Tag: distances and midpoints

Questions Related to distances and midpoints

The product of perpendiculars drawn from the point $(1,2)$ to the pair of lines $x^{2}+4xy+y^{2}=0$ is

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $\dfrac {9}{4}$

  2. $\dfrac {3}{4}$

  3. $\dfrac {9}{16}$

  4. $none\ of\ these$

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Correct Option: A

If the equation ${ ax }^{ 2 }-6xy+{ y }^{ 2 }+2gx+2fy+c=0$ represents  pair of lines whose slopes are m and ${ m }^{ 2 }$, then sum of all possible values of a is

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. 17

  2. -|19

  3. 19

  4. -17

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Correct Option: A

The angle between the pair of straight lines represented by the equation 
$x^{2}+\lambda xy+2y^{2}+3x-5y+2=0$, is $\tan^{-1}\left(\dfrac{1}{3}\right)$ where $'\lambda'$ is a non-negative real number then $\lambda$ is 

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $2$

  2. $0$

  3. $3$

  4. $1$

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Correct Option: C
Explanation:

Given equation of pair of straight lines be $x^2+\lambda x{y}+2{y^2}+3{x}-5{y}+2=0$

$\implies a=1,b=2,h=\dfrac{\lambda}{2}$
$\text{tan}^{-1} \bigg(2\dfrac{\sqrt{h^2-a{b}}}{a+b}\bigg)=\text{tan}^{-1}\bigg(\dfrac{1}{3}\bigg)$
$\dfrac{\lambda^2}{4}-2=\dfrac{1}{4}\implies \lambda= 3$

For the pair of lines represented by $ax^{2}+2hxy+by^{2}=0$ to be equally inclined to coordinates axes we have, 

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $h^2=ab$

  2. $h+a=0$

  3. $a=0$

  4. $h=0$

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Correct Option: D
Explanation:

$ax^{2} + 2hxy + by^{2} = 0$

Let the lines
$b(y - m _{1}x) y - m _{2}x) = ax^{2} + 2hxy + by^{2}$
$m _{1} + m _{2} = \dfrac {-2h}{b}$
and $m _{1}m _{2} = \dfrac {a}{b}$
If $m _{1} = m _{2}$ for equally inclined so
$\dfrac {-2h}{b} = 0\Rightarrow h = 0$.

Consider a general equation of degree $2$, as $\lambda x^{2}-10xy+12y^{2}+5x-16y-3=0$ For the value of $\lambda$ obtained for the given equation to be a pair of straight lines, if $\theta$ is the acute angle between $L _{1}=0$ and $L _{2}=0$ then $\theta$ lies in the interval

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $(45^{\circ},60^{\circ})$

  2. $(30^{\circ},45^{\circ})$

  3. $(15^{\circ},30^{\circ})$

  4. $(0^{\circ},15^{\circ})$

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Correct Option: D
Explanation:
Given pair of line 
$\lambda x^2-10xy+12y^2+5x-16y-3=0$
on comparing above eq with general form of pair of eq we get
$a=\lambda,h-5,b=12,g=\dfrac{5}{2},f=-8,c=-3$
Given eq is pair of eq so 
$abc+2fgh-af^2-bg^2-ch^2=0$
$\lambda \times 12\times (-3)+2(-8)\left ( \dfrac{5}{2} \right )\left ( -5 \right )-\lambda\times 64-12\left ( \dfrac{25}{4} \right )-(-3)(25)=0$
$-36\lambda +200-64\lambda-75+75=0$
$-100\lambda +200=0$
$\lambda=2$
eq of pair becomes 
$2x^2-10xy+12y^2+5x-16y-3=0$
$2x^2-(10y-5)x+(12y^2-16y-3)=0$
$x=\dfrac{10y-5\pm \sqrt{(10y-5)^2-8(12y^2-16y-3)}}{4}$
$4x=10y-5\pm \sqrt{100y^2+25-100y-96y^2+128y+24)}$
$4x=10y-5\pm \sqrt{4y^2+28y+49)}$
$4x=10y-5\pm \sqrt{(2y+7)^2}$
$4x=10y-5\pm (2y+7)$
$4x=10y-5+ 2y+7$ or $4x=10y-5- (2y+7)$
$4x-12y-2=0$ or $4x-8y+12=0$
$2x-6y-1=0$ or $2x-4y+6=0$
$L _{1} : 2x-6y-1=0$
$L _{2} : 2x-8y-6=0$
Slope of line $L _{1},L _{2}$ $m _{1}=\dfrac{1}{3}$ and  $m _{2}=\dfrac{1}{4}$
$\tan\theta=\left | \dfrac{m _{1}-m _{2}}{1+m _{1}m _{2}} \right |$
$\tan\theta=\left | \dfrac{\dfrac{1}{3}-\dfrac{1}{4}}{1+\dfrac{1}{3}\dfrac{1}{4}} \right |$
$\tan\theta=\dfrac{1}{13}$
$\therefore \theta \epsilon (0^0,15^0)$

By rotating the coordinates axes through $30^{o}$ in anticlockwise sense the equation $x^{2}+2\sqrt{3}xy-y^{2}=2a^{2}$ changes to

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $X^{2}-Y^{2}=3a^{2}$

  2. $X^{2}-Y^{2}=a$

  3. $X^{2}-Y^{2}=2a^{2}$

  4. $none of these$

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Correct Option: D
Explanation:

$ x=x'\cos  \theta -y'\sin  \theta =x'\left( { \dfrac { { \sqrt { 3 }  } }{ 2 }  } \right) -\frac { { y' } }{ 2 }  \ y=x'\sin  \theta +y'\cos  \theta =x'\left( { \dfrac { 1 }{ 2 }  } \right) +y'\left( { \dfrac { { \sqrt { 3 }  } }{ 2 }  } \right)  \ { x^{ 2 } }+2\sqrt { 3 } xy-{ y^{ 2 } }=2{ a^{ 2 } } \ \dfrac { { { { \left[ { \sqrt { 3 } x'-2y' } \right]  }^{ 2 } } } }{ 4 } -\dfrac { { { { \left[ { x'-\sqrt { 3 } y' } \right]  }^{ 2 } } } }{ 4 } +2\sqrt { 3 } \dfrac { { \left[ { \sqrt { 3 } x'-y' } \right]  } }{ 2 } \dfrac { { { { \left[ { x'-\sqrt { 3 } y' } \right]  }^{ 2 } } } }{ 2 } =2{ a^{ 2 } } \ \dfrac { { 2x{ '^{ 2 } }-2y{ '^{ 2 } } } }{ 4 } -\sqrt { 3 } x'y'+\dfrac { { \sqrt { 3 }  } }{ 2 } \left[ { \sqrt { 3 } x'-y' } \right] \left[ { x'+\sqrt { 3 } y } \right] =2{ a^{ 2 } } \ -\sqrt { 3 } x'y'+\dfrac { { \sqrt { 3 }  } }{ 2 } \left[ { \sqrt { 3 } x{ '^{ 2 } }-\sqrt { 3 } y{ '^{ 2 } }+2x'y' } \right] =2{ a^{ 2 } } \ 2x{ '^{ 2 } }-2y{ '^{ 2 } }=2{ a^{ 2 } } \ x{ '^{ 2 } }-2y{ '^{ 2 } }=2{ a^{ 2 } } $


$ Hence,\, the\, \, option\, \, D\, is\, the\, correct\, answer. $

The equation of pair of lines joining origin to the points of intersection of $x^{2}+y^{2}=9$ and $x+y=3$ is

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $x^{2}+(3-x)^{2}=9$

  2. $xy=0$

  3. $(3+y)^{2}+y^{2}=9$

  4. $(x-y)^{2}=9$

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Correct Option: A

The equation $x^{3}+y^{3}=0$ represents

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. three real straight lines

  2. three points

  3. the combined equation of a pair of straight lines

  4. none of these

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Correct Option: D
Explanation:

$x^{3}+y^{3}=(x+y)(x^{2}-xy+y^{2})$
$(x^{2}-xy+y^{2})$ does not factorize so this equation is not combined equation of line.

If the pair of lines ${ ax }^{ 2 }+2hxy+{ by }^{ 2 }+2gx+2fy+c=0$ intersect on the y-axis, then

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $2fgh={ bg }^{ 2 }+{ ch }^{ 2 }$

  2. ${ bg }^{ 2 }\neq { ch }^{ 2 }$

  3. $abc=2fgh$

  4. $2fgh=af+{ ch }^{ 2 }$

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Correct Option: A

The equation ${ x }^{ 2 }{ y }^{ 2 }-2x{ y }^{ 2 }-3{ y }^{ 2 }-4{ x }^{ 2 }y+8xy+12y=0$ represents 

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. a pair of straight lines

  2. a pair of straight lines and a circle

  3. a pair of straight lines and a parabola

  4. a set of four lines forming a square

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Correct Option: D
Explanation:

Given equation is 

${ x }^{ 2 }{ y }^{ 2 }-{ 2xy }^{ 2 }-{ 3y }^{ 2 }-{ 4x }^{ 2 }y+8xy+12y=0$
$\Rightarrow { y }^{ 2 }\left( { x }^{ 2 }-2x-3 \right) -4y\left( { x }^{ 2 }-2x-3 \right) =0$
$\Rightarrow { y }\left( { y }-4 \right) \left( { x }-3 \right) \left( x+1 \right) =0$
$\Rightarrow y=0,y=4,x=3,x=-1$
Hence, the equation represents four straight lines which evidently form a square.