Tag: two dimensional analytical geometry

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Questions Related to two dimensional analytical geometry

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

The difference of the slopes of the lines $x ^ { 2 } \left( \sec ^ { 2 } \theta - \sin ^ { 2 } \theta \right) - ( 2 \tan \theta ) x y + y ^ { 2 } \sin ^ { 2 } \theta = 0$

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

According to the question..........

$\begin{array}{l} Let,\, { m _{ 1\,  } }& \, { m _{ 2 } } \ sum\, of\, the\, slope:\, { m _{ 1 } }+{ m _{ 2 } }=\dfrac { { -2h } }{ b } ----(i) \ and,\,  \ product\, of\, slope:{ m _{ 1 } }.\, { m _{ 2 } }=\dfrac { a }{ b } -----(ii) \ Here, \ a{ x^{ 2 } }+2hxy+b{ y^{ 2 } }=0.........(general\, equ\, of\, straight\, line.) \ cofficient\, of: \ a={ \sec ^{ 2 }  }\theta -{ \sin ^{ 2 }  }\theta  \ h=-\tan  \theta  \ b={ \sin ^{ 2 }  }\theta  \ Now,\, value\, put\, { { into } } \ sum\, of\, the\, slope:\, { m _{ 1 } }+{ m _{ 2 } }=\dfrac { { -2h } }{ b } ----(i) \ \Rightarrow { m _{ 1 } }+{ m _{ 2 } }=\dfrac { { -2(-tan\theta ) } }{ { { { \sin   }^{ 2 } }\theta  } } =\dfrac { { 2\sin  \theta \times 2 } }{ { 2{ { \sin   }^{ 2 } }\theta \, .\, \cos  \theta  } } =\dfrac { 4 }{ { 2sin\theta \cos  \theta  } } =\dfrac { 4 }{ { \sin  2\theta  } }  \ and, \ product\, of\, slope:{ m _{ 1 } }+{ m _{ 2 } }=\dfrac { a }{ b } -----(ii) \ \Rightarrow { m _{ 1 } }.\, { m _{ 2 } }=\dfrac { { { { \sec   }^{ 2 } }\theta -{ { \sin   }^{ 2 } }\theta  } }{ { { { \sin   }^{ 2 } }\theta  } } =\dfrac { 1 }{ { { { \sin   }^{ 2 } }\theta \, .\, { { \cos   }^{ 2 } }\theta  } } -1\, \, \, \, \, \, \, \, \left[ { divide\, by\, 4 } \right.  \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\dfrac { 4 }{ { 4{ { \sin   }^{ 2 } }\theta \, .\, { { \cos   }^{ 2 } }\theta  } } -1\, \, =\dfrac { 4 }{ { { { (\sin  2\theta ) }^{ 2 } }\,  } } -1\,  \ \, \, \, Now,find\, difference: \ \, \, \, \, \, \, \, \, \, { ({ m _{ 1 } }-{ m _{ 2 } })^{ 2 } }={ ({ m _{ 1 } }+{ m _{ 2 } })^{ 2 } }-4{ m _{ 1 } }.\, { m _{ 2 } } \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ={ \left( { \dfrac { 4 }{ { \sin  2\theta  } }  } \right) ^{ 2 } }-4\left( { \dfrac { 4 }{ { ({ { \sin   }^{ 2 } }2\theta )\,  } } -1\,  } \right)  \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\dfrac { { 16 } }{ { ({ { \sin   }^{ 2 } }2\theta )\,  } } -\, \dfrac { { 16 } }{ { ({ { \sin   }^{ 2 } }2\theta )\,  } } +4 \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \Rightarrow \, \, { ({ m _{ 1 } }-{ m _{ 2 } })^{ 2 } }\, \, \, =4 \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \Rightarrow \, ({ m _{ 1 } }-{ m _{ 2 } })=+\sqrt { 4 } =2 \ \, \, \, \therefore \, \, \, the\, \, differece\, of\, slope\, \, is\, 2. \ So,\, that\, the\, correct\, option\, is\, B.\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,  \end{array}$

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

Curves $a{ x }^{ 2 }+2hxy-2gx-2fy+c=0$ and $a'{ x }^{ 2 }-2hxy+(a'+a-b){ y }^{ 2 }-2g'x-2f'y+c=0\quad $ intersect at four concyclic points $A,B,C$ and $D$. If $P$ is the point $\left( \cfrac { g'+g }{ a'+a } ,\cfrac { f'+f }{ a'+a }  \right) $, then which of the following is/are true

  1. $P$ is also concyclic with points $A,B,C,D$
  2. $PA,PB,PC$ in G.P
  3. ${ PA }^{ 2 }+{ PB }^{ 2 }+{ PC }^{ 2 }=3{ PD }^{ 2 }\quad $
  4. $PA,PB,PC$ in AP
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The intersection of two conics at four concyclic points is a standard result in geometry. The point P defined by the given ratios is the center of the circle passing through these four points.

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

The four straight lines given by the equations $12x^2+7xy-12y^2=0$ and $12x^2+7xy-12y^2-x+7y-1=0$ lie along the sides of a 

  1. Square

  2. Rhombus

  3. Rectangle

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

taking $y$ is constant and finding the value of$x$ by roots formula.


$12x^2+(7y)x-12y^2=0$

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}=\dfrac{(7y)-\sqrt{(7y)^2-4(-12)y^212}}{2\times 12}$

$\dfrac{\Rightarrow x=-7y\pm\sqrt{49y^2+576y^2}{}}{2\times 12}$

$\Rightarrow 24x=-7y\pm \sqrt{625}y$

$\Rightarrow 24xy=-7y\pm 25y$.........(1)

Again $12x^2+(7y-1)x+(-12y^2-1+7y)=0$

$\therefore x=\dfrac{-(7y-1)\pm\sqrt{(7y-1)^2-4}(12)(7y-1-12y^2)}{24}$

$\Rightarrow x=\dfrac{1-7y\pm\sqrt{49y^2+1-14y-336y+48+576y^2}}{24}$

$\Rightarrow 24x=1-7y\pm\sqrt{625y^2-350y+49}$

$\Rightarrow 24x=1-7y\pm (25y-7)$

$\Rightarrow x=\dfrac{1-7y\pm 25y-7}{24}$........(ii)

from (i) and (ii) we can clearly see co.efficient of $x$ and $y$ are same so slope are sample $m _1=\dfrac{24}{18},m _2=-\dfrac{24}{32}$

So $m _1m _2= -1$

$\therefore $ it is a square

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

$3x^2+8xy-3y^2=0$ represents a pair of lines AB and BC, whereas the equation $3x^2+8xy-3y^2+2x=4y-1=0$ represents two lines CD and DA.
Answer the given question.
The equation of the CD is,

  1. $x+3y+1=0$
  2. $x+3y-1=0$
  3. $x-3y+1=0$
  4. $x+y+1=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation 3x^2 + 8xy - 3y^2 + 2x - 4y - 1 = 0 represents a pair of lines. By factoring this quadratic, we can identify the individual linear equations for the sides CD and DA.

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

The value of $k$ so that the equation $12{x}^{2}-10{y}^{2}+11x-5y+k=0$ may represent a pair of straight lines is

  1. $k=\dfrac{91}{48}$
  2. $k=\dfrac{94}{43}$
  3. $k=\dfrac{83}{23}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\Rightarrow$  The given equation is $12x^2-10y^2+11x-5y+k=0$

$\Rightarrow$  Comparing it with $ax^2+2hxy+by^2+2gx+2fy+c=0$
$\Rightarrow$  $a=12,\,b=-10,\,h=o,\,g=\dfrac{11}{2},\,f=\dfrac{-5}{2},\,c=k$
$\Rightarrow$  The condition is $abc+2fgh-af^2-bg^2-ch^2=0$
$\Rightarrow$  $12\times (-10)\times k+2\times (\dfrac{-5}{2})\times\dfrac{11}{2}\times 0-12\times (\dfrac{-5}{2})^2-(-10)\times (\dfrac{11}{2})^2-k\times (0)^2=0$
$\Rightarrow$  $-120k+0-75+\dfrac{605}{2}-0=0$
$\Rightarrow$  $\dfrac{-240k-150+605}{2}=0$
$\Rightarrow$  $-240k+455=0$
$\Rightarrow$  $k=\dfrac{455}{240}$
$\therefore$   $k=\dfrac{91}{48}$

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

$\begin{array}{l}\\left( {3x - 2y} \right)\left( {2x + y} \right)=\end{array}$

  1. $ =6{{x}^{2}}-xy+2{{y}^{2}}\,\, $
  2. $ =6{{x}^{2}}+xy-2{{y}^{2}}\,\, $
  3. $ =6{{x}^{2}}-xy-2{{y}^{2}}\,\, $
  4. $ =6{{x}^{2}}+xy+2{{y}^{2}}\,\, $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Part (1)

$ \left( a-2 \right)\left( a+2 \right) $

${{a}^{2}}-4$ 

Part (2)

$ \left( 3x-2y \right)\left( 2x+y \right) $

$ =6{{x}^{2}}+3xy-4xy-2{{y}^{2}} $

$ =6{{x}^{2}}-xy-2{{y}^{2}}\,\, $

Hence, this is the answer.

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

If the pair of lines ${x^2}\, + \,2xy\, + \,a{y^2}\, = \,0$ and $a{x^2}\, + \,2xy\, + \,{y^2}\, = \,0$ have exactly one line in common, then joint equation of the other two lines is given by

  1. $3{x^2}\, + \,8xy\, - 3\,{y^2}\, = \,0$
  2. $3{x^2}\, + \,10xy\, + 3\,{y^2}\, = \,0$
  3. ${y^2}\, + \,2xy\, - 3\,{x^2}\, = \,0$
  4. ${x^2}\, + \,2xy\, - 3\,{y^2}\, = \,0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given the two pairs of lines, we find the common line by solving the intersection of the quadratic forms. The joint equation of the remaining two lines is found by dividing the combined equation by the common factor.

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

The lines $2x^2+6xy+y^2=0$ are equally inclined to the lines $4x^2+18xy+by^2=0$ when $b=1$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Consider the equation $2{x}^{2}+6xy+{y}^{2}=0$
Equation of angle bisector is 
$\dfrac{{x}^{2}-{y}^{2}}{2-1}=\dfrac{xy}{3}$      
$\dfrac{{x}^{2}-{y}^{2}}{1}=\dfrac{xy}{3}$
$\dfrac{{x}^{2}-{y}^{2}}{xy}=\dfrac{1}{3}$         .......$(1)$
Consider the equation $4{x}^{2}+18xy+b{y}^{2}=0$
Equation of angle bisector is 
$\dfrac{{x}^{2}-{y}^{2}}{4-b}=\dfrac{xy}{9}$ 
$\dfrac{{x}^{2}-{y}^{2}}{xy}=\dfrac{4-b}{9}$      .......$(2)$
From $(1)$  and $(2)$ we have
$\dfrac{{x}^{2}-{y}^{2}}{xy}=\dfrac{1}{3}=\dfrac{4-b}{9}$
$\Rightarrow\,\dfrac{1}{3}=\dfrac{4-b}{9}$
$\Rightarrow\,4-b=\dfrac{9}{3}=3$
$\Rightarrow\,-b=3-4$
$\Rightarrow\,b=1$
$\therefore\,b=1$

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

Find the equations of the two straight lines drawn through the point $(0,a)$ on which the perpendicular let fall from the point $(2a,2a)$ are each of length $a$.
 then equation of the straight line joining the feet of these perpendiculars is $y+2x=5a$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

By calculating the lines passing through (0, a) and applying the condition that the perpendicular distance from (2a, 2a) is 'a', we can verify the equation of the line joining the feet of the perpendiculars.

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

If a pair of perpendicular straight lines drawn through the origin forms an isosceles triangle with the line $2x+3y=6$, then area of the triangle so formed is?

  1. $36/13$
  2. $12/17$
  3. $13/5$
  4. $17/3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As the lines are perpendicular and form an isosceles triangle the other two angles must be $45^\circ$

Let the slope of the line be m
$tan 45^\circ = \Bigg|\cfrac{-\cfrac{2}{3}-m}{1-\cfrac{2m}{3}}\Bigg| = 1$
$m = -5$ and the other line slope =$\cfrac{1}{5}$
Lines 
$y +5x= 0$ and $5y =x$
Intersection points $(0,0)$ , $(-\cfrac{6}{13} , \cfrac{30}{13})$ and $(\cfrac{30}{13} , \cfrac{6}{13})$
Perpendicular distance from origin to line $2x+3y=6$ is $\cfrac{|0+0-6|}{\sqrt{2^2+3^2}} = \cfrac{6}{\sqrt{13}}$
Distance between the points are $\sqrt{\Bigg(\cfrac{36}{13}\Bigg)^2+\Bigg(\cfrac{24}{13}\Bigg)^2} = \cfrac{\sqrt{1872}}{13} = \cfrac{12\sqrt{13}}{13}$
Area = $\cfrac{1}{2} \times \cfrac{12\sqrt{13}}{13} \times \cfrac{6}{\sqrt{13}} = \cfrac{36}{13}$