NCERT Solutions · Class 11 Mathematics Straight Lines

Intercept form. A line with intercepts \(\displaystyle a\) on the \(\displaystyle x\)-axis and \(\displaystyle b\) on the \(\displaystyle y\)-axis (both non-zero) is \(\displaystyle \dfrac{x}{a}+\dfrac{y}{b}=1\).The intercepts are equal, so \(\displaystyle b=a\) and the line is \[\frac{x}{a}+\frac{y}{a}=1\quad\Longrightarrow\quad x+y=a. \]It passes through $\displaystyle (2, 3)$, so \[a=2+3=5. \]Check: \(\displaystyle 2+3=5\). ✓The required line is \(\displaystyle x+y=5\).

Intercept form with a constraint. Let the intercepts be \(\displaystyle a\) and \(\displaystyle b\), so the line is \(\displaystyle \dfrac{x}{a}+\dfrac{y}{b}=1\).Given \(\displaystyle a+b=9\). Since the line passes through $\displaystyle (2, 2)$, \[\frac{2}{a}+\frac{2}{b}=1\quad\Longrightarrow\quad 2(a+b)=ab\quad\Longrightarrow\quad ab=2(9)=18. \]So \(\displaystyle a\) and \(\displaystyle b\) are the roots of \[t^{2}-(a+b)t+ab=0\quad\Longrightarrow\quad t^{2}-9t+18=0\quad\Longrightarrow\quad (t-3)(t-6)=0, \] giving \(\displaystyle t=3\) or \(\displaystyle t=6\). Hence \(\displaystyle (a,b)=(3,6)\) or \(\displaystyle (6,3)\).For \(\displaystyle (a,b)=(3,6)\): \(\displaystyle \dfrac{x}{3}+\dfrac{y}{6}=1\Rightarrow 2x+y=6\).For \(\displaystyle (a,b)=(6,3)\): \(\displaystyle \dfrac{x}{6}+\dfrac{y}{3}=1\Rightarrow x+2y=6\).Both pass through $\displaystyle (2, 2)$: \(\displaystyle 2(2)+2=6\) and \(\displaystyle 2+2(2)=6\). ✓The lines are \(\displaystyle 2x+y-6=0\) and \(\displaystyle x+2y-6=0\).

Slope-intercept form. The slope of a line making an angle \(\displaystyle \theta\) with the positive \(\displaystyle x\)-axis is \(\displaystyle m=\tan\theta\).Here \(\displaystyle \theta=\dfrac{2\pi}{3}=120^{\circ}\), so \[m=\tan 120^{\circ}=-\tan 60^{\circ}=-\sqrt{3}. \]The line passes through $\displaystyle (0, 2)$, which is its \(\displaystyle y\)-intercept, so by \(\displaystyle y=mx+c\): \[y=-\sqrt{3}\,x+2\quad\Longrightarrow\quad \sqrt{3}\,x+y-2=0. \]A parallel line has the same slope \(\displaystyle -\sqrt{3}\). Crossing the \(\displaystyle y\)-axis $\displaystyle 2$ units below the origin means it passes through \(\displaystyle (0,-2)\), so \(\displaystyle c=-2\): \[y=-\sqrt{3}\,x-2\quad\Longrightarrow\quad \sqrt{3}\,x+y+2=0. \]The line is \(\displaystyle \sqrt{3}\,x+y-2=0\), and the parallel line is \(\displaystyle \sqrt{3}\,x+y+2=0\).

Perpendicular from the origin. Let the foot of the perpendicular be A $\displaystyle (-2, 9)$. The segment OA is perpendicular to the required line, so their slopes multiply to \(\displaystyle -1\).Slope of OA, with O $\displaystyle (0, 0)$ and A $\displaystyle (-2, 9)$: \[m_{\mathrm{OA}}=\frac{9-0}{-2-0}=-\frac{9}{2}. \]Hence the slope of the line is \[m=-\frac{1}{m_{\mathrm{OA}}}=\frac{2}{9}. \]The line passes through A $\displaystyle (-2, 9)$, so by the point-slope form: \[y-9=\frac{2}{9}(x+2)\quad\Longrightarrow\quad 9y-81=2x+4. \]The required line is \(\displaystyle 2x-9y+85=0\).

Two-point form for a linear relation. Since L is a linear function of C, the points \(\displaystyle (\mathrm{C},\mathrm{L})\) lie on a straight line through \(\displaystyle (20,\,124.942)\) and \(\displaystyle (110,\,125.134)\).Slope: \[m=\frac{125.134-124.942}{110-20}=\frac{0.192}{90}=\frac{4}{1875}\approx 0.002133. \]Two-point (point-slope) form using \(\displaystyle (20,\,124.942)\): \[\mathrm{L}-124.942=\frac{4}{1875}(\mathrm{C}-20), \] \[\mathrm{L}=124.942+\frac{4}{1875}\mathrm{C}-\frac{80}{1875}=\frac{4}{1875}\mathrm{C}+124.8993\overline{3}. \]Check: at \(\displaystyle \mathrm{C}=110\), \(\displaystyle \frac{4}{1875}(110)+124.89933\ldots=0.234\overline{6}+124.89933\ldots=125.134\). ✓\(\displaystyle \mathrm{L}=\dfrac{4}{1875}\,\mathrm{C}+124.8993\ldots\), i.e. \(\displaystyle \mathrm{L}=124.942+\dfrac{4}{1875}(\mathrm{C}-20)\approx 0.002133\,\mathrm{C}+124.899\).

Two-point form of a linear model. Let \(\displaystyle x\) be the selling price in rupees per litre and \(\displaystyle y\) the weekly demand in litres. The data give the points \(\displaystyle (14,\,980)\) and \(\displaystyle (16,\,1220)\).Slope: \[m=\frac{1220-980}{16-14}=\frac{240}{2}=120. \]Point-slope form through \(\displaystyle (14,\,980)\): \[y-980=120(x-14)\quad\Longrightarrow\quad y=120x-1680+980=120x-700. \]Check: at \(\displaystyle x=16\), \(\displaystyle y=120(16)-700=1920-700=1220\). ✓At \(\displaystyle x=17\): \[y=120(17)-700=2040-700=1340. \]He could sell $\displaystyle 1340$ litres of milk weekly at Rs $\displaystyle 17$/litre.

Intercept form with the mid-point formula. Picture the line cutting the axes so that it forms a right triangle with them: it meets the \(\displaystyle x\)-axis at A \(\displaystyle (p,0)\) and the \(\displaystyle y\)-axis at B \(\displaystyle (0,q)\), and AB is the "segment between the axes".P \(\displaystyle (a,b)\) is the mid-point of AB, so by the mid-point formula \[\left(\frac{p+0}{2},\ \frac{0+q}{2}\right)=(a,b)\quad\Longrightarrow\quad \frac{p}{2}=a,\ \ \frac{q}{2}=b, \] \[p=2a,\qquad q=2b. \]The intercept form of the line with intercepts \(\displaystyle p\) and \(\displaystyle q\) is \(\displaystyle \dfrac{x}{p}+\dfrac{y}{q}=1\), so \[\frac{x}{2a}+\frac{y}{2b}=1. \]Multiplying both sides by $\displaystyle 2$: \[\frac{x}{a}+\frac{y}{b}=2. \]Hence the equation of the line is \(\displaystyle \dfrac{x}{a}+\dfrac{y}{b}=2\), as required.

Intercept form with the section formula. Picture the line cutting the axes at A \(\displaystyle (p,0)\) on the \(\displaystyle x\)-axis and B \(\displaystyle (0,q)\) on the \(\displaystyle y\)-axis; AB is the "segment between the axes", and R lies on it nearer to A.R \(\displaystyle (h,k)\) divides AB in the ratio \(\displaystyle 1:2\), so by the section formula \[(h,k)=\left(\frac{1\cdot 0+2\cdot p}{1+2},\ \frac{1\cdot q+2\cdot 0}{1+2}\right)=\left(\frac{2p}{3},\ \frac{q}{3}\right). \]Therefore \[h=\frac{2p}{3}\ \Longrightarrow\ p=\frac{3h}{2},\qquad k=\frac{q}{3}\ \Longrightarrow\ q=3k. \]Intercept form \(\displaystyle \dfrac{x}{p}+\dfrac{y}{q}=1\) gives \[\frac{x}{3h/2}+\frac{y}{3k}=1\quad\Longrightarrow\quad \frac{2x}{3h}+\frac{y}{3k}=1. \]Multiplying by $\displaystyle 3$: \[\frac{2x}{h}+\frac{y}{k}=3. \]Check with \(\displaystyle h=3,\,k=2\): then \(\displaystyle p=\tfrac92,\ q=6\), and R \(\displaystyle =\left(\tfrac{2}{3}\cdot\tfrac92,\ \tfrac63\right)=(3,2)\). ✓The required line is \(\displaystyle \dfrac{2x}{h}+\dfrac{y}{k}=3\), that is, \(\displaystyle 2kx+hy=3hk\).

Equation of a line through two points. Three points are collinear if the third point satisfies the equation of the line through the first two.Line through A $\displaystyle (3, 0)$ and B $\displaystyle (-2, -2)$. Slope: \[m=\frac{-2-0}{-2-3}=\frac{-2}{-5}=\frac{2}{5}. \]Point-slope form through A $\displaystyle (3, 0)$: \[y-0=\frac{2}{5}(x-3)\quad\Longrightarrow\quad 5y=2x-6\quad\Longrightarrow\quad 2x-5y-6=0. \]Now substitute C $\displaystyle (8, 2)$ into the left-hand side: \[2(8)-5(2)-6=16-10-6=0. \]Since C satisfies the equation, it lies on the line AB.Hence the points $\displaystyle (3, 0)$, $\displaystyle (-2, -2)$ and $\displaystyle (8, 2)$ are collinear.