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Straight Lines Test 27

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Straight Lines Test 27
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  • Question 1
    1 / -0
    The coordinates of the foot of perpendicular drawn from the point $$A(1,\ 8,\ 4)$$ to the line joining the points $$B(0,\ -1,\ 3)$$ and $$C(2,\ -3,\ -1)$$. 
    Solution
    Let $$P$$ be the foot of the perpendicular drawn from point $$A$$ on the line joining points $$B$$ and $$C$$.
    Let $$P (a,b,c)$$ be the coordinates of image of point $$A$$.
    Equation of line $$BC$$ is given by:
    $$\dfrac{x-x_{1}}{x_{2}-x_{1}}=\dfrac{y-y_{1}}{y_{2}-y_{1}}=\dfrac{z-z_{1}}{z_{2}-z_{1}}$$
    $$\dfrac{x-0}{2}=\dfrac{y+1}{-2}=\dfrac{z-3}{-4}=\lambda$$
    General coordinates of $$P$$ is $$(2\lambda, -2\lambda-1, -4\lambda+3)$$
    Direction ratios of $$AP(2\lambda+1, -2\lambda-9, -4\lambda-1)$$
    $$AP\bot BC$$
    $$2(2\lambda+1)-2(2\lambda-9)-4(-4\lambda-1)=0$$
    $$4\lambda+2+4\lambda+18+16\lambda+4=0$$
    $$24+24\lambda=0$$
    $$\lambda=-1$$
    $$P(-2, 1,7)$$
    Coordinates of foot of perpendicular is $$(-2, 1, 7)$$
    Coordinates of image of $$A$$ is $$P (a,b,c)$$ is 
    $$\dfrac{a-1}{2}=-2, a=-3$$
    $$\dfrac{b+8}{2}=1, b=-6$$
    $$\dfrac{c+4}{2}=7, c=10$$
    $$P(-3, -6, 10)$$




  • Question 2
    1 / -0
    Two points A$$\left ( x_{1},y_{1} \right )andB\left ( x_{2},y_{2} \right )$$ are chosen on the graph of $$f(x)=inx with 0<x_{1}<x_{2}.$$ the point C and D trisect line segment AB with AC<CB. through Ca horizontal line is drawn to cut the curve at$$E\left ( x_{3},y_{3} \right ).if x_{1}=1000$$then the value of$$x_{3}$$ equals
    Solution

    We have,

    Since, $$C$$ trisects $$AB$$

    Now,

    $$ {{x}_{1}}=1,\,\,{{y}_{1}}=\ln 1=0 $$

    $$ {{x}_{2}}=1000,\,\,{{y}_{2}}=\ln 1000=3 $$

    Now, Ratio$${{m}_{1}}:{{m}_{2}}=1:2$$

    $$ a=\dfrac{{{m}_{1}}{{x}_{2}}+{{m}_{2}}{{x}_{1}}}{{{m}_{1}}+{{m}_{2}}} $$

    $$ =\dfrac{1\times 1000+2\times 1}{1+2} $$

    $$ =\dfrac{1002}{3} $$

    $$ a=334 $$

    $$ b=\dfrac{{{m}_{1}}{{y}_{2}}+{{m}_{2}}{{y}_{1}}}{{{m}_{1}}+{{m}_{2}}} $$

    $$ =\dfrac{1\times \ln 1000+2\times 0}{1+2} $$

    $$ b=\dfrac{\ln 1000}{3}\,\,\,\,.......\,\,\left( 1 \right) $$

    Given the curve

    $$y=\ln x\,\,\,......\,\,\left( 2 \right)$$

    From equation (1) and (2) to,

    $$ \ln x=\dfrac{\ln 1000}{3} $$

    $$ 3\ln x=\ln 1000 $$

    $$ \ln x={{\left( \ln 1000 \right)}^{\dfrac{1}{3}}} $$

    $$ x={{\left( 1000 \right)}^{\dfrac{1}{3}}} $$

    $$ x=10 $$

    Hence,$${{x}_{3}}=10$$

    This is the answer.
  • Question 3
    1 / -0
    Find the relation between $$x$$ and $$y$$ such that the point $$P(x, y)$$ is equidistant from the points $$A(1, 4)$$ and $$B(-1, 2)$$.
    Solution
    Let $$P\left(x,y\right)$$ be equidistance from the points $$A\left(1,4\right)$$ and $$B\left(−1,2\right)$$
    Given:$$AP=BP$$
    $${AP}^{2}={BP}^{2}$$
    By distance formula,
    $${\left(x-1\right)}^{2}+{\left(y-4\right)}^{2}={\left(x+1\right)}^{2}+{\left(y-2\right)}^{2}$$
    $$\Rightarrow\,{x}^{2}-2x+1+{y}^{2}-8y+16={x}^{2}+2x+1+{y}^{2}-4y+4$$
    $$\Rightarrow\,-2x+1-8y+16-2x-1+4y-4=0$$
    $$\Rightarrow\,-4x-4y+12=0$$
    $$\Rightarrow\,x+y-3=0$$ gives the relation between $$x$$ and $$y$$.
  • Question 4
    1 / -0
    If the line $$\left( \cfrac { x }{ 2 } +\cfrac { y }{ 3 } -1 \right) +\lambda(2x+y-1)=0$$ is parallel to x-axis then $$\lambda=$$
    Solution
    $$\left( \dfrac{x}{2}+ \dfrac{y}{3}-1 \right)+ \lambda (2x+y-1)=0$$
    $$(3x+2y-6)+ 6 \lambda (2x+y-1)=0$$
    $$(3+12 \lambda)x+ (2+6\lambda)y- (6+6\lambda)=0$$
    $$m=0$$
    $$-\left( \dfrac{3+12 \lambda}{2+6 \lambda} \right)=0$$
    $$3+12 \lambda =0$$
    $$\lambda=\dfrac{-3}{12}= \dfrac{-1}{4}$$
  • Question 5
    1 / -0
    The length of the median from the vertex $$A$$ of a triangle whose vertices are $$A ( - 1,3 ) , B ( 1 , - 1 )$$ and $$C ( 5,1 )$$ is
    Solution
    Midpoint of $$BC$$ is $$D\equiv\left(\dfrac{1+5}{2},\,\dfrac{-1+1}{2}\right)\equiv\left(3,0\right)$$
    Length of the median from the vertex $$A$$ is
    $$AD=\sqrt{{\left(3+1\right)}^{2}+{\left(0-3\right)}^{2}}=\sqrt{16+9}=\sqrt{25}=5$$ units.

  • Question 6
    1 / -0
    The midpoints of the sides of a triangle are $$\left(1,1\right),\left(4,3\right)$$ and $$\left(3,5\right)$$. The area of the triangle is ___ square units.
    Solution

  • Question 7
    1 / -0
    The angle between the lines $$x\cos{30}^{o}+y\sin{30}^{o}=3$$
    $$x\cos{60}^{o}+y\sin{60}^{o}=5$$ is
    Solution
    $$ \displaystyle x cos30^{\circ}+y sin 30^{\circ} = 3$$
    $$ y = -xcot30^{\circ}+3cosec 30^{\circ}$$
    $$ m_{1} = -cot 30^{\circ} = -\sqrt{3}$$
    $$ x cos60^{\circ}+y sin60^{\circ} = 5 $$
    $$ y = -xcot60^{\circ}+5cosec60^{\circ}$$
    $$ m_{2} = -cot60^{\circ} = -1/\sqrt{3}$$
    $$\displaystyle  tan \theta  = \left | \dfrac{m_{1}-m_{2}}{1+m_{1}m_{2}} \right |$$
    $$ = \left | \dfrac{-3+1/\sqrt{3}}{1+(-1/\sqrt{3})(-1/\sqrt{3})} \right |$$
    $$ = \displaystyle \left | \dfrac{-3+1}{\sqrt{3}(1+1)} \right | = \left | \dfrac{1}{\sqrt{3}} \right |$$
    $$ \theta  = 30^{\circ}$$

  • Question 8
    1 / -0
    The points$$ (0,1), (-2,3), (6,7), (8,3)$$ form 
    Solution
    $$(0, 1)$$    $$(-2, 3)$$     $$(6, 7)$$      $$(5, 3)$$
      $$A$$          $$B$$          $$C$$             $$D$$
    $$AB=\sqrt{(-2)^{2}+(-3-1)^{2}}=2\sqrt{2}$$
    $$BC=\sqrt{(6+2)^{2}+(7-3)^{2}}=4\sqrt{5}$$
    $$CD=\sqrt{2^{2}+4^{2}}=2\sqrt{5}$$
    $$DA=\sqrt{8^{2}+2^{2}}=2\sqrt{17}$$
    $$AC=\sqrt{6^{2}+6^{2}}=2\sqrt{6}$$
    $$BD=\sqrt{10^{2}}=10$$
    Neither of the sides are equal. then this is a general quadrilateral
  • Question 9
    1 / -0
    If $$ B (1, 3) $$ is equidistant from $$ A (6,  -1) $$ and $$ C (x, 8) $$, then  $$ x =$$
    Solution
    $$\textbf{Step 1: Apply distance formula between the points.}$$

                    $$\text{Given that, B(1,3) is equidistant from the points }\mathrm{A(6,-1)\text{ and }C(x,8).}$$

                    $$\text{We know, distance between }\mathrm{(x_1,y_1)\ and\ (x_2,y_2)=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$

                    $$\text{Distance between point B and point A,}$$
                    $$\Rightarrow\mathrm{BA=}\sqrt{(6-1)^2+(-1-3)^2}$$
                                 $$=\sqrt{5^2+(-4)^2}$$
                                 $$=\sqrt{25+16}$$
                                 $$=\sqrt{41}.......(1)$$

                    $$\text{Distance between point B and point C,}$$
                    $$\Rightarrow\mathrm{BC=}\sqrt{(x-1)^2+(8-3)^2}$$
                                 $$=\sqrt{x^2+1-2x+25}$$
                                 $$=\sqrt{x^2-2x+26}......(2)$$

    $$\textbf{Step 2: Apply suitable condition.}$$

                    $$\text{Since, B(1,3) is equidistant from the points }\mathrm{A(6,-1)\text{ and }C(x,8).}$$

                    $$\mathrm{\therefore BC=BA}$$

                    $$\mathrm{\Rightarrow BC^2=BA^2}$$

                    $$\Rightarrow x^2-2x+26=41$$               $$\textbf{[From (1) and (2)]}$$

                    $$\Rightarrow x^2-2x-15=0$$   

                    $$\Rightarrow x^2-5x+3x-15=0$$ 

                    $$\Rightarrow (x-5)(x+3)=0$$       

                    $$\Rightarrow x=5\text{ or }x=-3$$

    $$\textbf{Hence, the correct option is B.}$$
  • Question 10
    1 / -0
    If $$x+4y=7$$, where $$y\ \in\ N$$, then the minimum positive value of $$x+y$$ equals
    Solution

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