Class 11: The Straight Line

$\displaystyle \textbf{\hspace{0.2cm} DISTANCE\ BETWEEN\ PARALLEL\ LINES}$
$\displaystyle \text{If two lines are parallel, then they have the same distance between them throughout. Therefore,}$
$\displaystyle \text{to find the distance between two parallel lines choose an arbitrary point on one of them and find}$
$\displaystyle \text{the length of the perpendicular on the other. In order to choose a point on a line, we give an}$
$\displaystyle \text{arbitrary value to }x\text{ or }y\text{ and find the value of the other variable.}$
$\displaystyle \text{We may use the following algorithm to find the distance between two parallel lines.}$

$\displaystyle \textbf{ALGORITHM}$
$\displaystyle \textbf{STEP I}\qquad \text{Choose a point on any one of the two lines by giving a particular value to }x$
$\displaystyle \text{or }y\text{ of your choice.}$
$\displaystyle \textbf{STEP II}\qquad \text{Find the length of the perpendicular from the chosen point in Step I to the other line.}$
$\displaystyle \textbf{STEP III}\qquad \text{The length obtained in Step II is the required distance between the two lines.}$

$\displaystyle \textbf{THEOREM}$
$\displaystyle \text{Prove that the distance between two parallel lines }ax+by+c_1=0\text{ and }ax+by+c_2=0$
$\displaystyle \text{ is given by }d=\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}.$

$\displaystyle \textbf{PROOF}$
$\displaystyle \text{Given lines are}$
$\displaystyle ax+by+c_1=0\qquad\ldots\ldots\ldots\text{(i)}$
$\displaystyle ax+by+c_2=0\qquad\ldots\ldots\ldots\text{(ii)}$
$\displaystyle \text{Let }P(h,k)\text{ be a point on the line }ax+by+c_1=0.$
$\displaystyle \text{Then,}$
$\displaystyle ah+bk+c_1=0\qquad\ldots\ldots\ldots\text{(iii)}$
$\displaystyle \text{Clearly, the distance }d\text{ between parallel lines (i) and (ii) is equal to the length of the perpendicular from }P\text{ on line (ii).}$
$\displaystyle \therefore d=PM$
$\displaystyle =\left|\frac{ah+bk+c_2}{\sqrt{a^2+b^2}}\right|$
$\displaystyle =\left|\frac{-c_1+c_2}{\sqrt{a^2+b^2}}\right|\qquad\left[\text{From (iii): }ah+bk=-c_1\right]$
$\displaystyle =\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}.$
$\displaystyle \therefore \text{The distance between the parallel lines }ax+by+c_1=0\text{ and }ax+by+c_2=0$
$\displaystyle \text{ is }d=\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}.$
$\displaystyle \text{Q.E.D.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 1: }\text{Determine the distance between the following pair of parallel lines:}$

$\displaystyle \text{i) }4x-3y-9=0\text{ and }4x-3y-24=0$

$\displaystyle \text{Answer:}$
$\displaystyle \text{Given parallel lines }4x-3y-9=0\text{ and }4x-3y-24=0.$
$\displaystyle \text{Distance between two parallel lines }ax+by+c_1=0\text{ and }ax+by+c_2=0$
$\displaystyle \text{ is }\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}.$
$\displaystyle \therefore d=\frac{|-9-(-24)|}{\sqrt{4^2+(-3)^2}}=\frac{15}{5}=3\text{ units}.$

$\displaystyle \\$

$\displaystyle \text{ii) }8x+15y-34=0\text{ and }8x+15y+31=0$

$\displaystyle \text{Answer:}$
$\displaystyle \text{Given parallel lines }8x+15y-34=0\text{ and }8x+15y+31=0.$
$\displaystyle \therefore d=\frac{|-34-31|}{\sqrt{8^2+15^2}}=\frac{65}{17}\text{ units}.$

$\displaystyle \\$

$\displaystyle \text{iii) }y=mx+c\text{ and }y=mx+d$

$\displaystyle \text{Answer:}$
$\displaystyle \text{The given lines can be written as}$
$\displaystyle mx-y+c=0\text{ and }mx-y+d=0.$
$\displaystyle \therefore d=\frac{|c-d|}{\sqrt{m^2+(-1)^2}}=\frac{|c-d|}{\sqrt{m^2+1}}.$

$\displaystyle \\$

$\displaystyle \text{iv) }4x+3y-11=0\text{ and }8x+6y=15$

$\displaystyle \text{Answer:}$
$\displaystyle \text{The second line is }8x+6y-15=0,$
$\displaystyle \text{or }4x+3y-\frac{15}{2}=0.$
$\displaystyle \therefore d=\frac{\left|-11+\frac{15}{2}\right|}{\sqrt{4^2+3^2}}$
$\displaystyle =\frac{\frac{7}{2}}{5}=\frac{7}{10}\text{ units}.$

$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{The equations of two sides of a square are }5x-12y-65=0\text{ and}$
$\displaystyle 5x-12y+26=0.\text{ Find the area of the square.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given parallel lines }5x-12y-65=0\text{ and }5x-12y+26=0.$
$\displaystyle \text{Since these lines represent two opposite sides of the square, their distance is the side of the square.}$
$\displaystyle \therefore d=\left|\frac{-65-26}{\sqrt{5^2+(-12)^2}}\right|$
$\displaystyle =\frac{91}{\sqrt{25+144}}=\frac{91}{13}=7\text{ units}.$
$\displaystyle \therefore \text{Area of the square}=7^2=49\text{ square units}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Find the equations of two straight lines which are parallel to }x+7y+2=0$
$\displaystyle \text{and are at unit distance from the point }(1,-1).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given line is }x+7y+2=0.\qquad\ldots\ldots\ldots\text{(i)}$
$\displaystyle \text{The equation of a line parallel to line (i) is}$
$\displaystyle x+7y+\lambda=0.\qquad\ldots\ldots\ldots\text{(ii)}$
$\displaystyle \text{Line (ii) is at unit distance from the point }(1,-1).$
$\displaystyle \therefore \left|\frac{1+7(-1)+\lambda}{\sqrt{1^2+7^2}}\right|=1.$
$\displaystyle \left|\frac{\lambda-6}{\sqrt{50}}\right|=1.$
$\displaystyle |\lambda-6|=5\sqrt2.$
$\displaystyle \therefore \lambda-6=\pm5\sqrt2.$
$\displaystyle \therefore \lambda=6+5\sqrt2\text{ or }\lambda=6-5\sqrt2.$
$\displaystyle \therefore \text{The required lines are}$
$\displaystyle x+7y+6+5\sqrt2=0\text{ and }x+7y+6-5\sqrt2=0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Prove that the lines }2x+3y=19\text{ and }2x+3y+7=0\text{ are}$
$\displaystyle \text{equidistant from the line }2x+3y=6.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given lines can be written as}$
$\displaystyle 2x+3y-19=0,\qquad 2x+3y+7=0$
$\displaystyle \text{and}\qquad 2x+3y-6=0.$
$\displaystyle \text{Let }d_1\text{ be the distance between }2x+3y-19=0\text{ and }2x+3y-6=0.$
$\displaystyle \therefore d_1=\left|\frac{-19-(-6)}{\sqrt{2^2+3^2}}\right|$
$\displaystyle =\frac{13}{\sqrt{13}}=\sqrt{13}.$
$\displaystyle \text{Let }d_2\text{ be the distance between }2x+3y+7=0\text{ and }2x+3y-6=0.$
$\displaystyle \therefore d_2=\left|\frac{7-(-6)}{\sqrt{2^2+3^2}}\right|$
$\displaystyle =\frac{13}{\sqrt{13}}=\sqrt{13}.$
$\displaystyle \therefore d_1=d_2.$
$\displaystyle \therefore \text{The lines }2x+3y=19\text{ and }2x+3y+7=0\text{ are equidistant from}$
$\displaystyle \text{the line }2x+3y=6.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Find the equation of the line midway between the parallel lines}$
$\displaystyle 9x+6y-7=0\text{ and }3x+2y+6=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The first line can be written as}$
$\displaystyle 3x+2y-\frac{7}{3}=0.\qquad\ldots\ldots\ldots\text{(i)}$
$\displaystyle \text{The second line is}$
$\displaystyle 3x+2y+6=0.\qquad\ldots\ldots\ldots\text{(ii)}$
$\displaystyle \text{Let the equation of the line midway between lines (i) and (ii) be}$
$\displaystyle 3x+2y+\lambda=0.\qquad\ldots\ldots\ldots\text{(iii)}$
$\displaystyle \text{Since line (iii) is equidistant from lines (i) and (ii),}$
$\displaystyle \left|\frac{-\frac{7}{3}-\lambda}{\sqrt{3^2+2^2}}\right|$
$\displaystyle =\left|\frac{6-\lambda}{\sqrt{3^2+2^2}}\right|.$
$\displaystyle \left|-\frac{7}{3}-\lambda\right|=|6-\lambda|.$
$\displaystyle \therefore -\frac{7}{3}-\lambda=-(6-\lambda).$
$\displaystyle -\frac{7}{3}-\lambda=-6+\lambda.$
$\displaystyle 2\lambda=\frac{11}{3}.$
$\displaystyle \therefore \lambda=\frac{11}{6}.$
$\displaystyle \therefore \text{The equation of the required line is}$
$\displaystyle 3x+2y+\frac{11}{6}=0.$
$\displaystyle \therefore 18x+12y+11=0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Find the ratio in which the line }3x+4y+2=0\text{ divides the distance}$
$\displaystyle \text{between the lines }3x+4y+5=0\text{ and }3x+4y-5=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since the coefficients of }x\text{ and }y\text{ are the same, the three lines are parallel.}$
$\displaystyle \text{Let }d_1\text{ be the distance between }3x+4y+2=0\text{ and }3x+4y+5=0.$
$\displaystyle \therefore d_1=\left|\frac{2-5}{\sqrt{3^2+4^2}}\right|$
$\displaystyle =\frac{3}{5}.$
$\displaystyle \text{Let }d_2\text{ be the distance between }3x+4y+2=0\text{ and }3x+4y-5=0.$
$\displaystyle \therefore d_2=\left|\frac{2-(-5)}{\sqrt{3^2+4^2}}\right|$
$\displaystyle =\frac{7}{5}.$
$\displaystyle \therefore \frac{d_1}{d_2}=\frac{\frac{3}{5}}{\frac{7}{5}}=\frac{3}{7}.$
$\displaystyle \therefore \text{The line }3x+4y+2=0\text{ divides the distance between the given lines in the ratio }3:7.$
$\displaystyle \\$