NCERT Solutions · Class 11 Mathematics Straight Lines

Angle between two lines.Let \(\displaystyle m\) be the slope of the other line. With \(\displaystyle m_1=2\) and the angle between the lines equal to \(\displaystyle 60^{\circ}\), \[\tan 60^{\circ}=\left|\frac{m-2}{1+2m}\right| \;\Rightarrow\; \frac{m-2}{1+2m}=\pm\sqrt3. \]Taking \(\displaystyle +\sqrt3\): \[m-2=\sqrt3+2\sqrt3\,m \;\Rightarrow\; m(1-2\sqrt3)=2+\sqrt3 \;\Rightarrow\; m=\frac{2+\sqrt3}{1-2\sqrt3}=-\frac{8+5\sqrt3}{11}, \] after multiplying numerator and denominator by \(\displaystyle 1+2\sqrt3\).Taking \(\displaystyle -\sqrt3\): \[m-2=-\sqrt3-2\sqrt3\,m \;\Rightarrow\; m(1+2\sqrt3)=2-\sqrt3 \;\Rightarrow\; m=\frac{2-\sqrt3}{1+2\sqrt3}=\frac{5\sqrt3-8}{11}. \]Each line passes through \(\displaystyle (2,3)\), so \(\displaystyle y-3=m(x-2) \):For \(\displaystyle m=-\dfrac{8+5\sqrt3}{11} \): \[11(y-3)=-(8+5\sqrt3)(x-2) \;\Rightarrow\; (8+5\sqrt3)x+11y-(49+10\sqrt3)=0. \]For \(\displaystyle m=\dfrac{5\sqrt3-8}{11} \): \[11(y-3)=(5\sqrt3-8)(x-2) \;\Rightarrow\; (5\sqrt3-8)x-11y+(49-10\sqrt3)=0. \]\(\displaystyle y-3=\dfrac{2+\sqrt3}{1-2\sqrt3}(x-2) \) or \(\displaystyle y-3=\dfrac{2-\sqrt3}{1+2\sqrt3}(x-2) \); equivalently \(\displaystyle (8+5\sqrt3)x+11y-(49+10\sqrt3)=0 \) or \(\displaystyle (5\sqrt3-8)x-11y+(49-10\sqrt3)=0 \).

Perpendicular bisector: midpoint plus negative reciprocal slope.The right bisector of a segment passes through its midpoint and is perpendicular to it.Midpoint of \(\displaystyle (3,4)\) and \(\displaystyle (-1,2)\): \[\left(\frac{3+(-1)}{2},\;\frac{4+2}{2}\right)=(1,3). \]Slope of the segment: \[m=\frac{2-4}{-1-3}=\frac{-2}{-4}=\frac{1}{2}. \]Slope of the bisector \(\displaystyle =-\dfrac{1}{m}=-2 \). Through \(\displaystyle (1,3)\): \[y-3=-2(x-1) \;\Rightarrow\; y-3=-2x+2. \]\(\displaystyle 2x+y-5=0\)

Drop a perpendicular, then solve the two lines simultaneously.The foot of the perpendicular is the point where the given line meets the perpendicular drawn to it from \(\displaystyle (-1,3)\).Slope of \(\displaystyle 3x-4y-16=0 \) is \(\displaystyle \dfrac{3}{4} \), so the perpendicular from \(\displaystyle (-1,3)\) has slope \(\displaystyle -\dfrac{4}{3} \): \[y-3=-\frac{4}{3}(x+1) \;\Rightarrow\; 3y-9=-4x-4 \;\Rightarrow\; 4x+3y-5=0. \]Solve \(\displaystyle 3x-4y=16 \) and \(\displaystyle 4x+3y=5 \). Multiply the first by $\displaystyle 3$ and the second by $\displaystyle 4$ and add: \[9x-12y=48,\quad 16x+12y=20 \;\Rightarrow\; 25x=68 \;\Rightarrow\; x=\frac{68}{25}. \] Then \(\displaystyle 3y=5-4\cdot\frac{68}{25}=\frac{125-272}{25}=-\frac{147}{25} \), so \(\displaystyle y=-\dfrac{49}{25} \).Check: \(\displaystyle 3\!\left(\tfrac{68}{25}\right)-4\!\left(-\tfrac{49}{25}\right)=\tfrac{204+196}{25}=16 \). ✓The foot of the perpendicular is \(\displaystyle \left(\dfrac{68}{25},\,-\dfrac{49}{25}\right) \).

Perpendicularity plus substitution of the point.Let \(\displaystyle \mathrm{P}(-1,2) \) be the foot of the perpendicular from the origin \(\displaystyle \mathrm{O}(0,0) \).Slope of \(\displaystyle \mathrm{OP} \): \[m_{\mathrm{OP}}=\frac{2-0}{-1-0}=-2. \]Since \(\displaystyle \mathrm{OP} \) is perpendicular to the line \(\displaystyle y=mx+c \), \[m\cdot(-2)=-1 \;\Rightarrow\; m=\frac{1}{2}. \]\(\displaystyle \mathrm{P}(-1,2) \) lies on the line, so \[2=\frac{1}{2}(-1)+c \;\Rightarrow\; c=2+\frac{1}{2}=\frac{5}{2}. \]\(\displaystyle m=\dfrac{1}{2} \), \(\displaystyle c=\dfrac{5}{2} \)

Distance of a line from the origin.The distance of \(\displaystyle \mathrm{A}x+\mathrm{B}y+\mathrm{C}=0 \) from the origin is \(\displaystyle \dfrac{|\mathrm{C}|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}} \).First line: \(\displaystyle x\cos\theta-y\sin\theta-k\cos 2\theta=0 \). Here \(\displaystyle \sqrt{\cos^2\theta+\sin^2\theta}=1 \), so \[p=\frac{|k\cos 2\theta|}{1}=|k\cos 2\theta| \;\Rightarrow\; p^{2}=k^{2}\cos^{2}2\theta. \]Second line: \(\displaystyle x\sec\theta+y\,\mathrm{cosec}\,\theta-k=0 \). Now \[\sec^{2}\theta+\mathrm{cosec}^{2}\theta=\frac{1}{\cos^{2}\theta}+\frac{1}{\sin^{2}\theta}=\frac{\sin^{2}\theta+\cos^{2}\theta}{\sin^{2}\theta\cos^{2}\theta}=\frac{1}{\sin^{2}\theta\cos^{2}\theta}, \] so \(\displaystyle \sqrt{\sec^{2}\theta+\mathrm{cosec}^{2}\theta}=\dfrac{1}{|\sin\theta\cos\theta|} \) and \[q=\frac{|k|}{1/|\sin\theta\cos\theta|}=|k\sin\theta\cos\theta|=\frac{|k\sin 2\theta|}{2}, \] using \(\displaystyle 2\sin\theta\cos\theta=\sin 2\theta \). Hence \(\displaystyle 4q^{2}=k^{2}\sin^{2}2\theta \).Adding, \[p^{2}+4q^{2}=k^{2}\cos^{2}2\theta+k^{2}\sin^{2}2\theta=k^{2}\left(\cos^{2}2\theta+\sin^{2}2\theta\right)=k^{2}. \]Hence \(\displaystyle p^{2}+4q^{2}=k^{2} \).

Altitude = perpendicular from a vertex to the opposite side.Side BC. Slope of BC through \(\displaystyle \mathrm{B}(4,-1) \) and \(\displaystyle \mathrm{C}(1,2) \): \[m_{\mathrm{BC}}=\frac{2-(-1)}{1-4}=\frac{3}{-3}=-1. \] Equation of BC: \(\displaystyle y+1=-1(x-4) \Rightarrow x+y-3=0 \).Altitude from A. It is perpendicular to BC, so its slope is \(\displaystyle -\dfrac{1}{-1}=1 \). Through \(\displaystyle \mathrm{A}(2,3) \): \[y-3=1\cdot(x-2) \;\Rightarrow\; x-y+1=0. \]Its length is the distance from \(\displaystyle \mathrm{A}(2,3) \) to the line \(\displaystyle x+y-3=0 \): \[d=\frac{|2+3-3|}{\sqrt{1^{2}+1^{2}}}=\frac{2}{\sqrt2}=\sqrt2. \]Altitude from A: \(\displaystyle x-y+1=0 \); its length is \(\displaystyle \sqrt{2} \) units.

Intercept form plus the distance formula.A line with intercepts \(\displaystyle a\) and \(\displaystyle b\) on the axes is \[\frac{x}{a}+\frac{y}{b}=1 \;\Rightarrow\; bx+ay-ab=0. \]Its distance from the origin is \[p=\frac{|b\cdot 0+a\cdot 0-ab|}{\sqrt{a^{2}+b^{2}}}=\frac{|ab|}{\sqrt{a^{2}+b^{2}}}. \]Squaring and taking reciprocals, \[p^{2}=\frac{a^{2}b^{2}}{a^{2}+b^{2}} \;\Rightarrow\; \frac{1}{p^{2}}=\frac{a^{2}+b^{2}}{a^{2}b^{2}}=\frac{a^{2}}{a^{2}b^{2}}+\frac{b^{2}}{a^{2}b^{2}}=\frac{1}{b^{2}}+\frac{1}{a^{2}}. \]Hence \(\displaystyle \dfrac{1}{p^{2}}=\dfrac{1}{a^{2}}+\dfrac{1}{b^{2}} \).