\displaystyle \textbf{ANGLE \ BETWEEN \ TWO \ STRAIGHT \ LINES \ WHEN \ THEIR \ EQUATIONS \ ARE \ GIVEN}

\displaystyle \textbf{THEOREM}

\displaystyle \text{Prove that the acute angle } \theta \text{ between the lines }
\displaystyle a_1x+b_1y+c_1=0 \text{ and } a_2x+b_2y+c_2=0
\displaystyle \text{ is given by }

\displaystyle \tan\theta=\left|\frac{a_2b_1-a_1b_2}{a_1a_2+b_1b_2}\right|

\displaystyle \textbf{PROOF}

\displaystyle \text{Let }m_1\text{ and }m_2\text{ be the slopes of the lines}
\displaystyle a_1x+b_1y+c_1=0\text{ and }a_2x+b_2y+c_2=0.

\displaystyle \text{Then,}

\displaystyle m_1=-\frac{a_1}{b_1}\text{ and }m_2=-\frac{a_2}{b_2}

\displaystyle \text{Now,}

\displaystyle \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|

\displaystyle =\left|\frac{-\frac{a_1}{b_1}+\frac{a_2}{b_2}}{1+\left(-\frac{a_1}{b_1}\right)\left(-\frac{a_2}{b_2}\right)}\right|

\displaystyle \Rightarrow \tan\theta=\left|\frac{a_2b_1-a_1b_2}{a_1a_2+b_1b_2}\right|

\displaystyle \Rightarrow \theta=\tan^{-1}\left|\frac{a_2b_1-a_1b_2}{a_1a_2+b_1b_2}\right|

\displaystyle \therefore \text{Hence proved.}

\displaystyle \textbf{CONDITION \ FOR \ THE \ LINES \ TO \ BE \ PARALLEL}

\displaystyle \text{If the lines }a_1x+b_1y+c_1=0\text{ and }a_2x+b_2y+c_2=0
\displaystyle \text{ are parallel, then}

\displaystyle m_1=m_2

\displaystyle \Rightarrow -\frac{a_1}{b_1}=-\frac{a_2}{b_2}

\displaystyle \Rightarrow \frac{a_1}{a_2}=\frac{b_1}{b_2}

\displaystyle \textbf{CONDITION \ FOR \ THE \ LINES \ TO \ BE \ PERPENDICULAR}

\displaystyle \text{If the lines }a_1x+b_1y+c_1=0\text{ and }a_2x+b_2y+c_2=0
\displaystyle \text{ are perpendicular, then}

\displaystyle m_1m_2=-1

\displaystyle \Rightarrow -\frac{a_1}{b_1}\times-\frac{a_2}{b_2}=-1

\displaystyle \Rightarrow a_1a_2+b_1b_2=0

\displaystyle \text{It follows from the above discussion that the lines}
\displaystyle a_1x+b_1y+c_1=0\text{ and }a_2x+b_2y+c_2=0\text{ are:}

\displaystyle \text{(i) Coincident, if }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}

\displaystyle \text{(ii) Parallel, if }\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}

\displaystyle \text{(iii) Intersecting, if }\frac{a_1}{a_2}\ne\frac{b_1}{b_2}

\displaystyle \text{(iv) Perpendicular, if }a_1a_2+b_1b_2=0

\displaystyle \\

—————

\displaystyle \textbf{Question 1: }\text{Find the acute angle between each of the following pairs of straight lines:}
\displaystyle \text{(i) }3x+y+12=0\text{ and }x+2y-1=0.
\displaystyle \text{(ii) }3x-y+5=0\text{ and }x-3y+1=0.
\displaystyle \text{(iii) }3x+4y-7=0\text{ and }4x-3y+5=0.
\displaystyle \text{(iv) }x-4y=3\text{ and }6x-y=11.
\displaystyle \text{(v) }(m^2-mn)y=(mn+n^2)x+n^3\text{ and}
\displaystyle (mn+m^2)y=(mn-n^2)x+m^3.
\displaystyle \textbf{Answer:}
\displaystyle \text{We know that the acute angle }\theta\text{ between the lines}
\displaystyle a_1x+b_1y+c_1=0\text{ and }a_2x+b_2y+c_2=0\text{ is given by}
\displaystyle \tan\theta=\left|\frac{a_2b_1-a_1b_2}{a_1a_2+b_1b_2}\right|.

\displaystyle \text{(i) The given lines are}
\displaystyle 3x+y+12=0\text{ and }x+2y-1=0.
\displaystyle \therefore a_1=3,\quad b_1=1,\quad a_2=1,\quad b_2=2.
\displaystyle \tan\theta=\left|\frac{1(1)-3(2)}{3(1)+1(2)}\right|
\displaystyle =\left|\frac{-5}{5}\right|=1.
\displaystyle \Rightarrow \theta=45^\circ.
\displaystyle \therefore \text{The acute angle between the lines is }45^\circ.

\displaystyle \text{(ii) The given lines are}
\displaystyle 3x-y+5=0\text{ and }x-3y+1=0.
\displaystyle \therefore a_1=3,\quad b_1=-1,\quad a_2=1,\quad b_2=-3.
\displaystyle \tan\theta=\left|\frac{1(-1)-3(-3)}{3(1)+(-1)(-3)}\right|
\displaystyle =\left|\frac{-1+9}{3+3}\right|=\frac{4}{3}.
\displaystyle \Rightarrow \theta=\tan^{-1}\left(\frac{4}{3}\right).
\displaystyle \therefore \text{The acute angle between the lines is }\tan^{-1}\left(\frac{4}{3}\right).

\displaystyle \text{(iii) The given lines are}
\displaystyle 3x+4y-7=0\text{ and }4x-3y+5=0.
\displaystyle \therefore a_1=3,\quad b_1=4,\quad a_2=4,\quad b_2=-3.
\displaystyle a_1a_2+b_1b_2=3(4)+4(-3)=12-12=0.
\displaystyle \therefore \text{The lines are perpendicular.}
\displaystyle \therefore \text{The angle between the lines is }90^\circ.

\displaystyle \text{(iv) The given lines are}
\displaystyle x-4y-3=0\text{ and }6x-y-11=0.
\displaystyle \therefore a_1=1,\quad b_1=-4,\quad a_2=6,\quad b_2=-1.
\displaystyle \tan\theta=\left|\frac{6(-4)-1(-1)}{1(6)+(-4)(-1)}\right|
\displaystyle =\left|\frac{-24+1}{6+4}\right|=\frac{23}{10}.
\displaystyle \Rightarrow \theta=\tan^{-1}\left(\frac{23}{10}\right).
\displaystyle \therefore \text{The acute angle between the lines is }\tan^{-1}\left(\frac{23}{10}\right).

\displaystyle \text{(v) Writing the given lines in general form, we get}
\displaystyle -(mn+n^2)x+(m^2-mn)y-n^3=0
\displaystyle \text{and}
\displaystyle -(mn-n^2)x+(mn+m^2)y-m^3=0.
\displaystyle \therefore a_1=-n(m+n),\quad b_1=m(m-n),
\displaystyle a_2=-n(m-n),\quad b_2=m(m+n).
\displaystyle a_2b_1-a_1b_2
\displaystyle =-mn(m-n)^2+mn(m+n)^2
\displaystyle =mn\left[(m+n)^2-(m-n)^2\right]
\displaystyle =4m^2n^2.
\displaystyle a_1a_2+b_1b_2
\displaystyle =n^2(m+n)(m-n)+m^2(m-n)(m+n)
\displaystyle =(m^2+n^2)(m^2-n^2)=m^4-n^4.
\displaystyle \therefore \tan\theta=\left|\frac{4m^2n^2}{m^4-n^4}\right|.
\displaystyle \Rightarrow \theta=\tan^{-1}\left|\frac{4m^2n^2}{m^4-n^4}\right|.
\displaystyle \text{If }m^4=n^4,\text{ then the lines are perpendicular and }\theta=90^\circ.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the acute angle between the lines }2x-y+3=0\text{ and }x+y+2=0.
\displaystyle \textbf{Answer:}
\displaystyle \text{For the lines }a_1x+b_1y+c_1=0\text{ and }a_2x+b_2y+c_2=0,
\displaystyle \tan\theta=\left|\frac{a_2b_1-a_1b_2}{a_1a_2+b_1b_2}\right|.
\displaystyle \text{Here }a_1=2,\ b_1=-1,\ a_2=1,\ b_2=1.
\displaystyle \therefore \tan\theta=\left|\frac{1(-1)-2(1)}{2(1)+(-1)(1)}\right|
\displaystyle =\left|\frac{-1-2}{2-1}\right|=3.
\displaystyle \Rightarrow \theta=\tan^{-1}(3).
\displaystyle \therefore \text{The acute angle between the given lines is }\tan^{-1}(3).
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Prove that the points }(2,-1),(0,2),(2,3)\text{ and }(4,0)
\displaystyle \text{are the vertices of a parallelogram and find the angle between its diagonals.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let }A(2,-1),\ B(0,2),\ C(2,3)\text{ and }D(4,0).
\displaystyle \text{The equation of }AB\text{ is}
\displaystyle \frac{x-2}{0-2}=\frac{y+1}{2+1}.
\displaystyle \Rightarrow 3x+2y-4=0.
\displaystyle \text{The equation of }CD\text{ is}
\displaystyle \frac{x-2}{4-2}=\frac{y-3}{0-3}.
\displaystyle \Rightarrow 3x+2y-12=0.
\displaystyle \therefore AB\parallel CD.
\displaystyle \text{The equation of }BC\text{ is}
\displaystyle \frac{x-0}{2-0}=\frac{y-2}{3-2}.
\displaystyle \Rightarrow x-2y+4=0.
\displaystyle \text{The equation of }DA\text{ is}
\displaystyle \frac{x-4}{2-4}=\frac{y-0}{-1-0}.
\displaystyle \Rightarrow x-2y-4=0.
\displaystyle \therefore BC\parallel DA.
\displaystyle \text{Since both pairs of opposite sides are parallel, }ABCD\text{ is a parallelogram.}
\displaystyle \text{Now, the equation of diagonal }AC\text{ is}
\displaystyle x-2=0.
\displaystyle \text{The equation of diagonal }BD\text{ is}
\displaystyle \frac{x-0}{4-0}=\frac{y-2}{0-2}.
\displaystyle \Rightarrow x+2y-4=0.
\displaystyle \text{For the lines }a_1x+b_1y+c_1=0\text{ and }a_2x+b_2y+c_2=0,
\displaystyle \tan\theta=\left|\frac{a_2b_1-a_1b_2}{a_1a_2+b_1b_2}\right|.
\displaystyle \text{For }AC:x-2=0,\quad a_1=1,\ b_1=0.
\displaystyle \text{For }BD:x+2y-4=0,\quad a_2=1,\ b_2=2.
\displaystyle \therefore \tan\theta=\left|\frac{1(0)-1(2)}{1(1)+0(2)}\right|=2.
\displaystyle \Rightarrow \theta=\tan^{-1}(2).
\displaystyle \therefore \text{The acute angle between the diagonals is }\tan^{-1}(2).
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the angle between the line joining the points }(2,0),(0,3)
\displaystyle \text{and the line }x+y=1.
\displaystyle \textbf{Answer:}
\displaystyle \text{The equation of the line joining }(2,0)\text{ and }(0,3)\text{ is}
\displaystyle \frac{x-2}{0-2}=\frac{y}{3}.
\displaystyle \Rightarrow 3x+2y-6=0.
\displaystyle \text{The second line is }x+y-1=0.
\displaystyle \text{For the lines }a_1x+b_1y+c_1=0\text{ and }a_2x+b_2y+c_2=0,
\displaystyle \tan\theta=\left|\frac{a_2b_1-a_1b_2}{a_1a_2+b_1b_2}\right|.
\displaystyle \text{Here }a_1=3,\ b_1=2,\ a_2=1,\ b_2=1.
\displaystyle \therefore \tan\theta=\left|\frac{1(2)-3(1)}{3(1)+2(1)}\right|
\displaystyle =\left|\frac{2-3}{3+2}\right|=\frac{1}{5}.
\displaystyle \Rightarrow \theta=\tan^{-1}\left(\frac{1}{5}\right).
\displaystyle \therefore \text{The acute angle between the given lines is }\tan^{-1}\left(\frac{1}{5}\right).
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }\theta\text{ is the angle which the straight line joining the points}
\displaystyle (x_1,y_1)\text{ and }(x_2,y_2)\text{ subtends at the origin, prove that}
\displaystyle \tan\theta=\frac{x_2y_1-x_1y_2}{x_1x_2+y_1y_2}\text{ and}
\displaystyle \cos\theta=\frac{x_1x_2+y_1y_2}{\sqrt{{x_1}^2+{y_1}^2}\sqrt{{x_2}^2+{y_2}^2}}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Let }O(0,0),\ A(x_1,y_1)\text{ and }B(x_2,y_2).
\displaystyle \text{The equations of }OA\text{ and }OB\text{ are respectively}
\displaystyle y_1x-x_1y=0
\displaystyle \text{and}
\displaystyle y_2x-x_2y=0.
\displaystyle \text{For the lines }a_1x+b_1y+c_1=0\text{ and }a_2x+b_2y+c_2=0,
\displaystyle \tan\theta=\left|\frac{a_2b_1-a_1b_2}{a_1a_2+b_1b_2}\right|.
\displaystyle \text{Here }a_1=y_1,\ b_1=-x_1,\ a_2=y_2,\ b_2=-x_2.
\displaystyle \therefore \tan\theta
\displaystyle =\left|\frac{y_2(-x_1)-y_1(-x_2)}{y_1y_2+(-x_1)(-x_2)}\right|
\displaystyle =\left|\frac{x_2y_1-x_1y_2}{x_1x_2+y_1y_2}\right|.
\displaystyle \text{Thus, for the directed angle }\theta,
\displaystyle \tan\theta=\frac{x_2y_1-x_1y_2}{x_1x_2+y_1y_2}.
\displaystyle \text{Also,}
\displaystyle OA=\sqrt{{x_1}^2+{y_1}^2},
\displaystyle OB=\sqrt{{x_2}^2+{y_2}^2}
\displaystyle \text{and}
\displaystyle AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.
\displaystyle \text{By the cosine rule in }\triangle AOB,
\displaystyle AB^2=OA^2+OB^2-2(OA)(OB)\cos\theta.
\displaystyle \therefore \cos\theta=\frac{OA^2+OB^2-AB^2}{2(OA)(OB)}.
\displaystyle =\frac{{x_1}^2+{y_1}^2+{x_2}^2+{y_2}^2-(x_2-x_1)^2-(y_2-y_1)^2}{2\sqrt{{x_1}^2+{y_1}^2}\sqrt{{x_2}^2+{y_2}^2}}
\displaystyle =\frac{2x_1x_2+2y_1y_2}{2\sqrt{{x_1}^2+{y_1}^2}\sqrt{{x_2}^2+{y_2}^2}}.
\displaystyle \therefore \cos\theta=\frac{x_1x_2+y_1y_2}{\sqrt{{x_1}^2+{y_1}^2}\sqrt{{x_2}^2+{y_2}^2}}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Prove that the straight lines}
\displaystyle (a+b)x+(a-b)y=2ab,\quad (a-b)x+(a+b)y=2ab
\displaystyle \text{and }x+y=0\text{ form an isosceles triangle whose vertex angle is}
\displaystyle 2\tan^{-1}\left(\frac{a}{b}\right),\text{ where }a>0\text{ and }b>0.
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the given lines be}
\displaystyle L_1:(a+b)x+(a-b)y-2ab=0,
\displaystyle L_2:(a-b)x+(a+b)y-2ab=0
\displaystyle \text{and}
\displaystyle L_3:x+y=0.
\displaystyle \text{For the lines }a_1x+b_1y+c_1=0\text{ and }a_2x+b_2y+c_2=0,
\displaystyle \tan\theta=\left|\frac{a_2b_1-a_1b_2}{a_1a_2+b_1b_2}\right|.
\displaystyle \text{Let }\theta_1\text{ be the angle between }L_1\text{ and }L_3.
\displaystyle \tan\theta_1
\displaystyle =\left|\frac{1(a-b)-(a+b)(1)}{(a+b)(1)+(a-b)(1)}\right|
\displaystyle =\left|\frac{-2b}{2a}\right|=\frac{b}{a}.
\displaystyle \therefore \theta_1=\tan^{-1}\left(\frac{b}{a}\right).
\displaystyle \text{Let }\theta_2\text{ be the angle between }L_2\text{ and }L_3.
\displaystyle \tan\theta_2
\displaystyle =\left|\frac{1(a+b)-(a-b)(1)}{(a-b)(1)+(a+b)(1)}\right|
\displaystyle =\left|\frac{2b}{2a}\right|=\frac{b}{a}.
\displaystyle \therefore \theta_2=\tan^{-1}\left(\frac{b}{a}\right).
\displaystyle \therefore \theta_1=\theta_2.
\displaystyle \text{Hence the two base angles are equal, so the triangle is isosceles.}
\displaystyle \text{Let }\theta\text{ be its vertex angle formed by }L_1\text{ and }L_2.
\displaystyle \therefore \theta+\theta_1+\theta_2=180^\circ.
\displaystyle \Rightarrow \theta=180^\circ-2\tan^{-1}\left(\frac{b}{a}\right).
\displaystyle \text{Since }a>0\text{ and }b>0,
\displaystyle \tan^{-1}\left(\frac{a}{b}\right)+\tan^{-1}\left(\frac{b}{a}\right)=90^\circ.
\displaystyle \therefore \theta
\displaystyle =2\left[90^\circ-\tan^{-1}\left(\frac{b}{a}\right)\right]
\displaystyle =2\tan^{-1}\left(\frac{a}{b}\right).
\displaystyle \therefore \text{The given lines form an isosceles triangle whose vertex angle is}
\displaystyle 2\tan^{-1}\left(\frac{a}{b}\right).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find the angle between the lines }x=a\text{ and }by+c=0.
\displaystyle \textbf{Answer:}
\displaystyle \text{The given lines are}
\displaystyle x-a=0
\displaystyle \text{and}
\displaystyle by+c=0.
\displaystyle \text{For the lines }a_1x+b_1y+c_1=0\text{ and }a_2x+b_2y+c_2=0,
\displaystyle \tan\theta=\left|\frac{a_2b_1-a_1b_2}{a_1a_2+b_1b_2}\right|.
\displaystyle \text{Here }a_1=1,\ b_1=0,\ a_2=0,\ b_2=b.
\displaystyle \therefore \tan\theta=\left|\frac{0(0)-1(b)}{1(0)+0(b)}\right|
\displaystyle =\left|\frac{-b}{0}\right|=\infty.
\displaystyle \therefore \theta=90^\circ.
\displaystyle \text{Hence, the angle between the given lines is }90^\circ.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find the tangent of the angle between the lines which have intercepts }3,4
\displaystyle \text{and }1,8\text{ on the axes respectively.}
\displaystyle \textbf{Answer:}
\displaystyle \text{The equation of the line with intercepts }3\text{ and }4\text{ is}
\displaystyle \frac{x}{3}+\frac{y}{4}=1
\displaystyle \Rightarrow 4x+3y-12=0.
\displaystyle \text{The equation of the line with intercepts }1\text{ and }8\text{ is}
\displaystyle \frac{x}{1}+\frac{y}{8}=1
\displaystyle \Rightarrow 8x+y-8=0.
\displaystyle \text{For the lines }a_1x+b_1y+c_1=0\text{ and }a_2x+b_2y+c_2=0,
\displaystyle \tan\theta=\left|\frac{a_2b_1-a_1b_2}{a_1a_2+b_1b_2}\right|.
\displaystyle \text{Here }a_1=4,\ b_1=3,\ a_2=8,\ b_2=1.
\displaystyle \therefore \tan\theta=\left|\frac{8(3)-4(1)}{4(8)+3(1)}\right|
\displaystyle =\left|\frac{24-4}{32+3}\right|
\displaystyle =\frac{20}{35}=\frac{4}{7}.
\displaystyle \therefore \text{The tangent of the angle between the given lines is }\frac{4}{7}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Show that the line }a^2x+ay+1=0\text{ is perpendicular to the line}
\displaystyle x-ay=1\text{ for all non-zero real values of }a.
\displaystyle \textbf{Answer:}
\displaystyle \text{The given lines are}
\displaystyle a^2x+ay+1=0
\displaystyle \text{and}
\displaystyle x-ay-1=0.
\displaystyle \text{For the lines }a_1x+b_1y+c_1=0\text{ and }a_2x+b_2y+c_2=0,
\displaystyle \text{if }a_1a_2+b_1b_2=0,\text{ then the lines are perpendicular.}
\displaystyle \text{Here }a_1=a^2,\ b_1=a,\ a_2=1,\ b_2=-a.
\displaystyle \therefore a_1a_2+b_1b_2=a^2(1)+a(-a)=a^2-a^2=0.
\displaystyle \therefore \text{The given lines are perpendicular for every non-zero real value of }a.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Show that the tangent of an angle between the lines}
\displaystyle \frac{x}{a}+\frac{y}{b}=1\text{ and }\frac{x}{a}-\frac{y}{b}=1\text{ is }\frac{2ab}{a^2-b^2}.
\displaystyle \textbf{Answer:}
\displaystyle \text{The given lines are}
\displaystyle \frac{x}{a}+\frac{y}{b}=1
\displaystyle \Rightarrow bx+ay-ab=0
\displaystyle \text{and}
\displaystyle \frac{x}{a}-\frac{y}{b}=1
\displaystyle \Rightarrow bx-ay-ab=0.
\displaystyle \text{For the lines }a_1x+b_1y+c_1=0\text{ and }a_2x+b_2y+c_2=0,
\displaystyle \tan\theta=\left|\frac{a_2b_1-a_1b_2}{a_1a_2+b_1b_2}\right|.
\displaystyle \text{Here }a_1=b,\ b_1=a,\ a_2=b,\ b_2=-a.
\displaystyle \therefore \tan\theta
\displaystyle =\left|\frac{b(a)-b(-a)}{b(b)+a(-a)}\right|
\displaystyle =\left|\frac{2ab}{b^2-a^2}\right|
\displaystyle =\left|\frac{2ab}{a^2-b^2}\right|.
\displaystyle \text{Thus, for a directed angle,}
\displaystyle \tan\theta=\frac{2ab}{a^2-b^2}.
\displaystyle \text{Hence proved.}
\displaystyle \\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.