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NCERT Exemplar · Class 11 Mathematics Straight Lines

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EXERCISE 10.3 41–50 (part 5 of 6)

  1. Choose the correct answer from the given four options in Exercises $\displaystyle 22$ to $\displaystyle 41$

    Exercise 41

    One vertex of the equilateral triangle with centroid at the origin and one side as x+y−2=0\displaystyle x+y-2=0 is
    (A)
    (−1,−1)\displaystyle (-1, -1)
    (B)
    (2,2)\displaystyle (2,2)
    (C)
    (−2,−2)\displaystyle (-2, -2)
    (D)
    (2,−2)\displaystyle (2,-2)
    [Hint: Let ABC be the equilateral triangle with vertex A(h,k)\displaystyle \mathrm{A}(h, k) and let D(α,β)\displaystyle \mathrm{D}(\alpha, \beta) be the point on BC. Then 2α+h3=0=2β+k3\displaystyle \frac{2 \alpha+h}{3}=0=\frac{2 \beta+k}{3}. Also α+β−2=0\displaystyle \alpha+\beta-2=0 and (k−0h−0)×(−1)=−1\displaystyle \left(\frac{k-0}{h-0}\right) \times(-1)=-1].

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    (C) \(\displaystyle (-2,-2)\)The centroid is also the incentre and circumcentre, so \(\displaystyle R=2r\).\[r=\frac{|0+0-2|}{\sqrt2}=\sqrt2,\qquad R=2\sqrt2 \]The opposite vertex lies on \(\displaystyle y=x\), on the side of the origin away from the given line:\[\sqrt{h^2+h^2}=2\sqrt2,\ h<0 \Rightarrow h=-2 \]NCERT_Solution_Class11_Maths_Exemplar_Ch10_Ex10-3_Q41
  2. Fill in the blank in Exercises $\displaystyle 42$ to 47.

    Exercise 42

    If a,b,c\displaystyle a, b, c are in A.P., then the straight lines ax+by+c=0\displaystyle a x+b y+c=0 will always pass through ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    \(\displaystyle (1,-2)\)
    \(\displaystyle (1,-2)\)\[a,\ b,\ c \text{ in A.P.} \Rightarrow 2b=a+c \]\[a\cdot 1+b\cdot(-2)+c=a-2b+c=0 \]So \(\displaystyle x=1,\ y=-2\) satisfies \(\displaystyle ax+by+c=0\) for every such \(\displaystyle a,b,c\).

  3. Exercise 43

    The line which cuts off equal intercept from the axes and pass through the point (1,−2)\displaystyle (1, -2) is ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    \(\displaystyle x+y+1=0\)
    \(\displaystyle x+y+1=0\)\[\frac{x}{a}+\frac{y}{a}=1 \Rightarrow x+y=a \]\[(1,-2):\quad 1-2=a \Rightarrow a=-1 \]\[x+y=-1 \]NCERT_Solution_Class11_Maths_Exemplar_Ch10_Ex10-3_Q43

  4. Exercise 44

    Equations of the lines through the point (3,2)\displaystyle (3,2) and making an angle of 45∘\displaystyle 45^{\circ} with the line x−2y=3\displaystyle x-2 y=3 are ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    \(\displaystyle 3 x-y-7=0, x+3 y-9=0\)
    \(\displaystyle 3x-y-7=0\) and \(\displaystyle x+3y-9=0\)Given line \(\displaystyle x-2y=3\) has slope \(\displaystyle \tfrac12\).\[\tan45^\circ=\left|\frac{m-\frac12}{1+\frac m2}\right|=1 \]\[m-\tfrac12=\pm\left(1+\tfrac m2\right) \Rightarrow m=3 \ \text{or}\ m=-\tfrac13 \]\[y-2=3(x-3) \Rightarrow 3x-y-7=0 \]\[y-2=-\tfrac13(x-3) \Rightarrow x+3y-9=0 \]NCERT_Solution_Class11_Maths_Exemplar_Ch10_Ex10-3_Q44

  5. Exercise 45

    The points (3,4)\displaystyle (3,4) and (2,−6)\displaystyle (2,-6) are situated on the ____\displaystyle \_\_\_\_ of the line 3x−4y−8=0\displaystyle 3 x-4 y-8=0.

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    NCERT’s answer
    opposite sides
    opposite sides\[f(x,y)=3x-4y-8 \]\[f(3,4)=9-16-8=-15<0 \]\[f(2,-6)=6+24-8=22>0 \]The signs differ, so the points lie on opposite sides of the line.NCERT_Solution_Class11_Maths_Exemplar_Ch10_Ex10-3_Q45

  6. Exercise 46

    A point moves so that square of its distance from the point (3,−2)\displaystyle (3, -2) is numerically equal to its distance from the line 5x−12y=3\displaystyle 5 x-12 y=3. The equation of its locus is ____\displaystyle \_\_\_\_.

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    \(\displaystyle 13x^2+13y^2-83x+64y+172=0\)For a point \(\displaystyle (x,y)\) of the locus:\[(x-3)^2+(y+2)^2=\frac{|5x-12y-3|}{13} \]\[13\left[(x-3)^2+(y+2)^2\right]=\pm(5x-12y-3) \]The sign \(\displaystyle -\) gives \(\displaystyle 13x^2+13y^2-73x+40y+166=0\), with \(\displaystyle r^2=-\tfrac{1703}{676}<0\): no real points. The sign \(\displaystyle +\) gives\[13x^2+13y^2-83x+64y+172=0,\qquad r^2=\frac{2041}{676}>0 \]NCERT_Solution_Class11_Maths_Exemplar_Ch10_Ex10-3_Q46

  7. Exercise 47

    Locus of the mid-points of the portion of the line xsin⁡θ+ycos⁡θ=p\displaystyle x \sin \theta+y \cos \theta=p intercepted between the axes is ____\displaystyle \_\_\_\_.

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    NCERT’s answer
    \(\displaystyle 4 x^2 y^2=p^2\left(x^2+y^2\right)\)
    \(\displaystyle 4x^2y^2=p^2(x^2+y^2)\), that is \(\displaystyle \dfrac1{x^2}+\dfrac1{y^2}=\dfrac4{p^2}\)\[A\left(\frac{p}{\sin\theta},0\right),\qquad B\left(0,\frac{p}{\cos\theta}\right) \]Mid-point \(\displaystyle (h,k)\) of \(\displaystyle AB\):\[h=\frac{p}{2\sin\theta},\qquad k=\frac{p}{2\cos\theta} \]\[\sin\theta=\frac{p}{2h},\qquad \cos\theta=\frac{p}{2k} \]\[\sin^2\theta+\cos^2\theta=1 \Rightarrow \frac{p^2}{4h^2}+\frac{p^2}{4k^2}=1 \]\[p^2(h^2+k^2)=4h^2k^2 \]NCERT_Solution_Class11_Maths_Exemplar_Ch10_Ex10-3_Q47
  8. State whether the statements in Exercises $\displaystyle 48$ to $\displaystyle 56$ are true or false. Justify.

    Exercise 48

    If the vertices of a triangle have integral coordinates, then the triangle can not be equilateral.

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    NCERT’s answer
    True
    True. For integral vertices the squared side is a positive integer:\[a^2=(x_2-x_1)^2+(y_2-y_1)^2 \]An equilateral triangle would have an irrational area:\[\Delta=\frac{\sqrt3}{4}\,a^2 \]but the coordinate formula gives a rational one:\[\Delta=\tfrac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right| \]Contradiction.

  9. Exercise 49

    The points A(−2,1),B(0,5),C(−1,2)\displaystyle \mathrm{A}(-2,1), \mathrm{B}(0,5), \mathrm{C}(-1,2) are collinear.

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    NCERT’s answer
    False
    False. The slopes of \(\displaystyle AB\) and \(\displaystyle AC\) differ, so the points are not collinear.\[m_{AB} = \frac{5-1}{0-(-2)} = 2 \]\[m_{AC} = \frac{2-1}{-1-(-2)} = 1 \]\[m_{AB} \ne m_{AC} \]

  10. Exercise 50

    Equation of the line passing through the point (acos⁡3θ,asin⁡3θ)\displaystyle \left(a \cos ^3 \theta, a \sin ^3 \theta\right) and perpendicular to the line xsec⁡θ+ycosec⁡θ=a\displaystyle x \sec \theta+y \operatorname{cosec} \theta=a is xcos⁡θ−ysin⁡θ=asin⁡2θ\displaystyle x \cos \theta-y \sin \theta=a \sin 2 \theta.

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    NCERT’s answer
    False
    False. The perpendicular through the point has right side \(\displaystyle a\cos 2\theta\), not \(\displaystyle a\sin 2\theta\).Slope of the given line:\[m = -\frac{\sec\theta}{\operatorname{cosec}\theta} = -\tan\theta \]Slope of the perpendicular:\[m' = \cot\theta = \frac{\cos\theta}{\sin\theta} \]Line through \(\displaystyle (a\cos^3\theta,\ a\sin^3\theta)\):\[y - a\sin^3\theta = \frac{\cos\theta}{\sin\theta}\,(x - a\cos^3\theta) \]\[x\cos\theta - y\sin\theta = a(\cos^4\theta - \sin^4\theta) \]\[= a(\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta) = a\cos 2\theta \]