Tag: two dimensional analytical geometry

Questions Related to two dimensional analytical geometry

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

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

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: B
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}$

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

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

  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

Show answer
Correct Option: A

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 

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

  1. Square

  2. Rhombus

  3. Rectangle

  4. None of these

Show answer
Correct Option: A
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

$3x^2+8xy-3y^2=0$ 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,

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

  1. $x+3y+1=0$

  2. $x+3y-1=0$

  3. $x-3y+1=0$

  4. $x+y+1=0$

Show answer
Correct Option: A

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

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

  1. $k=\dfrac{91}{48}$

  2. $k=\dfrac{94}{43}$

  3. $k=\dfrac{83}{23}$

  4. None of these

Show answer
Correct Option: A
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 $abc+2fgh-af^2-bg^2-ch^2=0$
$\Rightarrow$  $12\times (-10)\times k+2\times (\dfrac{-5}{2})\times\dfrac{11}{2}\times 0-12\times (\dfrac{-5}{2})^2-(-10)\times (\dfrac{11}{2})^2-k\times (0)^2=0$
$\Rightarrow$  $-120k+0-75+\dfrac{605}{2}-0=0$
$\Rightarrow$  $\dfrac{-240k-150+605}{2}=0$
$\Rightarrow$  $-240k+455=0$
$\Rightarrow$  $k=\dfrac{455}{240}$
$\therefore$   $k=\dfrac{91}{48}$

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

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

  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}}\,\, $

Show answer
Correct Option: C
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.

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

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

  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. ${x^2}\, + \,2xy\, - 3\,{y^2}\, = \,0$

Show answer
Correct Option: B

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

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

  1. True

  2. False

Show answer
Correct Option: A
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
$\dfrac{{x}^{2}-{y}^{2}}{xy}=\dfrac{1}{3}=\dfrac{4-b}{9}$
$\Rightarrow\,\dfrac{1}{3}=\dfrac{4-b}{9}$
$\Rightarrow\,4-b=\dfrac{9}{3}=3$
$\Rightarrow\,-b=3-4$
$\Rightarrow\,b=1$
$\therefore\,b=1$

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 $y+2x=5a$

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

  1. True

  2. False

Show answer
Correct Option: A

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?

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

  1. $36/13$

  2. $12/17$

  3. $13/5$

  4. $17/3$

Show answer
Correct Option: A
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 = $\cfrac{1}{2} \times \cfrac{12\sqrt{13}}{13} \times \cfrac{6}{\sqrt{13}} = \cfrac{36}{13}$