NCERT Solutions · Class 11 Mathematics Straight Lines

Slope-intercept form. Write each equation as \(\displaystyle y=mx+c\); then \(\displaystyle m\) is the slope and \(\displaystyle c\) is the \(\displaystyle y\)-intercept.(i) \(\displaystyle x+7y=0\) \[7y=-x\quad\Longrightarrow\quad y=-\frac{1}{7}x+0. \] Slope \(\displaystyle =-\dfrac{1}{7}\), \(\displaystyle y\)-intercept \(\displaystyle =0\).(ii) \(\displaystyle 6x+3y-5=0\) \[3y=-6x+5\quad\Longrightarrow\quad y=-2x+\frac{5}{3}. \] Slope \(\displaystyle =-2\), \(\displaystyle y\)-intercept \(\displaystyle =\dfrac{5}{3}\).(iii) \(\displaystyle y=0\) \[y=0\cdot x+0. \] Slope \(\displaystyle =0\), \(\displaystyle y\)-intercept \(\displaystyle =0\). (This is the \(\displaystyle x\)-axis itself.)Slopes and \(\displaystyle y\)-intercepts: (i) \(\displaystyle -\tfrac{1}{7}\) and \(\displaystyle 0\); (ii) \(\displaystyle -2\) and \(\displaystyle \tfrac{5}{3}\); (iii) \(\displaystyle 0\) and \(\displaystyle 0\).

Intercept form. Move the constant to the right and divide through so that the right-hand side becomes $\displaystyle 1$, giving \(\displaystyle \dfrac{x}{a}+\dfrac{y}{b}=1\); then \(\displaystyle a\) is the \(\displaystyle x\)-intercept and \(\displaystyle b\) the \(\displaystyle y\)-intercept.(i) \(\displaystyle 3x+2y-12=0\Rightarrow 3x+2y=12\). Dividing by $\displaystyle 12$: \[\frac{3x}{12}+\frac{2y}{12}=1\quad\Longrightarrow\quad \frac{x}{4}+\frac{y}{6}=1. \] \(\displaystyle x\)-intercept \(\displaystyle =4\), \(\displaystyle y\)-intercept \(\displaystyle =6\).(ii) \(\displaystyle 4x-3y=6\). Dividing by $\displaystyle 6$: \[\frac{4x}{6}-\frac{3y}{6}=1\quad\Longrightarrow\quad \frac{x}{3/2}+\frac{y}{-2}=1. \] \(\displaystyle x\)-intercept \(\displaystyle =\dfrac{3}{2}\), \(\displaystyle y\)-intercept \(\displaystyle =-2\).(iii) \(\displaystyle 3y+2=0\Rightarrow 3y=-2\). Dividing by \(\displaystyle -2\): \[\frac{y}{-2/3}=1. \] \(\displaystyle y\)-intercept \(\displaystyle =-\dfrac{2}{3}\). There is no \(\displaystyle x\) term, so the line is parallel to the \(\displaystyle x\)-axis and has no \(\displaystyle x\)-intercept.Intercepts: (i) \(\displaystyle 4\) and \(\displaystyle 6\); (ii) \(\displaystyle \tfrac{3}{2}\) and \(\displaystyle -2\); (iii) no \(\displaystyle x\)-intercept and \(\displaystyle -\tfrac{2}{3}\).

Perpendicular distance formula. The distance of \(\displaystyle (x_1,y_1)\) from \(\displaystyle Ax+By+C=0\) is \[d=\frac{|Ax_1+By_1+C|}{\sqrt{A^{2}+B^{2}}}. \]First put the line in general form: \[12(x+6)=5(y-2)\ \Longrightarrow\ 12x+72=5y-10\ \Longrightarrow\ 12x-5y+82=0. \]So \(\displaystyle A=12,\ B=-5,\ C=82\), and \(\displaystyle (x_1,y_1)=(-1,1)\): \[d=\frac{|12(-1)-5(1)+82|}{\sqrt{12^{2}+(-5)^{2}}}=\frac{|-12-5+82|}{\sqrt{144+25}}=\frac{65}{\sqrt{169}}=\frac{65}{13}=5. \]The distance is $\displaystyle 5$ units.

Perpendicular distance formula with an unknown point. Any point on the \(\displaystyle x\)-axis has the form \(\displaystyle (a,0)\).Put the line in general form: \[\frac{x}{3}+\frac{y}{4}=1\ \Longrightarrow\ 4x+3y=12\ \Longrightarrow\ 4x+3y-12=0. \]Distance of \(\displaystyle (a,0)\) from this line: \[d=\frac{|4a+3(0)-12|}{\sqrt{4^{2}+3^{2}}}=\frac{|4a-12|}{5}. \]Setting \(\displaystyle d=4\): \[|4a-12|=20\quad\Longrightarrow\quad 4a-12=20\ \text{or}\ 4a-12=-20, \] \[a=8\quad\text{or}\quad a=-2. \]Check: \(\displaystyle \dfrac{|4(8)-12|}{5}=\dfrac{20}{5}=4\) and \(\displaystyle \dfrac{|4(-2)-12|}{5}=\dfrac{20}{5}=4\). ✓The points are \(\displaystyle (8,0)\) and \(\displaystyle (-2,0)\).

Distance between parallel lines. For \(\displaystyle Ax+By+C_1=0\) and \(\displaystyle Ax+By+C_2=0\), \[d=\frac{|C_1-C_2|}{\sqrt{A^{2}+B^{2}}}. \](i) \(\displaystyle 15x+8y-34=0\) and \(\displaystyle 15x+8y+31=0\). Here \(\displaystyle A=15,\ B=8,\ C_1=-34,\ C_2=31\): \[d=\frac{|-34-31|}{\sqrt{15^{2}+8^{2}}}=\frac{65}{\sqrt{225+64}}=\frac{65}{\sqrt{289}}=\frac{65}{17}. \](ii) \(\displaystyle l(x+y)+p=0\) and \(\displaystyle l(x+y)-r=0\), i.e. \(\displaystyle lx+ly+p=0\) and \(\displaystyle lx+ly-r=0\). Here \(\displaystyle A=B=l,\ C_1=p,\ C_2=-r\): \[d=\frac{|p-(-r)|}{\sqrt{l^{2}+l^{2}}}=\frac{|p+r|}{|l|\sqrt{2}}. \]The distances are (i) \(\displaystyle \dfrac{65}{17}\) units and (ii) \(\displaystyle \dfrac{|p+r|}{\sqrt{2}\,|l|}\) units.

Family of parallel lines. Any line parallel to \(\displaystyle 3x-4y+2=0\) has the same coefficients of \(\displaystyle x\) and \(\displaystyle y\), so it is of the form \[3x-4y+k=0 \] for some constant \(\displaystyle k\) (this keeps the slope \(\displaystyle \tfrac34\) unchanged).It passes through \(\displaystyle (-2,3)\), so \[3(-2)-4(3)+k=0\quad\Longrightarrow\quad -6-12+k=0\quad\Longrightarrow\quad k=18. \]Check: \(\displaystyle 3(-2)-4(3)+18=-6-12+18=0\). ✓The required line is \(\displaystyle 3x-4y+18=0\).

Slope form, using the perpendicularity condition.Write the given line in slope-intercept form: \[x-7y+5=0 \;\Rightarrow\; y=\tfrac{1}{7}x+\tfrac{5}{7}, \qquad m_1=\tfrac{1}{7}. \]Two lines are perpendicular when \(\displaystyle m_1m_2=-1 \), so the required slope is \[m_2=-\frac{1}{m_1}=-7. \]An \(\displaystyle x\)-intercept of $\displaystyle 3$ means the line passes through \(\displaystyle (3,0)\). Using the point-slope form \(\displaystyle y-y_1=m(x-x_1) \): \[y-0=-7(x-3) \;\Rightarrow\; y=-7x+21. \]\(\displaystyle 7x+y-21=0\)

Angle between two lines.Put each line in slope form: \[\sqrt{3}\,x+y=1 \Rightarrow y=-\sqrt{3}\,x+1,\qquad m_1=-\sqrt{3}, \] \[x+\sqrt{3}\,y=1 \Rightarrow y=-\tfrac{1}{\sqrt3}x+\tfrac{1}{\sqrt3},\qquad m_2=-\tfrac{1}{\sqrt3}. \]If \(\displaystyle \theta\) is an angle between the lines, then \[\tan\theta=\left|\frac{m_2-m_1}{1+m_1m_2}\right|. \]Here \(\displaystyle m_1m_2=(-\sqrt3)\!\left(-\tfrac{1}{\sqrt3}\right)=1 \), and \(\displaystyle m_2-m_1=-\tfrac{1}{\sqrt3}+\sqrt3=\tfrac{-1+3}{\sqrt3}=\tfrac{2}{\sqrt3} \). So \[\tan\theta=\left|\frac{2/\sqrt3}{1+1}\right|=\frac{1}{\sqrt3} \;\Rightarrow\; \theta=30^{\circ}. \]The two lines cut each other in a pair of supplementary angles, so the other angle is \(\displaystyle 180^{\circ}-30^{\circ}=150^{\circ}\).The angles between the lines are \(\displaystyle 30^{\circ}\) and \(\displaystyle 150^{\circ}\).

Perpendicularity condition \(\displaystyle m_1m_2=-1\).Slope of the given line: \(\displaystyle 7x-9y-19=0 \Rightarrow y=\tfrac{7}{9}x-\tfrac{19}{9} \), so \(\displaystyle m_1=\dfrac{7}{9} \).Slope of the line through \(\displaystyle (h,3)\) and \(\displaystyle (4,1)\): \[m_2=\frac{1-3}{4-h}=\frac{-2}{4-h}. \]The two lines meet at right angles, so \(\displaystyle m_1m_2=-1 \): \[\frac{7}{9}\cdot\frac{-2}{4-h}=-1 \;\Rightarrow\; \frac{-14}{9(4-h)}=-1 \;\Rightarrow\; 14=9(4-h)=36-9h, \] \[9h=22 \;\Rightarrow\; h=\frac{22}{9}. \]Check: \(\displaystyle m_2=\dfrac{-2}{4-\frac{22}{9}}=\dfrac{-2}{14/9}=-\dfrac{9}{7} \), and \(\displaystyle \dfrac{7}{9}\times\left(-\dfrac{9}{7}\right)=-1 \). ✓\(\displaystyle h=\dfrac{22}{9} \)

Equal slopes for parallel lines.Case $\displaystyle 1$: \(\displaystyle \mathrm{B}\neq 0 \). Write \(\displaystyle \mathrm{A}x+\mathrm{B}y+\mathrm{C}=0 \) as \[y=-\frac{\mathrm{A}}{\mathrm{B}}x-\frac{\mathrm{C}}{\mathrm{B}}, \] so its slope is \(\displaystyle -\dfrac{\mathrm{A}}{\mathrm{B}} \). Parallel lines have equal slopes, so the required line through \(\displaystyle (x_1,y_1)\) has slope \(\displaystyle -\dfrac{\mathrm{A}}{\mathrm{B}} \) too. By the point-slope form, \[y-y_1=-\frac{\mathrm{A}}{\mathrm{B}}(x-x_1). \] Multiplying through by \(\displaystyle \mathrm{B}\) (allowed, since \(\displaystyle \mathrm{B}\neq 0\)): \[\mathrm{B}(y-y_1)=-\mathrm{A}(x-x_1) \;\Rightarrow\; \mathrm{A}(x-x_1)+\mathrm{B}(y-y_1)=0. \]Case $\displaystyle 2$: \(\displaystyle \mathrm{B}=0 \) (so \(\displaystyle \mathrm{A}\neq 0 \)). The given line is \(\displaystyle \mathrm{A}x+\mathrm{C}=0 \), a line parallel to the \(\displaystyle y\)-axis; any line parallel to it is \(\displaystyle x=x_1 \). And with \(\displaystyle \mathrm{B}=0\) the claimed equation reduces to \(\displaystyle \mathrm{A}(x-x_1)=0 \), i.e. \(\displaystyle x=x_1 \) — the same line.Hence in every case the line through \(\displaystyle (x_1,y_1)\) parallel to \(\displaystyle \mathrm{A}x+\mathrm{B}y+\mathrm{C}=0 \) is \(\displaystyle \mathrm{A}(x-x_1)+\mathrm{B}(y-y_1)=0 \).