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NCERT Exemplar · Class 11 Mathematics Straight Lines

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EXERCISE 10.3 1–10 (part 1 of 6)

  1. Exercise 1

    Find the equation of the straight line which passes through the point (1,−2)\displaystyle (1, -2) and cuts off equal intercepts from axes.

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    NCERT’s answer
    \(\displaystyle x+y+1=0\)
    Equal non-zero intercepts \(\displaystyle a\): \[\frac{x}{a}+\frac{y}{a}=1 \Rightarrow x+y=a \] The line passes through \(\displaystyle (1,-2)\): \[a=1+(-2)=-1 \] \[x+y+1=0 \]NCERT_Solution_Class11_Maths_Exemplar_Ch10_Ex10-3_Q1The line \(\displaystyle 2x+y=0\) through the origin cuts off no intercepts, so it is not counted.Answer: \(\displaystyle x+y+1=0\)

  2. Exercise 2

    Find the equation of the line passing through the point (5,2)\displaystyle (5,2) and perpendicular to the line joining the points (2,3)\displaystyle (2,3) and (3,−1)\displaystyle (3,-1).

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    NCERT’s answer
    \(\displaystyle x-4 y+3=0\)
    \[m_{AB}=\frac{-1-3}{3-2}=-4 \] Perpendicular slope: \[m=-\frac{1}{m_{AB}}=\frac14 \] Through \(\displaystyle P(5,2)\): \[y-2=\frac14(x-5) \] \[4y-8=x-5 \] \[x-4y+3=0 \]NCERT_Solution_Class11_Maths_Exemplar_Ch10_Ex10-3_Q2Answer: \(\displaystyle x-4y+3=0\)

  3. Exercise 3

    Find the angle between the lines y=(2−3)(x+5)\displaystyle y=(2-\sqrt{3})(x+5) and y=(2+3)(x−7)\displaystyle y=(2+\sqrt{3})(x-7).

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    NCERT’s answer
    \(\displaystyle 60^{\circ}\) or \(\displaystyle 120^{\circ}\)
    \[m_1=2-\sqrt3,\qquad m_2=2+\sqrt3 \] \[m_1m_2=4-3=1,\qquad m_2-m_1=2\sqrt3 \] Acute angle between the lines: \[\tan\theta=\left|\frac{m_2-m_1}{1+m_1m_2}\right|=\frac{2\sqrt3}{2}=\sqrt3 \] \[\theta=60^\circ \] The other angle between the lines is the supplement: \[180^\circ-60^\circ=120^\circ \]Answer: \(\displaystyle 60^\circ\) or \(\displaystyle 120^\circ\)

  4. Exercise 4

    Find the equation of the lines which passes through the point (3,4)\displaystyle (3,4) and cuts off intercepts from the coordinate axes such that their sum is 14.

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    NCERT’s answer
    \(\displaystyle x+y=7\) or \(\displaystyle \frac{x}{6}+\frac{y}{8}=1\)
    Intercepts \(\displaystyle a\) and \(\displaystyle b\), with \(\displaystyle a+b=14\): \[\frac{x}{a}+\frac{y}{14-a}=1 \] The line passes through \(\displaystyle (3,4)\): \[\frac{3}{a}+\frac{4}{14-a}=1 \] \[3(14-a)+4a=a(14-a) \] \[a^2-13a+42=0 \] \[(a-6)(a-7)=0 \Rightarrow a=6\ \text{or}\ 7 \] \[a=6,\ b=8:\quad \frac{x}{6}+\frac{y}{8}=1 \Rightarrow 4x+3y=24 \] \[a=7,\ b=7:\quad \frac{x}{7}+\frac{y}{7}=1 \Rightarrow x+y=7 \]NCERT_Solution_Class11_Maths_Exemplar_Ch10_Ex10-3_Q4Answer: \(\displaystyle 4x+3y-24=0\) and \(\displaystyle x+y-7=0\)

  5. Exercise 5

    Find the points on the line x+y=4\displaystyle x+y=4 which lie at a unit distance from the line 4x+3y=10\displaystyle 4 x+3 y=10.

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    NCERT’s answer
    \(\displaystyle (3,1),(-7,11)\)
    A point on \(\displaystyle x+y=4\) is \(\displaystyle (t,\,4-t)\). Its distance from \(\displaystyle 4x+3y-10=0\) is \(\displaystyle 1\): \[\frac{|4t+3(4-t)-10|}{\sqrt{4^2+3^2}}=1 \] \[\frac{|t+2|}{5}=1 \Rightarrow t+2=\pm5 \] \[t=3\ \text{or}\ t=-7 \]NCERT_Solution_Class11_Maths_Exemplar_Ch10_Ex10-3_Q5Answer: \(\displaystyle (3,1)\) and \(\displaystyle (-7,11)\)

  6. Exercise 6

    Show that the tangent of an angle between the lines xa+yb=1\displaystyle \frac{x}{a}+\frac{y}{b}=1 and xa−yb=1\displaystyle \frac{x}{a}-\frac{y}{b}=1 is 2aba2−b2\displaystyle \frac{2 a b}{a^2-b^2}.

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[\frac{x}{a}+\frac{y}{b}=1 \Rightarrow y=-\frac{b}{a}x+b,\qquad m_1=-\frac{b}{a} \] \[\frac{x}{a}-\frac{y}{b}=1 \Rightarrow y=\frac{b}{a}x-b,\qquad m_2=\frac{b}{a} \] Angle from the first line to the second: \[\tan\theta=\frac{m_2-m_1}{1+m_1m_2}=\frac{2b/a}{1-b^2/a^2} \] \[\tan\theta=\frac{2b}{a}\cdot\frac{a^2}{a^2-b^2}=\frac{2ab}{a^2-b^2} \]Answer: \(\displaystyle \tan\theta=\dfrac{2ab}{a^2-b^2}\)

  7. Exercise 7

    Find the equation of lines passing through (1,2)\displaystyle (1,2) and making angle 30\displaystyle 30° with y\displaystyle y-axis.

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    A line at \(\displaystyle 30^\circ\) to the \(\displaystyle y\)-axis is inclined at \(\displaystyle 60^\circ\) or \(\displaystyle 120^\circ\) to the \(\displaystyle x\)-axis: \[m=\tan60^\circ=\sqrt3\quad\text{or}\quad m=\tan120^\circ=-\sqrt3 \] Through \(\displaystyle (1,2)\): \[y-2=\sqrt3(x-1) \Rightarrow \sqrt3x-y+2-\sqrt3=0 \] \[y-2=-\sqrt3(x-1) \Rightarrow \sqrt3x+y-2-\sqrt3=0 \]NCERT_Solution_Class11_Maths_Exemplar_Ch10_Ex10-3_Q7Answer: \(\displaystyle \sqrt3x-y+2-\sqrt3=0\) and \(\displaystyle \sqrt3x+y-2-\sqrt3=0\)

  8. Exercise 8

    Find the equation of the line passing through the point of intersection of 2x+y=5\displaystyle 2 x+y=5 and x+3y+8=0\displaystyle x+3 y+8=0 and parallel to the line 3x+4y=7\displaystyle 3 x+4 y=7.

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    NCERT’s answer
    \(\displaystyle 3 x+4 y+3=0\)
    Point of intersection: \[2x+y=5 \Rightarrow y=5-2x \] \[x+3(5-2x)+8=0 \Rightarrow -5x+23=0 \] \[x=\frac{23}{5},\qquad y=-\frac{21}{5} \] A line parallel to \(\displaystyle 3x+4y=7\) has the form \(\displaystyle 3x+4y=k\): \[k=3\cdot\frac{23}{5}+4\left(-\frac{21}{5}\right)=\frac{69-84}{5}=-3 \] \[3x+4y+3=0 \]Answer: \(\displaystyle 3x+4y+3=0\)

  9. Exercise 9

    For what values of a\displaystyle a and b\displaystyle b the intercepts cut off on the coordinate axes by the line ax+by+8=0\displaystyle a x+b y+8=0 are equal in length but opposite in signs to those cut off by the line 2x−3y+6=0\displaystyle 2 x-3 y+6=0 on the axes.

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    NCERT’s answer
    \(\displaystyle a=\frac{-8}{3}, b=4\)
    Intercepts of \(\displaystyle 2x-3y+6=0\) (put \(\displaystyle y=0\), then \(\displaystyle x=0\)): \[x=-3,\qquad y=2 \] The required line has intercepts \(\displaystyle 3\) and \(\displaystyle -2\). For \(\displaystyle ax+by+8=0\): \[x\text{-intercept}=-\frac{8}{a}=3 \Rightarrow a=-\frac{8}{3} \] \[y\text{-intercept}=-\frac{8}{b}=-2 \Rightarrow b=4 \] Check: \[-\tfrac{8}{3}x+4y+8=0 \iff 2x-3y-6=0 \] NCERT_Solution_Class11_Maths_Exemplar_Ch10_Ex10-3_Q9 Answer: \(\displaystyle a=-\dfrac{8}{3},\ b=4\)

  10. Exercise 10

    If the intercept of a line between the coordinate axes is divided by the point (−5,4)\displaystyle (-5, 4) in the ratio 1:2\displaystyle 1: 2, then find the equation of the line.

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    NCERT’s answer
    \(\displaystyle 8 x-5 y+60=0\)
    Let the line meet the axes at \(\displaystyle A(a,0)\) and \(\displaystyle B(0,b)\). \(\displaystyle P(-5,4)\) divides \(\displaystyle AB\) in the ratio \(\displaystyle 1:2\): \[-5=\frac{1\cdot 0+2\cdot a}{1+2}\Rightarrow a=-\frac{15}{2} \] \[4=\frac{1\cdot b+2\cdot 0}{1+2}\Rightarrow b=12 \] \[\frac{x}{-15/2}+\frac{y}{12}=1 \] \[-8x+5y=60 \] NCERT_Solution_Class11_Maths_Exemplar_Ch10_Ex10-3_Q10 Answer: \(\displaystyle 8x-5y+60=0\) (if the ratio \(\displaystyle 1:2\) is counted from the \(\displaystyle y\)-axis end instead, \(\displaystyle a=-15,\ b=6\) and the line is \(\displaystyle 2x-5y+30=0\))