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Straight Lines Test 29

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Straight Lines Test 29
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  • Question 1
    1 / -0
    If area of a triangle is $$35$$ square units with vertices $$\left( {2, - 6} \right),\,\,\left( {5,\,\,4} \right)$$ and $$({k},\,\,4)$$ then $${k}$$ is :
    Solution

  • Question 2
    1 / -0
    Points A & B are in the first quadrant:Point 'o' isthe origin. If the slope of OA is 1, slope of OB is 7 and OA=OB, then the slope of AB is-
    Solution
    Let the point $$A\left(a,b\right)$$ and $$B(c,d)$$ be in first quadrant.
    Slope of $$OA=\dfrac{{y}_{2}-{y}_{2}}{{x}_{2}-{x}_{1}}=\dfrac{b-0}{a-0}=1$$
    $$\Rightarrow\,\dfrac{b}{a}=1$$
    $$\Rightarrow\,b=a$$ or $$a=b$$

    Slope of $$OB=\dfrac{{y}_{2}-{y}_{2}}{{x}_{2}-{x}_{1}}=\dfrac{d-0}{c-0}=7$$
    $$\Rightarrow\,\dfrac{d}{c}=7$$
    $$\Rightarrow\,d=7c$$

    Now,$$OA=OB$$
    $$\sqrt{{\left(a-0\right)}^{2}+{\left(b-0\right)}^{2}}=\sqrt{{\left(c-0\right)}^{2}+{\left(d-0\right)}^{2}}$$
    $${a}^{2}+{b}^{2}={c}^{2}+{d}^{2}$$
    $${a}^{2}+{a}^{2}={c}^{2}+{49c}^{2}$$ since $$a=b,\,d=7c$$
    $$2{a}^{2}=50{c}^{2}$$
    $${a}^{2}=25{c}^{2}$$
    $$\Rightarrow\,a=-5c,5c$$

    Since $$A$$ and $$B$$ are in first quadrant.
    $$a=5c$$
    Slope of $$AB=\dfrac{d-b}{c-a}$$
    $$=\dfrac{7c-a}{c-a}$$ since $$b=a$$

    $$=\dfrac{7c-5c}{c-5c}$$ since $$a=5c$$

    $$=\dfrac{2c}{-4c}=-\dfrac{1}{2}$$

    $$\therefore\,$$Slope of $$AB$$ is $$-\dfrac{1}{2}$$
  • Question 3
    1 / -0
    The area of a triangle with vertices $$A(5,0),B(8,0)$$ and $$C(8,4)$$ in square units is 
    Solution
    Area of Triangle using coordinates of A, B and C
    Area $$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]$$--------(1)
    Here in our case $$ A(5,0) = (x_1, y_1)$$
                                 $$ A(8,0) = (x_2, y_2)$$
                                 $$ A(8,4) = (x_3, y_3)$$

    Substituting above values in Eq - 1, we get
    Area $$ = 0.5 [ 5(0-4) + 8(4-0) + 8(0-0)]$$
    Area $$ = 0.5 [ -20 + 32]$$
    Area $$ = 0.5 \times (12)=6$$
    $$ Area =  6 $$ square units.
  • Question 4
    1 / -0
    Area of the triangle with vertices $$(-2,2), (1,5)$$ and $$(6,-1)$$ is 
    Solution
    Area of triangle having vertices $$(x_1,y_1), (x_2,y_2)$$ and $$(x_3,y_3)$$ is given by
    Area $$ = \dfrac{1}{2} \times [ x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]$$

    $$=\dfrac{1}{2} [-2(5+1) + 1(-1-2)+6(2-5)]$$

    $$=\dfrac{1}{2}[-12-3-18]$$

    $$=-\dfrac{33}{2}$$

    $$\therefore$$ Area $$=\dfrac{33}{2}~sq.\, units$$
  • Question 5
    1 / -0
    The vertices of a $$\triangle ABC$$ are $$A(3,8),B(-4,2)$$ and $$C(5,-1)$$. The area of $$\triangle ABC$$ is
    Solution
    Here $$\left( { x }_{ 1 }=3,{ y }_{ 1 }=8 \right) ;\left( { x }_{ 2 }=-4,{ y }_{ 2 }=2 \right) ;\left( { x }_{ 3 }=5,{ y }_{ 3 }=-1 \right) $$
    $$Ar\left( \triangle ABC \right) =\cfrac { 1 }{ 2 } \left[ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right)  \right] $$
    $$=\cfrac { 1 }{ 2 } \left[ 3(2+1)-4(1-8)+5(8-2) \right] =\cfrac { 1 }{ 2 } (9+36+30)=\cfrac { 75 }{ 2 } =37\cfrac { 1 }{ 2 } $$
  • Question 6
    1 / -0
    If $$A$$ and $$B$$ are two points having co-ordinates $$(3,4)$$ and $$(5,-2)$$ respectively and $$P$$ is a point such that $$PA=PB$$ and area of triangle $$PAB=10$$ square units, then the co-ordinates of $$P$$ are
    Solution

  • Question 7
    1 / -0
    the area of a triangle with vertices $$ (-3, 0),(3,0) $$ and $$ (0, k) $$ is $$ 9 $$ sq units. then the value of $$ k $$ will be 
    Solution
    We know that , area of a triangle with vertices $$ (a_1,y_1),(x_2,y_2) $$ and $$ (x_3,y_3) $$ is given by 
    $$ \Delta  = \frac {1}{2} \left| \begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right|  $$
    $$ \therefore  \Delta = \frac {1}{2} \left| \begin{matrix} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{matrix} \right|  $$
    Expanding along $$ R_1 $$
    $$ 9 = \frac {1}{2} [ -3 ( -k) - 0 +1 ( 3k) ] $$
    $$ \Rightarrow 18 = 3k + 3k = 6 k $$
    $$ \therefore K = \frac {18 }{6} = 3 $$
  • Question 8
    1 / -0
    Let $$PQR$$ be a right angled isosceles traingle, right angled at $$P(2,1).$$ If the equation of the line $$QR$$ is $$2x+y=3$$, then the equation represnting the pair of lines $$PQ$$ and $$PR$$ is
    Solution
    Let $$m$$ be the slope of $$PQ$$ them 
    $$\displaystyle\tan { { 45 }^{o}= } \left| \frac { m-\left( -2 \right)  }{ 1+m\left( -2 \right)  }  \right| $$
    $$\displaystyle\Rightarrow 1=\left| \frac { m+2 }{ 1-2m }  \right| \Rightarrow \pm 1=\frac { m+2 }{ 1-2m } $$ 
    $$\displaystyle\Rightarrow m+2=1-2m $$ or $$-1+2m=m+2$$
    $$\displaystyle m=-\frac { 1 }{ 3 } $$ or $$ m=3$$

    $$PQ$$ markes $$\displaystyle { 45 }^{ o }$$ with $$QR$$ 
    Equation of $$PQ$$ is
    $$\displaystyle y-1=-\frac { 1 }{ 3 } \left( x-2 \right) $$ $$\displaystyle\Rightarrow x+3y-5=0$$  

    Equation of $$PR$$ is 
    $$\displaystyle y-1=3\left( x-2 \right) $$ $$\displaystyle \Rightarrow 3x-y-5=0$$ 

    Combined equation of $$PQ$$ and $$PR$$ is
    $$\displaystyle \left( x+3y-5 \right) \left( 3x-y-5 \right) =0$$
    $$\displaystyle \Rightarrow 3{ x }^{ 2 }-3{ y }^{ 2 }+8xy-20x-10y+25=0$$

  • Question 9
    1 / -0
    If one of the diagonals of a square is along the line $$x=2y$$ and one of its vertices is $$\left(3,0\right)$$, then its sides through this vertex are given by the equations
    Solution
    Diagonal of the square is along $$x-2y=0$$   ...(1)
    The point $$\left( 3,0 \right) $$ does not lie on (1)
    Let the side through this vertex be $$y-0=m\left( x-3 \right) $$   .....(2)

    Angle between side (2) and diagonal (1) is $${ 45 }^{ 0 }$$
    $$\displaystyle \Rightarrow\left| \frac { m-\frac { 1 }{ 2 }  }{ 1+m\left( \frac { 1 }{ 2 }  \right)  }  \right| =\tan {{ 45 }^{ 0 } } $$

    $$\displaystyle\Rightarrow \frac { 2m-1 }{ 2+m } =\pm 1 \Rightarrow m=3,-\displaystyle\frac{1}{3}$$

    $$\displaystyle\therefore $$ From $$(2)$$, the required sides are $$\displaystyle y-3x+9=0$$ and $$\displaystyle3y+x-3=0$$

  • Question 10
    1 / -0
    Two medians drawn from acute angles of a right angled triangle intersect at an angle $$\dfrac\pi6$$. If the length of the hypotenuse of the triangle is $$3$$ units, then area of the triangle (in sq. units) is
    Solution
    $$G\equiv \left(\dfrac{b}{3},\dfrac{a}{3}\right) $$ 

    Slope of line AG $$=\dfrac{a-\dfrac{a}{3}}{0-\dfrac{b}{3}}=\dfrac{-2a}{b}$$

    Slope of MG $$=\dfrac{\dfrac{a}{2}-\dfrac{a}{3}}{0-\dfrac{b}{3}}=\dfrac{-a}{2b}$$

    $$\tan 30^o=\dfrac{\dfrac{-a}{2b}-(\dfrac{-2a}{b})}{1+\dfrac{a^2}{b^2}}$$

    $$\implies \dfrac{1}{\sqrt3}=\dfrac{3ab}{2(a^2+b^2)}$$

    $$\implies \dfrac{1}{2} ab=\dfrac{(a^2+b^2)}{3\sqrt3}$$

    In a right angled triangle $$AC^2+BC^2=AB^2$$ 
    $$\implies a^2+b^2=9$$

    $$\therefore Area =\dfrac{1}{2} ab=\dfrac{9}{3\sqrt3}=\sqrt3$$ sq.units

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