)}{a_1b_2-a_2b_1}\right|\text{ sq. units. Deduce the condition}$
$\displaystyle \text{for these lines to form a rhombus.}$
$\displaystyle \text{Answer:}$ $\displaystyle \text{Let }ABCD\text{ be the required parallelogram with sides}$
$\displaystyle AB:a_1x+b_1y+c_1=0,\qquad BC:a_2x+b_2y+c_2=0,$
$\displaystyle CD:a_1x+b_1y+d_1=0,\qquad AD:a_2x+b_2y+d_2=0.$
$\displaystyle \text{Let }p_1\text{ and }p_2\text{ be the distances between the pairs of parallel sides }$
$\displaystyle AB,CD\text{ and }BC,AD\text{ respectively.}$
$\displaystyle \text{If }\theta\text{ is the angle between the adjacent sides, then}$
$\displaystyle \sin\theta=\frac{p_1}{AD}=\frac{p_2}{AB}.$
$\displaystyle \therefore AD=\frac{p_1}{\sin\theta},\qquad AB=\frac{p_2}{\sin\theta}.$
$\displaystyle \therefore \text{Area of parallelogram }=AB\times p_1=\frac{p_1p_2}{\sin\theta}.$
$\displaystyle \text{Now }m_1=-\frac{a_1}{b_1},\qquad m_2=-\frac{a_2}{b_2}.$
$\displaystyle \tan\theta=\frac{m_1-m_2}{1+m_1m_2}=\frac{a_2b_1-a_1b_2}{a_1a_2+b_1b_2}.$
$\displaystyle \therefore \sin\theta=\frac{|a_1b_2-a_2b_1|}{\sqrt{(a_1^2+b_1^2)(a_2^2+b_2^2)}}.$
$\displaystyle \text{Also,}$
$\displaystyle p_1=\frac{|c_1-d_1|}{\sqrt{a_1^2+b_1^2}},\qquad$
$\displaystyle p_2=\frac{|c_2-d_2|}{\sqrt{a_2^2+b_2^2}}.$
$\displaystyle \therefore \text{Area}=\frac{p_1p_2}{\sin\theta}$
$\displaystyle =\frac{|c_1-d_1|}{\sqrt{a_1^2+b_1^2}}\times\frac{|c_2-d_2|}{\sqrt{a_2^2+b_2^2}}$
$\displaystyle \times\frac{\sqrt{(a_1^2+b_1^2)(a_2^2+b_2^2)}}{|a_1b_2-a_2b_1|}$
$\displaystyle =\frac{|c_1-d_1||c_2-d_2|}{|a_1b_2-a_2b_1|}$
$\displaystyle =\left|\frac{(d_1-c_1)(d_2-c_2)}{a_1b_2-a_2b_1}\right|.$
$\displaystyle \text{For a rhombus, }AB=AD.$
$\displaystyle \therefore \frac{p_2}{\sin\theta}=\frac{p_1}{\sin\theta}\Rightarrow p_1=p_2.$
$\displaystyle \therefore \frac{|c_1-d_1|}{\sqrt{a_1^2+b_1^2}}=\frac{|c_2-d_2|}{\sqrt{a_2^2+b_2^2}}.$
$\displaystyle \text{This is the required condition for the parallelogram to be a rhombus.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Prove that the area of the parallelogram formed by the lines}$
$\displaystyle 3x-4y+a=0,\quad 3x-4y+3a=0,\quad 4x-3y-a=0\text{ and }4x-3y-2a=0$
$\displaystyle \text{ is }\frac{2a^2}{7}\text{ sq. units.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given lines are}$
$\displaystyle 3x-4y+a=0\qquad\ldots\ldots\ldots\text{(i)}$
$\displaystyle 3x-4y+3a=0\qquad\ldots\ldots\ldots\text{(ii)}$
$\displaystyle 4x-3y-a=0\qquad\ldots\ldots\ldots\text{(iii)}$
$\displaystyle 4x-3y-2a=0\qquad\ldots\ldots\ldots\text{(iv)}$
$\displaystyle \text{Here, }a_1=3,\ b_1=-4,\ c_1=a,\ d_1=3a,$
$\displaystyle a_2=4,\ b_2=-3,\ c_2=-a,\ d_2=-2a.$
$\displaystyle \text{Area of the parallelogram}$
$\displaystyle =\frac{|c_1-d_1||c_2-d_2|}{|a_1b_2-a_2b_1|}$
$\displaystyle =\frac{|a-3a||-a-(-2a)|}{|3(-3)-4(-4)|}$
$\displaystyle =\frac{|-2a||a|}{|-9+16|}$
$\displaystyle =\frac{2|a|^2}{7}$
$\displaystyle =\frac{2a^2}{7}\text{ sq. units.}$
$\displaystyle \therefore \text{The area of the parallelogram is }\frac{2a^2}{7}\text{ sq. units.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Show that the diagonals of the parallelogram whose sides are}$
$\displaystyle lx+my+n=0,\quad lx+my+n'=0,\quad mx+ly+n=0\text{ and }mx+ly+n'=0$
$\displaystyle \text{intersect at an angle }\frac{\pi}{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given lines are}$
$\displaystyle lx+my+n=0\qquad\ldots\ldots\ldots\text{(i)}$
$\displaystyle lx+my+n'=0\qquad\ldots\ldots\ldots\text{(ii)}$
$\displaystyle mx+ly+n=0\qquad\ldots\ldots\ldots\text{(iii)}$
$\displaystyle mx+ly+n'=0\qquad\ldots\ldots\ldots\text{(iv)}$
$\displaystyle \text{Let }A,\ B,\ C\text{ and }D\text{ be the intersections of (i), (iii); (i), (iv); (ii), (iv)}$
$\displaystyle \text{and (ii), (iii), respectively.}$
$\displaystyle \text{Solving (i) and (iii),}$
$\displaystyle A=\left(-\frac{n}{l+m},-\frac{n}{l+m}\right).$
$\displaystyle \text{Solving (ii) and (iv),}$
$\displaystyle C=\left(-\frac{n'}{l+m},-\frac{n'}{l+m}\right).$
$\displaystyle \therefore \text{Slope of diagonal }AC$
$\displaystyle =\frac{-\frac{n'}{l+m}+\frac{n}{l+m}}{-\frac{n'}{l+m}+\frac{n}{l+m}}=1.$
$\displaystyle \text{Solving (i) and (iv),}$
$\displaystyle B=\left(\frac{mn'-ln}{l^2-m^2},\frac{mn-ln'}{l^2-m^2}\right).$
$\displaystyle \text{Solving (ii) and (iii),}$
$\displaystyle D=\left(\frac{mn-ln'}{l^2-m^2},\frac{mn'-ln}{l^2-m^2}\right).$
$\displaystyle \therefore \text{Slope of diagonal }BD$
$\displaystyle =\frac{\frac{mn'-ln}{l^2-m^2}-\frac{mn-ln'}{l^2-m^2}}{\frac{mn-ln'}{l^2-m^2}-\frac{mn'-ln}{l^2-m^2}}=-1.$
$\displaystyle \therefore m_{AC}\times m_{BD}=1\times(-1)=-1.$
$\displaystyle \therefore AC\perp BD.$
$\displaystyle \therefore \text{The diagonals intersect at an angle }\frac{\pi}{2}.$
$\displaystyle \\$