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NCERT Exemplar · Class 11 Mathematics Straight Lines

59 questions · 59 still being checked

EXERCISE 10.3 21–30 (part 3 of 6)

  1. Exercise 21

    If p\displaystyle p is the length of perpendicular from the origin on the line xa+yb=1\displaystyle \frac{x}{a}+\frac{y}{b}=1 and a2\displaystyle a^2, p2,b2\displaystyle p^2, b^2 are in A.P, then show that a4+b4=0\displaystyle a^4+b^4=0.

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    Distance of \(\displaystyle bx+ay-ab=0\) from the origin:\[p=\frac{|ab|}{\sqrt{a^2+b^2}} \Rightarrow p^2=\frac{a^2b^2}{a^2+b^2} \]\(\displaystyle a^2,p^2,b^2\) in A.P.:\[2p^2=a^2+b^2 \]\[\frac{2a^2b^2}{a^2+b^2}=a^2+b^2 \]\[2a^2b^2=a^4+2a^2b^2+b^4 \]\[a^4+b^4=0 \]Answer: \(\displaystyle a^4+b^4=0\)
  2. Choose the correct answer from the given four options in Exercises $\displaystyle 22$ to $\displaystyle 41$

    Exercise 22

    A line cutting off intercept - 3\displaystyle 3 from the y\displaystyle y-axis and the tengent at angle to the x\displaystyle x-axis is 35\displaystyle \frac{3}{5}, its equation is
    (A)
    5y−3x+15=0\displaystyle 5 y-3 x+15=0
    (B)
    3y−5x+15=0\displaystyle 3 y-5 x+15=0
    (C)
    5y−3x−15=0\displaystyle 5 y-3 x-15=0
    (D)
    None of these

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    NCERT’s answer
    A
    (A) \(\displaystyle 5y-3x+15=0\)Slope \(\displaystyle \tfrac35\), \(\displaystyle y\)-intercept \(\displaystyle -3\):\[y=\tfrac35x-3 \]\[5y=3x-15 \Rightarrow 5y-3x+15=0 \]

  3. Exercise 23

    Slope of a line which cuts off intercepts of equal lengths on the axes is
    (A)
    1\displaystyle 1
    (B) - 0\displaystyle 0 (C) 2\displaystyle 2 (D) 3\displaystyle \sqrt{3}

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    NCERT’s answer
    A
    (A) \(\displaystyle -1\)Equal intercepts \(\displaystyle a\) on both axes:\[\frac xa+\frac ya=1 \]\[y=-x+a \Rightarrow m=-1 \]

  4. Exercise 24

    The equation of the straight line passing through the point (3,2)\displaystyle (3,2) and perpendicular to the line y=x\displaystyle y=x is
    (A)
    x−y=5\displaystyle x-y=5
    (B)
    x+y=5\displaystyle x+y=5
    (C)
    x+y=1\displaystyle x+y=1
    (D)
    x−y=1\displaystyle x-y=1

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    NCERT’s answer
    B
    (B) \(\displaystyle x+y=5\)The line \(\displaystyle y=x\) has slope \(\displaystyle 1\), so the perpendicular has slope \(\displaystyle -1\):\[y-2=-(x-3) \]\[x+y=5 \]

  5. Exercise 25

    The equation of the line passing through the point (1,2)\displaystyle (1,2) and perpendicular to the line x+y+1=0\displaystyle x+y+1=0 is
    (A)
    y−x+1=0\displaystyle y-x+1=0
    (B)
    y−x−1=0\displaystyle y-x-1=0
    (C)
    y−x+2=0\displaystyle y-x+2=0
    (D)
    y−x−2=0\displaystyle y-x-2=0

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    NCERT’s answer
    B
    (B) \(\displaystyle y-x-1=0\)The perpendicular's slope is the negative reciprocal of the given slope.\[x+y+1=0 \Rightarrow m_1=-1 \] \[m_1m_2=-1 \Rightarrow m_2=1 \] \[y-2=1\cdot(x-1) \] \[y-x-1=0 \]

  6. Exercise 26

    The tangent of angle between the lines whose intercepts on the axes are a,−b\displaystyle a,-b and b,−a\displaystyle b,-a, respectively, is
    (A)
    a2−b2ab\displaystyle \frac{a^2-b^2}{a b}
    (B)
    b2−a22\displaystyle \frac{b^2-a^2}{2}
    (C)
    b2−a22ab\displaystyle \frac{b^2-a^2}{2 a b}
    (D)
    None of these

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    NCERT’s answer
    C
    (C) \(\displaystyle \dfrac{b^2-a^2}{2ab}\)Slopes from the intercept forms:\[\frac{x}{a}+\frac{y}{-b}=1 \Rightarrow m_1=\frac{b}{a} \] \[\frac{x}{b}+\frac{y}{-a}=1 \Rightarrow m_2=\frac{a}{b} \] \[\tan\theta=\frac{m_1-m_2}{1+m_1m_2}=\frac{\dfrac{b}{a}-\dfrac{a}{b}}{1+1}=\frac{b^2-a^2}{2ab} \]

  7. Exercise 27

    If the line xa+yb=1\displaystyle \frac{x}{a}+\frac{y}{b}=1 passes through the points (2,−3)\displaystyle (2, -3) and (4,−5)\displaystyle (4, -5), then (a,b)\displaystyle (a, b) is
    (A)
    (1,1)\displaystyle (1,1)
    (B)
    (−1,1)\displaystyle (- 1, 1)
    (C)
    (1,−1)\displaystyle (1,-1)
    (D)
    (−1,−1)\displaystyle (- 1, -1)

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    NCERT’s answer
    D
    (D) \(\displaystyle (-1,-1)\)Substitute both points:\[\frac{2}{a}-\frac{3}{b}=1 \qquad \frac{4}{a}-\frac{5}{b}=1 \]Second minus twice the first:\[\left(\frac{4}{a}-\frac{5}{b}\right)-2\left(\frac{2}{a}-\frac{3}{b}\right)=1-2 \Rightarrow \frac{1}{b}=-1 \] \[b=-1 \] \[\frac{2}{a}+3=1 \Rightarrow a=-1 \]

  8. Exercise 28

    The distance of the point of intersection of the lines 2x−3y+5=0\displaystyle 2 x-3 y+5=0 and 3x+4y=0\displaystyle 3 x+4 y=0 from the line 5x−2y=0\displaystyle 5 x-2 y=0 is
    (A)
    1301729\displaystyle \frac{130}{17 \sqrt{29}}
    (B)
    13729\displaystyle \frac{13}{7 \sqrt{29}}
    (C)
    1307\displaystyle \frac{130}{7}
    (D)
    None of these

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    NCERT’s answer
    A
    (A) \(\displaystyle \dfrac{130}{17\sqrt{29}}\)Intersection, using \(\displaystyle x=-\tfrac{4y}{3}\) from the second line:\[2\left(-\frac{4y}{3}\right)-3y+5=0 \Rightarrow y=\frac{15}{17},\quad x=-\frac{20}{17} \]Distance from \(\displaystyle 5x-2y=0\):\[d=\frac{\left|5\left(-\dfrac{20}{17}\right)-2\left(\dfrac{15}{17}\right)\right|}{\sqrt{5^2+2^2}}=\frac{130/17}{\sqrt{29}}=\frac{130}{17\sqrt{29}} \]

  9. Exercise 29

    The equations of the lines which pass through the point (3,−2)\displaystyle (3, -2) and are inclined at 60∘\displaystyle 60^{\circ} to the line 3x+y=1\displaystyle \sqrt{3} x+y=1 is
    (A)
    y+2=0,3x−y−2−33=0\displaystyle y+2=0, \sqrt{3} x-y-2-3 \sqrt{3}=0
    (B)
    x−2=0,3x−y+2+33=0\displaystyle x-2=0, \sqrt{3} x-y+2+3 \sqrt{3}=0
    (C)
    3x−y−2−33=0\displaystyle \sqrt{3} x-y-2-3 \sqrt{3}=0
    (D)
    None of these

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    NCERT’s answer
    A
    (A) \(\displaystyle y+2=0,\ \sqrt{3}x-y-2-3\sqrt{3}=0\)The given line has slope \(\displaystyle -\sqrt3\). Let the required slope be \(\displaystyle m\).\[\tan60^\circ=\left|\frac{m+\sqrt3}{1-\sqrt3\,m}\right|=\sqrt3 \] \[\frac{m+\sqrt3}{1-\sqrt3\,m}=\sqrt3 \Rightarrow m=0 \] \[\frac{m+\sqrt3}{1-\sqrt3\,m}=-\sqrt3 \Rightarrow m=\sqrt3 \]Lines through \(\displaystyle (3,-2)\):\[y+2=0 \] \[y+2=\sqrt3(x-3) \Rightarrow \sqrt3x-y-2-3\sqrt3=0 \]

  10. Exercise 30

    The equations of the lines passing through the point (1,0)\displaystyle (1,0) and at a distance 32\displaystyle \frac{\sqrt{3}}{2} from the origin, are
    (A)
    3x+y−3=0,3x−y−3=0\displaystyle \sqrt{3} x+y-\sqrt{3}=0, \sqrt{3} x-y-\sqrt{3}=0
    (B)
    3x+y+3=0,3x−y+3=0\displaystyle \sqrt{3} x+y+\sqrt{3}=0, \sqrt{3} x-y+\sqrt{3}=0
    (C)
    x+3y−3=0,x−3y−3=0\displaystyle x+\sqrt{3} y-\sqrt{3}=0, x-\sqrt{3} y-\sqrt{3}=0
    (D)
    None of these.

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    NCERT’s answer
    A
    (A) \(\displaystyle \sqrt{3}x+y-\sqrt{3}=0,\ \sqrt{3}x-y-\sqrt{3}=0\)The vertical line \(\displaystyle x=1\) is at distance \(\displaystyle 1\ne\tfrac{\sqrt3}{2}\), so use a slope \(\displaystyle m\).\[y=m(x-1)\Rightarrow mx-y-m=0 \] \[\frac{|m|}{\sqrt{m^2+1}}=\frac{\sqrt3}{2} \Rightarrow 4m^2=3m^2+3 \Rightarrow m=\pm\sqrt3 \] \[y=\pm\sqrt3\,(x-1) \] \[\sqrt3x-y-\sqrt3=0,\qquad \sqrt3x+y-\sqrt3=0 \]