NCERT Solutions · Class 11 Mathematics Straight Lines

Point of intersection, then the equal-intercept form.Solve \(\displaystyle 4x+7y-3=0\) and \(\displaystyle 2x-3y+1=0\). Doubling the second gives \(\displaystyle 4x-6y+2=0\); subtracting it from the first, \[13y-5=0 \;\Longrightarrow\; y=\frac{5}{13}, \qquad x=\frac{3y-1}{2}=\frac{\tfrac{15}{13}-1}{2}=\frac{1}{13}.\] So the lines meet at \(\displaystyle \left(\tfrac{1}{13},\tfrac{5}{13}\right)\).A line with equal intercepts \(\displaystyle a\;(a\neq 0)\) on the axes has the intercept form \[\frac{x}{a}+\frac{y}{a}=1, \qquad \text{i.e.} \qquad x+y=a.\]Since the line passes through \(\displaystyle \left(\tfrac{1}{13},\tfrac{5}{13}\right)\), \[a=\frac{1}{13}+\frac{5}{13}=\frac{6}{13}.\]\(\displaystyle x+y=\dfrac{6}{13}\), i.e. \(\displaystyle 13x+13y-6=0\)

Angle between two lines.Any line through the origin has the form \(\displaystyle y=Mx\), that is \(\displaystyle \dfrac{y}{x}=M\). It makes an angle \(\displaystyle \theta\) with \(\displaystyle y=mx+c\), whose slope is \(\displaystyle m\), so \[\tan\theta=\left|\frac{M-m}{1+mM}\right| \;\Longrightarrow\; \frac{M-m}{1+mM}=\pm\tan\theta.\]Taking the \(\displaystyle +\) sign: \[M-m=\tan\theta+mM\tan\theta \;\Longrightarrow\; M(1-m\tan\theta)=m+\tan\theta \;\Longrightarrow\; M=\frac{m+\tan\theta}{1-m\tan\theta}.\]Taking the \(\displaystyle -\) sign: \[M-m=-\tan\theta-mM\tan\theta \;\Longrightarrow\; M(1+m\tan\theta)=m-\tan\theta \;\Longrightarrow\; M=\frac{m-\tan\theta}{1+m\tan\theta}.\]The two cases pair the upper signs together and the lower signs together.\(\displaystyle \dfrac{y}{x}=\dfrac{m\pm\tan\theta}{1\mp m\tan\theta}\)

Section formula with an unknown ratio. If the point divides the join of \(\displaystyle (x_1,y_1)\) and \(\displaystyle (x_2,y_2)\) in the ratio \(\displaystyle k:1\), its coordinates are \(\displaystyle \left(\dfrac{kx_2+x_1}{k+1},\dfrac{ky_2+y_1}{k+1}\right)\).Let \(\displaystyle x+y=4\) cut the segment joining \(\displaystyle (-1,1)\) and \(\displaystyle (5,7)\) in the ratio \(\displaystyle k:1\). The point of division is \[\left(\frac{5k-1}{k+1},\;\frac{7k+1}{k+1}\right).\]This point lies on \(\displaystyle x+y=4\): \[\frac{5k-1}{k+1}+\frac{7k+1}{k+1}=4 \;\Longrightarrow\; \frac{12k}{k+1}=4 \;\Longrightarrow\; 12k=4k+4 \;\Longrightarrow\; k=\frac12.\]Since \(\displaystyle k>0\) the division is internal, and the point of division is \(\displaystyle (1,3)\), which indeed satisfies \(\displaystyle x+y=4\).The line divides the segment internally in the ratio \(\displaystyle 1:2\).

Distance measured along a given line. The distance is not the perpendicular distance here — it must be measured along \(\displaystyle 2x-y=0\).First note that \(\displaystyle (1,2)\) lies on \(\displaystyle 2x-y=0\), since \(\displaystyle 2(1)-2=0\). So travel along this line from \(\displaystyle (1,2)\) until it meets \(\displaystyle 4x+7y+5=0\).Put \(\displaystyle y=2x\) into \(\displaystyle 4x+7y+5=0\): \[4x+14x+5=0 \;\Longrightarrow\; 18x=-5 \;\Longrightarrow\; x=-\frac{5}{18}, \qquad y=-\frac{5}{9}.\]So the lines meet at \(\displaystyle Q\left(-\tfrac{5}{18},-\tfrac{5}{9}\right)\). The required distance is \[PQ=\sqrt{\left(1+\frac{5}{18}\right)^{2}+\left(2+\frac{5}{9}\right)^{2}}=\sqrt{\left(\frac{23}{18}\right)^{2}+\left(\frac{23}{9}\right)^{2}}=\frac{23}{18}\sqrt{1+4}.\]\(\displaystyle \dfrac{23\sqrt{5}}{18}\) units \(\displaystyle (\approx 2.86)\)

Parametric (distance) form of a line. A line through \(\displaystyle (x_1,y_1)\) inclined at \(\displaystyle \theta\) to the \(\displaystyle x\)-axis has points \[x=x_1+r\cos\theta, \qquad y=y_1+r\sin\theta,\] where \(\displaystyle |r|\) is the distance of the point from \(\displaystyle (x_1,y_1)\).With \(\displaystyle (x_1,y_1)=(-1,2)\) and \(\displaystyle r=3\), the point of intersection is \(\displaystyle (-1+3\cos\theta,\;2+3\sin\theta)\). It lies on \(\displaystyle x+y=4\): \[(-1+3\cos\theta)+(2+3\sin\theta)=4 \;\Longrightarrow\; 3(\cos\theta+\sin\theta)=3 \;\Longrightarrow\; \cos\theta+\sin\theta=1.\]Dividing by \(\displaystyle \sqrt{2}\), \[\cos\theta\cos 45^{\circ}+\sin\theta\sin 45^{\circ}=\frac{1}{\sqrt2} \;\Longrightarrow\; \cos(\theta-45^{\circ})=\cos 45^{\circ},\] so \(\displaystyle \theta-45^{\circ}=\pm 45^{\circ}\), giving \(\displaystyle \theta=90^{\circ}\) or \(\displaystyle \theta=0^{\circ}\).Check: \(\displaystyle \theta=0^{\circ}\) gives the meeting point \(\displaystyle (2,2)\) and \(\displaystyle \theta=90^{\circ}\) gives \(\displaystyle (-1,5)\); both lie on \(\displaystyle x+y=4\) and both are \(\displaystyle 3\) units from \(\displaystyle (-1,2)\).The line must be drawn at \(\displaystyle 0^{\circ}\) or \(\displaystyle 90^{\circ}\) to the \(\displaystyle x\)-axis, i.e. along \(\displaystyle y=2\) (parallel to the \(\displaystyle x\)-axis) or along \(\displaystyle x=-1\) (parallel to the \(\displaystyle y\)-axis).

A leg parallel to an axis fixes one coordinate of the right-angle vertex.Let the hypotenuse have ends \(\displaystyle A(1,3)\) and \(\displaystyle B(-4,1)\), and let the right angle be at \(\displaystyle C\). One leg is parallel to the \(\displaystyle x\)-axis and the other to the \(\displaystyle y\)-axis, so the vertical leg through \(\displaystyle C\) also passes through one end and the horizontal leg through \(\displaystyle C\) passes through the other. That leaves exactly two positions for \(\displaystyle C\):Case $\displaystyle 1$: \(\displaystyle C=(1,1)\). The leg \(\displaystyle AC\) joins \(\displaystyle (1,3)\) and \(\displaystyle (1,1)\) — a vertical line \(\displaystyle x=1\). The leg \(\displaystyle BC\) joins \(\displaystyle (-4,1)\) and \(\displaystyle (1,1)\) — a horizontal line \(\displaystyle y=1\).Case $\displaystyle 2$: \(\displaystyle C=(-4,3)\). The leg through \(\displaystyle B(-4,1)\) and \(\displaystyle C\) is \(\displaystyle x=-4\); the leg through \(\displaystyle A(1,3)\) and \(\displaystyle C\) is \(\displaystyle y=3\).In each case the two legs are perpendicular, so the angle at \(\displaystyle C\) is indeed a right angle.A quick sketch of \(\displaystyle A\), \(\displaystyle B\) and the two candidate corners makes it clear why there are exactly two such triangles.The legs are \(\displaystyle x=1\) and \(\displaystyle y=1\), or else \(\displaystyle x=-4\) and \(\displaystyle y=3\).

Image in a mirror line. The image \(\displaystyle Q(h,k)\) of \(\displaystyle P(3,8)\) is fixed by two conditions: \(\displaystyle PQ\) is perpendicular to the mirror, and the midpoint of \(\displaystyle PQ\) lies on the mirror.The mirror \(\displaystyle x+3y=7\) has slope \(\displaystyle -\tfrac13\), so \(\displaystyle PQ\) has slope \(\displaystyle 3\): \[\frac{k-8}{h-3}=3 \;\Longrightarrow\; k=3h-1. \tag{1}\]The midpoint \(\displaystyle \left(\dfrac{3+h}{2},\dfrac{8+k}{2}\right)\) lies on \(\displaystyle x+3y=7\): \[\frac{3+h}{2}+3\cdot\frac{8+k}{2}=7 \;\Longrightarrow\; 3+h+24+3k=14 \;\Longrightarrow\; h+3k=-13. \tag{2}\]Substituting ($\displaystyle 1$) into ($\displaystyle 2$): \[h+3(3h-1)=-13 \;\Longrightarrow\; 10h=-10 \;\Longrightarrow\; h=-1, \qquad k=3(-1)-1=-4.\]Check: the midpoint is \(\displaystyle (1,2)\) and \(\displaystyle 1+3(2)=7\). ✓The image is \(\displaystyle (-1,-4)\).

Equally inclined means the two angles are equal in magnitude.The slopes are \(\displaystyle m_1=3\) for \(\displaystyle y=3x+1\), and \(\displaystyle m_2=\tfrac12\) for \(\displaystyle 2y=x+3\).Angle of \(\displaystyle y=mx+4\) with the first line: \(\displaystyle \tan\alpha=\left|\dfrac{3-m}{1+3m}\right|\). Angle with the second: \(\displaystyle \tan\beta=\left|\dfrac{\tfrac12-m}{1+\tfrac{m}{2}}\right|=\left|\dfrac{1-2m}{2+m}\right|\).Setting \(\displaystyle \tan\alpha=\tan\beta\), the line \(\displaystyle y=mx+4\) lies between the other two, so the two ratios carry opposite signs: \[\frac{3-m}{1+3m}=-\frac{1-2m}{2+m}.\]Cross-multiplying, \[(3-m)(2+m)=-(1-2m)(1+3m),\] \[6+m-m^{2}=-(1+m-6m^{2})=-1-m+6m^{2},\] \[7m^{2}-2m-7=0.\]By the quadratic formula, \[m=\frac{2\pm\sqrt{4+196}}{14}=\frac{2\pm 10\sqrt2}{14}=\frac{1\pm 5\sqrt2}{7}.\](The same-sign choice gives \(\displaystyle 6+m-m^{2}=1+m-6m^{2}\), i.e. \(\displaystyle m^{2}=-1\), which has no real solution.)\(\displaystyle m=\dfrac{1\pm 5\sqrt2}{7}\)

Perpendicular distance formula. The distance of \(\displaystyle (x_1,y_1)\) from \(\displaystyle Ax+By+C=0\) is \(\displaystyle \dfrac{|Ax_1+By_1+C|}{\sqrt{A^{2}+B^{2}}}\).For \(\displaystyle \mathrm{P}(x,y)\), \[d_1=\frac{|x+y-5|}{\sqrt{1^{2}+1^{2}}}=\frac{|x+y-5|}{\sqrt2}, \qquad d_2=\frac{|3x-2y+7|}{\sqrt{3^{2}+(-2)^{2}}}=\frac{|3x-2y+7|}{\sqrt{13}},\] and we are given \(\displaystyle d_1+d_2=10\).The two given lines cut the plane into four regions. Inside any one region the signs of \(\displaystyle x+y-5\) and of \(\displaystyle 3x-2y+7\) do not change, so the modulus signs can be dropped with fixed signs. Take the region where both are positive: \[\frac{x+y-5}{\sqrt2}+\frac{3x-2y+7}{\sqrt{13}}=10.\]Multiplying by \(\displaystyle \sqrt{26}=\sqrt2\cdot\sqrt{13}\), \[\sqrt{13}\,(x+y-5)+\sqrt{2}\,(3x-2y+7)=10\sqrt{26},\] \[\left(\sqrt{13}+3\sqrt2\right)x+\left(\sqrt{13}-2\sqrt2\right)y+\left(7\sqrt2-5\sqrt{13}-10\sqrt{26}\right)=0.\]This is of the form \(\displaystyle Ax+By+C=0\) with \(\displaystyle A,B,C\) constants and \(\displaystyle A,B\) not both zero, which is the general equation of a straight line. Choosing the other three sign combinations changes only the constants \(\displaystyle A\), \(\displaystyle B\), \(\displaystyle C\) — every one of them is again an equation of the first degree in \(\displaystyle x\) and \(\displaystyle y\).Hence the locus of P is a straight line, i.e. P must move on a line.

Mid-line of two parallel lines. Two parallel lines can be written with identical \(\displaystyle x\) and \(\displaystyle y\) coefficients; the line equidistant from them is then the one whose constant term is the average of theirs.Multiply \(\displaystyle 3x+2y+6=0\) by \(\displaystyle 3\) so that both lines carry the same coefficients: \[9x+6y-7=0 \qquad \text{and} \qquad 9x+6y+18=0.\]A line equidistant from two parallel lines is itself parallel to them, say \(\displaystyle 9x+6y+\lambda=0\). The distance between two parallel lines \(\displaystyle Ax+By+C_1=0\) and \(\displaystyle Ax+By+C_2=0\) is \(\displaystyle \dfrac{|C_1-C_2|}{\sqrt{A^{2}+B^{2}}}\), and \(\displaystyle \sqrt{9^{2}+6^{2}}=\sqrt{117}\), so \[\frac{|\lambda-(-7)|}{\sqrt{117}}=\frac{|\lambda-18|}{\sqrt{117}} \;\Longrightarrow\; |\lambda+7|=|\lambda-18|.\]\(\displaystyle \lambda+7=\lambda-18\) is impossible, so \(\displaystyle \lambda+7=-(\lambda-18)\), giving \(\displaystyle 2\lambda=11\) and \(\displaystyle \lambda=\tfrac{11}{2}\).Thus \(\displaystyle 9x+6y+\tfrac{11}{2}=0\); multiplying by \(\displaystyle 2\) clears the fraction.Check with the point \(\displaystyle \left(-\tfrac{11}{18},0\right)\), which lies on it. Its distance from \(\displaystyle 9x+6y-7=0\) is \(\displaystyle \dfrac{\left|-\tfrac{11}{2}-7\right|}{\sqrt{117}}=\dfrac{12.5}{10.817}\approx 1.156\), and from \(\displaystyle 3x+2y+6=0\) it is \(\displaystyle \dfrac{\left|-\tfrac{11}{6}+6\right|}{\sqrt{13}}=\dfrac{4.167}{3.606}\approx 1.156\). Equal. ✓\(\displaystyle 18x+12y+11=0\)

Reflection at a mirror: the reflected ray comes from the image of the source.The mirror is the \(\displaystyle x\)-axis, so the image of the source \(\displaystyle P(1,2)\) is \(\displaystyle P'(1,-2)\). The reflected ray, produced backwards, passes through \(\displaystyle P'\); since it also passes through \(\displaystyle (5,3)\), the point A is where the line \(\displaystyle P'(1,-2)\,(5,3)\) meets the \(\displaystyle x\)-axis.Slope of that line: \[m=\frac{3-(-2)}{5-1}=\frac{5}{4}.\]Its equation, through \(\displaystyle (1,-2)\): \[y+2=\frac{5}{4}(x-1).\]Put \(\displaystyle y=0\): \[2=\frac{5}{4}(x-1) \;\Longrightarrow\; x-1=\frac{8}{5} \;\Longrightarrow\; x=\frac{13}{5}.\]Check: the incident ray \(\displaystyle P(1,2)\to A\) has slope \(\displaystyle \dfrac{0-2}{\tfrac{13}{5}-1}=-\dfrac{5}{4}\), the exact negative of the reflected ray's slope — the two rays make equal angles with the \(\displaystyle x\)-axis, as reflection requires. ✓A sketch showing \(\displaystyle P\), its mirror image \(\displaystyle P'\) below the \(\displaystyle x\)-axis, and the straight line from \(\displaystyle P'\) through \(\displaystyle (5,3)\) makes this construction easy to see.A \(\displaystyle =\left(\dfrac{13}{5},\,0\right)\)

Perpendicular distance formula. Write the line in the form \(\displaystyle Ax+By+C=0\): \[\frac{\cos\theta}{a}\,x+\frac{\sin\theta}{b}\,y-1=0.\]Put \(\displaystyle c=\sqrt{a^{2}-b^{2}}\); the two points are \(\displaystyle (c,0)\) and \(\displaystyle (-c,0)\), and let \[D=\sqrt{\frac{\cos^{2}\theta}{a^{2}}+\frac{\sin^{2}\theta}{b^{2}}}.\]The two perpendicular lengths are \[p_1=\frac{\left|\dfrac{c\cos\theta}{a}-1\right|}{D}, \qquad p_2=\frac{\left|\dfrac{-c\cos\theta}{a}-1\right|}{D}=\frac{\left|\dfrac{c\cos\theta}{a}+1\right|}{D}.\]Their product uses \(\displaystyle (1-t)(1+t)=1-t^{2}\) with \(\displaystyle t=\dfrac{c\cos\theta}{a}\): \[p_1p_2=\frac{\left|1-\dfrac{c^{2}\cos^{2}\theta}{a^{2}}\right|}{D^{2}}=\frac{\dfrac{a^{2}-(a^{2}-b^{2})\cos^{2}\theta}{a^{2}}}{\dfrac{\cos^{2}\theta}{a^{2}}+\dfrac{\sin^{2}\theta}{b^{2}}}.\](The modulus may be dropped because \(\displaystyle c^{2}\cos^{2}\theta\le a^{2}\).) Now simplify the numerator using \(\displaystyle a^{2}-a^{2}\cos^{2}\theta=a^{2}\sin^{2}\theta\): \[a^{2}-(a^{2}-b^{2})\cos^{2}\theta=a^{2}\sin^{2}\theta+b^{2}\cos^{2}\theta,\] and the denominator: \[\frac{\cos^{2}\theta}{a^{2}}+\frac{\sin^{2}\theta}{b^{2}}=\frac{b^{2}\cos^{2}\theta+a^{2}\sin^{2}\theta}{a^{2}b^{2}}.\]Therefore \[p_1p_2=\frac{a^{2}\sin^{2}\theta+b^{2}\cos^{2}\theta}{a^{2}}\cdot\frac{a^{2}b^{2}}{b^{2}\cos^{2}\theta+a^{2}\sin^{2}\theta}=b^{2}.\]Hence the product of the two perpendicular lengths is \(\displaystyle b^{2}\), independent of \(\displaystyle \theta\).

Least time means least distance, and the shortest route to a line is the perpendicular to it.Step $\displaystyle 1$ — find the junction. Solve \(\displaystyle 2x-3y+4=0\) and \(\displaystyle 3x+4y-5=0\). Multiply the first by \(\displaystyle 4\) and the second by \(\displaystyle 3\): \[8x-12y+16=0, \qquad 9x+12y-15=0.\] Adding, \(\displaystyle 17x+1=0\), so \(\displaystyle x=-\dfrac{1}{17}\). Then from \(\displaystyle 2x-3y+4=0\), \[3y=2\!\left(-\tfrac{1}{17}\right)+4=\frac{66}{17} \;\Longrightarrow\; y=\frac{22}{17}.\] The person stands at \(\displaystyle \left(-\tfrac{1}{17},\tfrac{22}{17}\right)\).Step $\displaystyle 2$ — the perpendicular direction. The path \(\displaystyle 6x-7y+8=0\) has slope \(\displaystyle \tfrac67\), so the perpendicular to it has slope \(\displaystyle -\tfrac76\).Step $\displaystyle 3$ — the required line. \[y-\frac{22}{17}=-\frac{7}{6}\left(x+\frac{1}{17}\right).\] Multiplying by \(\displaystyle 102\): \[102y-132=-119x-7 \;\Longrightarrow\; 119x+102y-125=0.\]Check: \(\displaystyle 119\!\left(-\tfrac{1}{17}\right)+102\!\left(\tfrac{22}{17}\right)-125=-7+132-125=0\). ✓\(\displaystyle 119x+102y-125=0\)