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NCERT Exemplar · Class 11 Mathematics Straight Lines

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EXERCISE 10.3 51–59 (part 6 of 6)

  1. State whether the statements in Exercises $\displaystyle 48$ to $\displaystyle 56$ are true or false. Justify.

    Exercise 51

    The straight line 5x+4y=0\displaystyle 5 x+4 y=0 passes through the point of intersection of the straight lines x+2y−10=0\displaystyle x+2 y-10=0 and 2x+y+5=0\displaystyle 2 x+y+5=0.

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    NCERT’s answer
    True
    True. Find the point where the two lines meet and test it in \(\displaystyle 5x+4y=0\).\[x + 2y = 10, \qquad 2x + y = -5 \]\[2(x+2y) - (2x+y) = 20 + 5 \Rightarrow 3y = 25 \]\[y = \frac{25}{3}, \qquad x = 10 - 2y = -\frac{20}{3} \]\[5x + 4y = -\frac{100}{3} + \frac{100}{3} = 0 \]

  2. Exercise 52

    The vertex of an equilateral triangle is (2,3)\displaystyle (2,3) and the equation of the opposite side is x+y=2\displaystyle x+y=2. Then the other two sides are y−3=(2±3)(x−2)\displaystyle y-3=(2 \pm \sqrt{3})(x-2).

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    NCERT’s answer
    True
    True. Each of the other sides makes \(\displaystyle 60^\circ\) with \(\displaystyle x+y=2\), whose slope is \(\displaystyle -1\).NCERT_Solution_Class11_Maths_Exemplar_Ch10_Ex10-3_Q52\[\left|\frac{m-(-1)}{1+(-1)m}\right| = \tan 60^\circ \]\[\frac{m+1}{1-m} = \pm\sqrt3 \]\[m = \frac{\sqrt3-1}{\sqrt3+1} = 2-\sqrt3 \quad\text{or}\quad m = \frac{\sqrt3+1}{\sqrt3-1} = 2+\sqrt3 \]Through \(\displaystyle (2,3)\):\[y - 3 = (2\pm\sqrt3)(x-2) \]

  3. Exercise 53

    The equation of the line joining the point (3,5)\displaystyle (3,5) to the point of intersection of the lines 4x+y−1=0\displaystyle 4 x+y-1=0 and 7x−3y−35=0\displaystyle 7 x-3 y-35=0 is equidistant from the points (0,0)\displaystyle (0,0) and (8,34)\displaystyle (8,34).

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    NCERT’s answer
    True
    True. The line through \(\displaystyle (3,5)\) and the meeting point of the two lines is at equal distances from \(\displaystyle (0,0)\) and \(\displaystyle (8,34)\).\[3(4x+y-1) + (7x-3y-35) = 0 \Rightarrow 19x = 38 \]\[x = 2, \qquad y = 1 - 4x = -7 \]\[m = \frac{5-(-7)}{3-2} = 12 \]\[y - 5 = 12(x-3) \Rightarrow 12x - y - 31 = 0 \]\[d_{(0,0)} = \frac{|-31|}{\sqrt{12^2+1}} = \frac{31}{\sqrt{145}} \]\[d_{(8,34)} = \frac{|96-34-31|}{\sqrt{145}} = \frac{31}{\sqrt{145}} \]

  4. Exercise 54

    The line xa+yb=1\displaystyle \frac{x}{a}+\frac{y}{b}=1 moves in such a way that 1a2+1b2=1c2\displaystyle \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}, where c\displaystyle c is a constant. The locus of the foot of the perpendicular from the origin on the given line is x2+y2=c2\displaystyle x^2+y^2=c^2.

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    NCERT’s answer
    True
    True. Let \(\displaystyle M(h,k)\) be the foot of the perpendicular from \(\displaystyle O\). Then \(\displaystyle OM\) is the distance of \(\displaystyle O\) from \(\displaystyle \frac{x}{a}+\frac{y}{b}-1=0\).NCERT_Solution_Class11_Maths_Exemplar_Ch10_Ex10-3_Q54\[OM = \frac{|-1|}{\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}}} = \frac{1}{\sqrt{1/c^2}} = c \]\[h^2 + k^2 = OM^2 = c^2 \]So \(\displaystyle M\) lies on \(\displaystyle x^2+y^2=c^2\).

  5. Exercise 55

    The lines ax+2y+1=0,bx+3y+1=0\displaystyle a x+2 y+1=0, b x+3 y+1=0 and cx+4y+1=0\displaystyle c x+4 y+1=0 are concurrent if a,b,c\displaystyle a, b, c are in G.P.

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    NCERT’s answer
    False
    False. The lines are concurrent when \(\displaystyle a, b, c\) are in A.P., not G.P.\[\begin{vmatrix} a & 2 & 1 \\ b & 3 & 1 \\ c & 4 & 1 \end{vmatrix} = a(3-4) - 2(b-c) + (4b-3c) = 2b - a - c \]\[2b = a + c \quad \text{(A.P.)} \]Counter-example: \(\displaystyle a, b, c = 1, 2, 4\) are in G.P. but \(\displaystyle 2b = 4 \ne 5 = a + c\). The first two lines meet at \(\displaystyle (1,-1)\):\[4(1) + 4(-1) + 1 = 1 \ne 0 \]

  6. Exercise 56

    Line joining the points (3,−4)\displaystyle (3,-4) and (−2,6)\displaystyle (-2,6) is perpendicular to the line joining the points (−3,6)\displaystyle (-3, 6) and (9,−18)\displaystyle (9, -18).

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    NCERT’s answer
    False
    False. Both slopes equal \(\displaystyle -2\), so the lines are parallel.\[m_1 = \frac{6-(-4)}{-2-3} = -2 \]\[m_2 = \frac{-18-6}{9-(-3)} = -2 \]\[m_1 m_2 = 4 \ne -1 \]
  7. Match the questions given under Column \(\displaystyle \mathrm{C}_1\) with their appropriate answers given under the Column \(\displaystyle \mathrm{C}_2\) in Exercises $\displaystyle 57$ to 59.

    Exercise 57

    Column C1\displaystyle \mathbf{C}_{\mathbf{1}}Column C2\displaystyle \mathbf{C}_{\mathbf{2}}
    (a) The coordinates of the points P and Q on the line x+5y=13\displaystyle x+5 y=13 which are at a distance of 2\displaystyle 2 units from the line 12x−5y+26=0\displaystyle 12 x-5 y+26=0 are(i) (3,1)\displaystyle (3, 1), (−7,11)\displaystyle (-7, 11)
    (b) The coordinates of the point on the line x+y=4\displaystyle x+y=4, which are at a unit distance from the line 4x+3y−10=0\displaystyle 4 x+3 y-10=0 are(ii) (−13,113),(43,73)\displaystyle \left(-\frac{1}{3}, \frac{11}{3}\right),\left(\frac{4}{3}, \frac{7}{3}\right)
    (c) The coordinates of the point on the line joining A(−2,5)\displaystyle \mathrm{A}(-2,5) and B(3,1)\displaystyle \mathrm{B}(3,1) such that AP=PQ=QB\displaystyle \mathrm{AP}=\mathrm{PQ}=\mathrm{QB} are(iii) (1,125),(−3,165)\displaystyle \left(1, \frac{12}{5}\right),\left(-3, \frac{16}{5}\right)

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    NCERT’s answer
    (a)
    \(\displaystyle \leftrightarrow\) (iii)
    (b)
    \(\displaystyle \leftrightarrow\) (i) and
    (c)
    \(\displaystyle \leftrightarrow\) (ii)
    (a)-(iii), (b)-(i), (c)-(ii)
    (a)
    Distance from \(\displaystyle 12x-5y+26=0\) is $\displaystyle 2$, with \(\displaystyle x=13-5y\):
    \[\frac{|12x-5y+26|}{13}=2 \]
    \[|182-65y|=26 \Rightarrow y=\tfrac{12}{5},\ \tfrac{16}{5} \]
    \[(x,y)=\left(1,\tfrac{12}{5}\right),\ \left(-3,\tfrac{16}{5}\right) \]
    (b)
    Distance from \(\displaystyle 4x+3y-10=0\) is $\displaystyle 1$, with \(\displaystyle y=4-x\):
    \[\frac{|4x+3y-10|}{5}=1 \]
    \[|x+2|=5 \Rightarrow x=3,\ -7 \]
    \[(3,1),\ (-7,11) \]
    (c)
    P and Q trisect AB (section formula):
    \[P=\left(\frac{2(-2)+3}{3},\ \frac{2(5)+1}{3}\right)=\left(-\frac13,\ \frac{11}{3}\right) \]
    \[Q=\left(\frac{-2+2(3)}{3},\ \frac{5+2(1)}{3}\right)=\left(\frac43,\ \frac73\right) \]
    Answer: (a)-(iii), (b)-(i), (c)-(ii)

  8. Exercise 58

    The value of the λ\displaystyle \lambda, if the lines (2x+3y+4)+λ(6x−y+12)=0\displaystyle (2 x+3 y+4)+\lambda(6 x-y+12)=0 are
    Column C1\displaystyle \mathbf{C}_{\mathbf{1}}Column C2\displaystyle \mathbf{C}_{\mathbf{2}}
    (a) parallel to y\displaystyle y-axis is(i) λ=−34\displaystyle \lambda=-\frac{3}{4}
    (b) perpendicular to 7x+y−4=0\displaystyle 7 x+y-4=0 is(ii) λ=−13\displaystyle \lambda=-\frac{1}{3}
    (c) passes through (1,2)\displaystyle (1,2) is(iii) λ=−1741\displaystyle \lambda=-\frac{17}{41}
    (d) parallel to x\displaystyle x axis is(iv) λ=3\displaystyle \lambda=3

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    NCERT’s answer
    (a)
    \(\displaystyle \leftrightarrow\) (iv)
    (b)
    \(\displaystyle \leftrightarrow\) (iii)
    (c)
    \(\displaystyle \leftrightarrow\) (i),
    (d)
    \(\displaystyle \leftrightarrow\) (ii)
    (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
    The family is
    \[(2+6\lambda)x+(3-\lambda)y+(4+12\lambda)=0 \]
    (a)
    Parallel to the \(\displaystyle y\)-axis: coefficient of \(\displaystyle y\) is zero.
    \[3-\lambda=0 \Rightarrow \lambda=3 \]
    (b)
    Perpendicular to \(\displaystyle 7x+y-4=0\) (slope \(\displaystyle -7\)): slope must be \(\displaystyle \tfrac17\).
    \[-\frac{2+6\lambda}{3-\lambda}=\frac17 \]
    \[-14-42\lambda=3-\lambda \Rightarrow \lambda=-\frac{17}{41} \]
    (c)
    Passes through \(\displaystyle (1,2)\):
    \[(2+6\lambda)+2(3-\lambda)+(4+12\lambda)=0 \]
    \[12+16\lambda=0 \Rightarrow \lambda=-\frac34 \]
    (d)
    Parallel to the \(\displaystyle x\)-axis: coefficient of \(\displaystyle x\) is zero.
    \[2+6\lambda=0 \Rightarrow \lambda=-\frac13 \]
    Answer: (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)

  9. Exercise 59

    The equation of the line through the intersection of the lines 2x−3y=0\displaystyle 2 x-3 y=0 and 4x−5y=2\displaystyle 4 x-5 y=2 and
    Column C1\displaystyle \mathbf{C}_{\mathbf{1}}Column C2\displaystyle \mathbf{C}_{\mathbf{2}}
    (a) through the point (2,1)\displaystyle (2,1) is(i) 2x−y=4\displaystyle 2 x-y=4
    (b) perpendicular to the line x+2y+1=0\displaystyle x+2 y+1=0 is(ii) x+y−5=0\displaystyle x+y-5=0
    (c) parallel to the line 3x−4y+5=0\displaystyle 3 x-4 y+5=0 is(iii) x−y−1=0\displaystyle x-y-1=0
    (d) equally inclined to the axes is(iv) 3x−4y−1=0\displaystyle 3 x-4 y-1=0

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    NCERT’s answer
    (a)
    \(\displaystyle \leftrightarrow\) (iii)
    (b)
    \(\displaystyle \leftrightarrow\) (i)
    (c)
    \(\displaystyle \leftrightarrow\) (iv),
    (d)
    \(\displaystyle \leftrightarrow\) (ii)
    (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
    Intersection of the given lines:
    \[x=\tfrac32 y,\quad 6y-5y=2 \Rightarrow y=2,\ x=3 \]
    \[(3,2) \](a) Through \(\displaystyle (3,2)\) and \(\displaystyle (2,1)\):
    \[m=\frac{2-1}{3-2}=1 \]
    \[y-1=x-2 \Rightarrow x-y-1=0 \]
    (b)
    Perpendicular to \(\displaystyle x+2y+1=0\) (slope \(\displaystyle -\tfrac12\)), so \(\displaystyle m=2\):
    \[y-2=2(x-3) \Rightarrow 2x-y=4 \]
    (c)
    Parallel to \(\displaystyle 3x-4y+5=0\): take \(\displaystyle 3x-4y+k=0\).
    \[3(3)-4(2)+k=0 \Rightarrow k=-1 \]
    \[3x-4y-1=0 \]
    (d)
    Equally inclined to the axes: \(\displaystyle m=\pm1\). \(\displaystyle m=1\) is already (a), so \(\displaystyle m=-1\):
    \[y-2=-(x-3) \Rightarrow x+y-5=0 \]
    Answer: (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)