Tag: distance formula in 2d

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

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 

  1. $2$
  2. $0$
  3. $3$
  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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

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

  1. $h^2=ab$
  2. $h+a=0$
  3. $a=0$
  4. $h=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
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
$\dfrac {-2h}{b} = 0\Rightarrow h = 0$.

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

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

  1. $(45^{\circ},60^{\circ})$
  2. $(30^{\circ},45^{\circ})$
  3. $(15^{\circ},30^{\circ})$
  4. $(0^{\circ},15^{\circ})$
Reveal answer Fill a bubble to check yourself
D Correct answer
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  $m _{2}=\dfrac{1}{4}$
$\tan\theta=\left | \dfrac{m _{1}-m _{2}}{1+m _{1}m _{2}} \right |$
$\tan\theta=\left | \dfrac{\dfrac{1}{3}-\dfrac{1}{4}}{1+\dfrac{1}{3}\dfrac{1}{4}} \right |$
$\tan\theta=\dfrac{1}{13}$
$\therefore \theta \epsilon (0^0,15^0)$
Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

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

  1. $X^{2}-Y^{2}=3a^{2}$
  2. $X^{2}-Y^{2}=a$
  3. $X^{2}-Y^{2}=2a^{2}$
  4. $none of these$
Reveal answer Fill a bubble to check yourself
D Correct answer
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. $

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

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

  1. $2fgh={ bg }^{ 2 }+{ ch }^{ 2 }$
  2. ${ bg }^{ 2 }\neq { ch }^{ 2 }$
  3. $abc=2fgh$
  4. $2fgh=af+{ ch }^{ 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the general second-degree equation to represent lines intersecting on the y-axis, the intersection point (0, y0) must satisfy the equation. Setting x=0 gives by^2 + 2fy + c = 0. The condition for the lines to intersect on the y-axis is derived from the general condition for intersection and the specific coordinate constraints.

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

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

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.