NCERT Solutions · Class 11 Mathematics Straight Lines

Read off the coefficients of the general equation.For \(\displaystyle \mathrm{A}x+\mathrm{B}y+\mathrm{C}=0 \): the line is parallel to the \(\displaystyle x\)-axis when \(\displaystyle \mathrm{A}=0 \) (and \(\displaystyle \mathrm{B}\neq0 \)), parallel to the \(\displaystyle y\)-axis when \(\displaystyle \mathrm{B}=0 \) (and \(\displaystyle \mathrm{A}\neq0 \)), and passes through the origin when \(\displaystyle \mathrm{C}=0 \).Here \(\displaystyle \mathrm{A}=k-3 \), \(\displaystyle \mathrm{B}=-(4-k^{2}) \), \(\displaystyle \mathrm{C}=k^{2}-7k+6 \).(a) Parallel to the \(\displaystyle x\)-axis: \(\displaystyle k-3=0 \Rightarrow k=3 \). Then \(\displaystyle \mathrm{B}=-(4-9)=5\neq0 \), and the line becomes \(\displaystyle 5y-6=0 \), i.e. \(\displaystyle y=\tfrac{6}{5} \) — indeed horizontal. So \(\displaystyle k=3 \).(b) Parallel to the \(\displaystyle y\)-axis: \(\displaystyle 4-k^{2}=0 \Rightarrow k=\pm 2 \). For \(\displaystyle k=2\): \(\displaystyle -x-4=0 \), i.e. \(\displaystyle x=-4 \). For \(\displaystyle k=-2\): \(\displaystyle -5x+24=0 \), i.e. \(\displaystyle x=\tfrac{24}{5} \). Both are vertical, so \(\displaystyle k=2 \) or \(\displaystyle k=-2 \).(c) Passing through the origin: \(\displaystyle k^{2}-7k+6=0 \Rightarrow (k-1)(k-6)=0 \Rightarrow k=1 \) or \(\displaystyle k=6 \). For \(\displaystyle k=1\) the line is \(\displaystyle -2x-3y=0 \) and for \(\displaystyle k=6\) it is \(\displaystyle 3x+32y=0 \); both are genuine lines through \(\displaystyle (0,0)\).(a) \(\displaystyle k=3\) (b) \(\displaystyle k=\pm 2\) (c) \(\displaystyle k=1\) or \(\displaystyle k=6\)

Intercept form, with the intercepts as roots of a quadratic.Let the intercepts be \(\displaystyle a\) and \(\displaystyle b\), so the line is \(\displaystyle \dfrac{x}{a}+\dfrac{y}{b}=1 \).Given \(\displaystyle a+b=1 \) and \(\displaystyle ab=-6 \), the numbers \(\displaystyle a,b\) are the roots of \[t^{2}-(a+b)t+ab=0 \;\Rightarrow\; t^{2}-t-6=0 \;\Rightarrow\; (t-3)(t+2)=0, \] so \(\displaystyle \{a,b\}=\{3,-2\} \).If \(\displaystyle a=3,\;b=-2\): \[\frac{x}{3}+\frac{y}{-2}=1 \;\Rightarrow\; 2x-3y=6. \]If \(\displaystyle a=-2,\;b=3\): \[\frac{x}{-2}+\frac{y}{3}=1 \;\Rightarrow\; -3x+2y=6 \;\Rightarrow\; 3x-2y+6=0. \]\(\displaystyle 2x-3y-6=0 \) and \(\displaystyle 3x-2y+6=0 \)

Distance of a point from a line.Clear the fractions in \(\displaystyle \dfrac{x}{3}+\dfrac{y}{4}=1 \) by multiplying by $\displaystyle 12$: \[4x+3y-12=0. \]A point on the \(\displaystyle y\)-axis has the form \(\displaystyle (0,k) \). Its distance from the line is \[\frac{|4(0)+3k-12|}{\sqrt{4^{2}+3^{2}}}=\frac{|3k-12|}{5}. \]Setting this equal to $\displaystyle 4$: \[|3k-12|=20 \;\Rightarrow\; 3k-12=20 \ \text{ or }\ 3k-12=-20, \] \[k=\frac{32}{3} \quad\text{or}\quad k=-\frac{8}{3}. \]The points are \(\displaystyle \left(0,\dfrac{32}{3}\right) \) and \(\displaystyle \left(0,-\dfrac{8}{3}\right) \).

Two-point form, then the distance formula.Let \(\displaystyle \mathrm{P}(\cos\theta,\sin\theta) \) and \(\displaystyle \mathrm{Q}(\cos\phi,\sin\phi) \). Using the two-point form, \[y-\sin\theta=\frac{\sin\phi-\sin\theta}{\cos\phi-\cos\theta}\,(x-\cos\theta). \]Apply the sum-to-product identities, writing \(\displaystyle \mathrm{A}=\dfrac{\theta+\phi}{2} \): \[\sin\phi-\sin\theta=2\cos \mathrm{A}\,\sin\frac{\phi-\theta}{2},\qquad \cos\phi-\cos\theta=-2\sin \mathrm{A}\,\sin\frac{\phi-\theta}{2}, \] so the slope is \(\displaystyle -\cot \mathrm{A} \) (the chord is not vertical unless \(\displaystyle \sin \mathrm{A}=0 \), handled below).Substituting and multiplying through by \(\displaystyle \sin \mathrm{A} \): \[(y-\sin\theta)\sin \mathrm{A}=-\cos \mathrm{A}\,(x-\cos\theta) \] \[\Rightarrow\; x\cos \mathrm{A}+y\sin \mathrm{A}=\cos\theta\cos \mathrm{A}+\sin\theta\sin \mathrm{A}=\cos(\theta-\mathrm{A})=\cos\frac{\theta-\phi}{2}. \]Since \(\displaystyle \cos^{2}\mathrm{A}+\sin^{2}\mathrm{A}=1 \), the distance of this line from the origin is \[d=\frac{\left|\cos\dfrac{\theta-\phi}{2}\right|}{\sqrt{\cos^{2}\mathrm{A}+\sin^{2}\mathrm{A}}}=\left|\cos\frac{\theta-\phi}{2}\right|. \](If \(\displaystyle \sin \mathrm{A}=0 \) the chord is the vertical line \(\displaystyle x=\cos \mathrm{A}\cdot\cos\frac{\theta-\phi}{2} \), whose distance from the origin is again \(\displaystyle \left|\cos\frac{\theta-\phi}{2}\right| \), so the formula still holds.)The perpendicular distance is \(\displaystyle \left|\cos\dfrac{\theta-\phi}{2}\right| \).

Find the point of intersection, then use \(\displaystyle x=\) constant.Solve the two lines simultaneously. From \(\displaystyle 3x+y=0 \) we get \(\displaystyle y=-3x \). Substituting into \(\displaystyle x-7y+5=0 \): \[x-7(-3x)+5=0 \;\Rightarrow\; 22x+5=0 \;\Rightarrow\; x=-\frac{5}{22}, \] and then \(\displaystyle y=-3\left(-\tfrac{5}{22}\right)=\tfrac{15}{22} \). The point of intersection is \(\displaystyle \left(-\tfrac{5}{22},\tfrac{15}{22}\right) \).Every line parallel to the \(\displaystyle y\)-axis has the form \(\displaystyle x=c \). Passing through this point forces \(\displaystyle c=-\dfrac{5}{22} \).\(\displaystyle x=-\dfrac{5}{22} \), i.e. \(\displaystyle 22x+5=0 \)

Find where the line meets the \(\displaystyle y\)-axis, then use the perpendicular slope.The point. On the \(\displaystyle y\)-axis, \(\displaystyle x=0 \), so \(\displaystyle \dfrac{0}{4}+\dfrac{y}{6}=1 \Rightarrow y=6 \). The line meets the \(\displaystyle y\)-axis at \(\displaystyle (0,6) \).The slope. Multiply \(\displaystyle \dfrac{x}{4}+\dfrac{y}{6}=1 \) by $\displaystyle 12$: \(\displaystyle 3x+2y=12 \Rightarrow y=-\tfrac{3}{2}x+6 \), so its slope is \(\displaystyle -\dfrac{3}{2} \). The perpendicular slope is \[m=-\frac{1}{-3/2}=\frac{2}{3}. \]The line. Through \(\displaystyle (0,6) \) with slope \(\displaystyle \tfrac{2}{3} \): \[y-6=\frac{2}{3}x \;\Rightarrow\; 3y-18=2x. \]\(\displaystyle 2x-3y+18=0 \)

Find the vertices, then use the coordinate area formula.Take the lines in pairs (assume \(\displaystyle k\neq 0 \), otherwise all three lines are concurrent at the origin and no triangle is formed). A and B lie on the vertical line \(\displaystyle x=k \), so take AB as the base: \[\text{base}=|k-(-k)|=2|k|, \] and the height is the horizontal distance from \(\displaystyle \mathrm{O}(0,0) \) to \(\displaystyle x=k \), namely \(\displaystyle |k| \).\[\text{Area}=\frac{1}{2}\times 2|k|\times|k|=k^{2}. \]The area is \(\displaystyle k^{2} \) square units.

Concurrency through a computed point. Three lines pass through one point exactly when the point of intersection of two of them lies on the third.Take the two lines that do not involve \(\displaystyle p\): \[3x+y-2=0, \qquad 2x-y-3=0.\] Adding them eliminates \(\displaystyle y\): \(\displaystyle 5x-5=0\), so \(\displaystyle x=1\), and then \(\displaystyle y=2-3x=-1\). These two lines meet at \(\displaystyle (1,-1)\).For all three to be concurrent, \(\displaystyle (1,-1)\) must satisfy \(\displaystyle px+2y-3=0\): \[p(1)+2(-1)-3=0 \;\Longrightarrow\; p-5=0.\]\(\displaystyle p=5\)

Concurrency: the meeting point of two lines lies on the third.For the first two lines to meet at all we need \(\displaystyle m_1 \neq m_2\). Equating their \(\displaystyle y\)-values, \[m_1x+c_1=m_2x+c_2 \;\Longrightarrow\; x=\frac{c_2-c_1}{m_1-m_2},\] and then \[y=m_1\!\left(\frac{c_2-c_1}{m_1-m_2}\right)+c_1=\frac{m_1c_2-m_1c_1+c_1m_1-c_1m_2}{m_1-m_2}=\frac{m_1c_2-m_2c_1}{m_1-m_2}.\]The three lines are concurrent, so this point also lies on \(\displaystyle y=m_3x+c_3\): \[\frac{m_1c_2-m_2c_1}{m_1-m_2}=m_3\!\left(\frac{c_2-c_1}{m_1-m_2}\right)+c_3.\]Multiplying throughout by \(\displaystyle m_1-m_2\), \[m_1c_2-m_2c_1=m_3c_2-m_3c_1+c_3m_1-c_3m_2.\]Bringing everything to the left and collecting the terms in \(\displaystyle m_1\), \(\displaystyle m_2\), \(\displaystyle m_3\), \[(m_1c_2-m_1c_3)+(m_2c_3-m_2c_1)+(m_3c_1-m_3c_2)=0.\]\(\displaystyle m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0\), as required.

Angle between two lines. If a line of slope \(\displaystyle m\) makes an angle \(\displaystyle \theta\) with a line of slope \(\displaystyle m_1\), then \[\tan\theta=\left|\frac{m-m_1}{1+mm_1}\right|.\]Here \(\displaystyle x-2y=3\) gives \(\displaystyle y=\tfrac{1}{2}x-\tfrac{3}{2}\), so \(\displaystyle m_1=\tfrac{1}{2}\), and \(\displaystyle \theta=45^{\circ}\), \(\displaystyle \tan\theta=1\): \[\left|\frac{m-\tfrac12}{1+\tfrac{m}{2}}\right|=1 \;\Longrightarrow\; \frac{2m-1}{2+m}=\pm 1.\]Taking \(\displaystyle +1\): \(\displaystyle 2m-1=2+m\), so \(\displaystyle m=3\). Taking \(\displaystyle -1\): \(\displaystyle 2m-1=-(2+m)\), so \(\displaystyle 3m=-1\) and \(\displaystyle m=-\tfrac13\).Now use the point-slope form through \(\displaystyle (3,2)\).For \(\displaystyle m=3\): \(\displaystyle y-2=3(x-3)\), i.e. \(\displaystyle 3x-y-7=0\). For \(\displaystyle m=-\tfrac13\): \(\displaystyle y-2=-\tfrac13(x-3)\), i.e. \(\displaystyle 3y-6=-x+3\), i.e. \(\displaystyle x+3y-9=0\).\(\displaystyle 3x-y-7=0\) and \(\displaystyle x+3y-9=0\)