Tag: pair of bisectors of angles

Questions Related to pair of bisectors of angles

$\displaystyle ax^{2}+2hxy+by^{2}=0$ represents a pair of straight lines through origin & angle between them is given by
$\displaystyle \tan \theta=\frac{2\sqrt{h^{2}-ab}}{a+b}$. If the lines are perpendicular then $\displaystyle a+b=0 $ and the equation of bisectors is given by  $\displaystyle \frac{x^{2}-y^{2}}{a-b}=\frac{xy}{h}$
The general equation of second degree given by
$\displaystyle ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ represent a pair of straight lines if $\displaystyle \triangle =0 $ or 
$ \displaystyle \begin{vmatrix}a&h  &g \\ h&b  &f \\ g&f  &c \end{vmatrix}=0 $ or $\displaystyle abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$
On the basis of above information answer the following question

Let $\displaystyle  f _{1}\left (x,y  \right )=ax^{2}+2hxy+by^{2}=0$ and let $\displaystyle  f _{i+1}\left (x,y  \right )=0 $ denotes the equation of bisectors of $\displaystyle  f _{i}\left (x,y  \right )=0 \forall $ $ i=1,2,3 $ then equation of $\displaystyle  f _{3}\left (x,y  \right )=0$ is

maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

  1. $\displaystyle \left (a-b \right )x^{2}-4hxy+\left (a-b \right )y^{2}=0 $

  2. $\displaystyle \left (a-b \right )x^{2}-4hxy-\left (a-b \right )y^{2}=0 $

  3. $\displaystyle \left (a-b \right )x^{2}+4hxy-\left (a-b \right )y^{2}=0 $

  4. $\displaystyle \left (a-b \right )x^{2}+4hxy+\left (a-b \right )y^{2}=0 $

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Correct Option: C

The sum and product of the slopes of a pair of straight lines are the arithmetic and the geometric means of 9 and 16 respectively. The equation of the bisectors of the angles between the lines through the origin are 

maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

  1. $24x^{2}-25xy+2y^{2}=0$

  2. $25x^{2}+44xy-25y^{2}=0$

  3. $11x^{2}-25xy-11y^{2}=0$

  4. none of these

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Correct Option: B
Explanation:

Let the equation of lines passing through origin be $y-m _1x=0,y-m _2x=0$
$\therefore$The combined equation of pair of straight lines $=(y-m _1x)(y-m _2x)=0$

$\Rightarrow m _1m _2x^2-(m _1+m _2)xy+y^2=0$

But given sum of the slopes ,$m _1+m _2=\frac{(9+16)}{2}=\frac{25}{2}$
product of the slopes ,$m _1m _2=\sqrt(9*16)=12$
On subtituting these values in the above equation.
$\Rightarrow 12x^2-\displaystyle\frac{25}{2}xy+y^2=0$

$\Rightarrow 24x^2-25xy+2y^2=0$ comparing with general equation of pair of straight lines passing through origin $ax^2+2hxy+by^2=0$
$\Rightarrow a=224,h=\displaystyle\frac{-25}{2},b=2$
If $ax^2+2hxy+cy^2=0$ is pair of equation of line passing through origin then pair of equation of the angular bisector of these pair of lines is obtained by
 $h(x^2-y^2)=(a-b)xy$

$\therefore$ The required pair of equation of angular bisector is $h(x^2-y^2)=(a-b)xy$
$\Rightarrow \displaystyle\frac{-25}{2}(x^2-y^2)=(24-2)xy$
$\Rightarrow \displaystyle\frac{25}{2}*(x^2-y^2)=-22xy$

$\Rightarrow (25x^2-25y^2)=-44xy$

$\Rightarrow 25x^2+44xy-25y^2=0$

If $\displaystyle y=mx$ bisects the angle between the lines $\displaystyle x^{2}\left ( \tan ^{2}\theta +\cos ^{2}\theta  \right )+2xy\tan \theta -y^{2}\sin ^{2}\theta =0$  when $\displaystyle \theta =\dfrac\pi3$ the value of $m$ is

maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

  1. $\displaystyle \frac{-2- \sqrt 7}{ \sqrt 3}$

  2. $\displaystyle \frac{ \sqrt 7-2}{ \sqrt 3}$

  3. $\displaystyle 2 \sqrt 7 $

  4. $\displaystyle 2 \sqrt 3 $

Show answer
Correct Option: A,B
Explanation:
Equation of the bisectors of the angles between the given lines is

$\displaystyle \dfrac{x _2-y _2}{a-b}=\dfrac{xy}{h }$

Equation of the bisectors of the angles between the given lines is
$\displaystyle \frac{x^{2}-y^{2}}{\tan ^{2}\theta +\cos ^{2}\theta +\sin ^{2}\theta }=\frac{xy}{\tan \theta }$

$\displaystyle \Rightarrow \frac{x^{2}-y^{2}}{1+\tan ^{2}\theta  }=\frac{xy}{\tan \theta }$

$\displaystyle \Rightarrow \frac{x^{2}-y^{2}}{1+3  }=\frac{xy}{\sqrt 3 } \ when \ \ \theta=\pi/3$

Which satisfied by $y=mx $ if

$\displaystyle \frac{1-m^{2}}{4}=\frac{m}{\sqrt 3}$

$\displaystyle \Rightarrow \sqrt 3 m^{2}+4m-\sqrt 3=0$

$\displaystyle \Rightarrow m=\frac{-2\pm \sqrt 7}{\sqrt 3}$

If two of the lines represented by $ x^{4} + x^{3} y + cx^{2}y^{2} -xy^{3} + y^{4} =0$ bisect the angle between the other two, then the value of $c$ is

maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

  1. $0$

  2. $-1$

  3. $1$

  4. $-6$

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Correct Option: D
Explanation:

Since the product of the slopes of the four lines represented by the given equation is $1$ and a pair of lines represent the bisectors of the angles between the other two, the product of the slopes of each pair is $-1$. So let the equation of one pair be $ax^{2} + 2hxy -ay^{2} = 0$

The equation of its bisectors is $ \displaystyle \frac{x^{2}-y^{2}}{2a}=\frac{xy}{h} $

By hypothesis $ x^{4} +x^{3}y + cx^{2} y^{2}-xy^{3} + y^{4} $ $= (ax^{2} + 2hxy -ay^{2}) (hx^{2} -2axy -hy^{2})$ 

$ = ah(x^{4} + y^{4}) + 2(h^{2} -a^{ 2}) (x^{3}y- xy^{3}) -6ahx^{2}y^{2} $ 

Comparing the respective coefficients we get

$ah = 1 $ and $c = -6ah = -6$

The line $y=3x$ bisects the angle between the lines $ax^{2}+2axy+y^{2}=0$ if ${a}=$ 

maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

  1. $3$

  2. $11$

  3. $\displaystyle \frac{3}{11}$

  4. $\displaystyle \frac{11}{3}$

Show answer
Correct Option: C
Explanation:

Pair of angle bisectors represented by
$ax^2+2hxy+by^2=0$   is given by
$\dfrac{x^2-y^2}{a-b}=\dfrac{xy}{h}$
$\therefore ax^2+2axy+y^2=0$
pair of angle bisector is,
$\dfrac{x^2-y^2}{a-1}=\dfrac{xy}{a}$
$ax^2-ay^2=(a-1)xy$
Given  $ y=3x$  is one of angle bisector of given lines
$m=\dfrac{y}{x},$        $am^2+(a-1)x-a=0$
$m=3$ is satisfied to this equation
$9a+3a-3-a=0$
$11a=3$
$\therefore a=\dfrac{3}{11}$

The angle of intersection of the curves  $x ^ { 2 } + 4 y ^ { 2 } = 32$  and  $x ^ { 2 } - y ^ { 2 } = 12$  at any point of their intersection is

maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

  1. $\dfrac { \pi } { 6 }$

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

  3. $\dfrac { \pi } { 3 }$

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

Show answer
Correct Option: A
Explanation:
${ x }^{ 2 }+{ 4y }^{ 2 }=32\quad \longrightarrow \left( i \right) $
${ x }^{ 2 }-{ y }^{ 2 }=12\quad \longrightarrow \left( ii \right) $
Solving, $y=\pm 2$
              $x=\pm 4$
$\therefore$   Point of ${ X }^{ n }$ are $\left( 4,2 \right) ,\left( 4,-2 \right) ,\left( -4,2 \right) ,\left( -4,-2 \right) $
At $(4,2)$
${ m } _{ 1 }=\dfrac { -x }{ 4y } $    [differentiating $(i)$ wrt $x$]
$=\dfrac { 2 }{ -16 } =\dfrac { -1 }{ 8 } $
${ m } _{ 2 }=y/x$    [differentiating $(ii)$ wrt $x$]
$=\dfrac { 2 }{ 4 } =\dfrac { 1 }{ 2 } $
$\tan\theta =\dfrac { \left| { m } _{ 1 }-{ m } _{ 2 } \right|  }{ 1+{ m } _{ 1 }{ m } _{ 2 } } =\dfrac { \left| \dfrac { -1 }{ 8 } -\dfrac { 1 }{ 2 }  \right|  }{ 1-\dfrac { 1 }{ 16 }  } =\dfrac { \dfrac { 2+8 }{ 16 }  }{ \dfrac { 15 }{ 16 }  } =\dfrac { 10 }{ 15 } =\dfrac { 2 }{ 3 } $
$\Rightarrow \theta ={ \tan }^{ -1 }\left( 2/3 \right) \simeq \pi /6$              [A]

Family of lines represented by the equation $(\cos \theta)x+(\cos \theta -\sin \theta)y-3(3\cos \theta+\sin \theta)=0$ passes through a fixed point $M$ for all real value of $\theta$. Find $M$ 

maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

  1. $(6,3)$

  2. $(3,6)$

  3. $(-6,2)$

  4. $(3,-6)$

Show answer
Correct Option: A
Explanation:

Let us consider the problem:

$\left( {\left( {\cos \theta  + \sin \theta } \right)x + \cos \theta  - \sin \theta } \right)y - 3\left( {3\cos \theta  + \sin \theta } \right) = 0$
$ \Rightarrow \cos \theta \left( {x + y - 9} \right) + \sin \theta \left( {x - y - 3} \right) = 0$
$ \Rightarrow $ $\left( {x + y - 9} \right) + \tan \theta \left( {x - y - 3} \right) = 0$
${L _1} + K{L _2} = 0$(pass through intersection of ${L _1}$ and ${L _2}$ for all value of $K$)
$x+y-9=0$
$ \Rightarrow $ $x - y - 3 = 0$ 
Hence,
$x+y=9$
$x-y=3$
hence the intersection point is $(6,3)$

If the equation $a{x}^{2}+2hxy+b{y}^{2}=0$ represents a pair of lines then  the equation of the pair of lines of angular bisectors is $h({x}^{2}-{y}^{2})-(a-b)xy=0$

maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

  1. True

  2. False

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Correct Option: A

If the line $y = mx$ bisects the angle between the line $ax^2 + 2h\ xy + by^2 = 0$ then $m$ is a root of the quadratic equation :

maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

  1. $hx^2 + (a - b)x - h = 0$

  2. $x^2 +h(a - b)x - 1 = 0$

  3. $(a - b)x^2 + hx - (a - b) = 0$

  4. $(a - b)x^2 - hx - (a - b) = 0$

Show answer
Correct Option: A
Explanation:

Equation of bisectors of the pair of straight lines $ax^2+2hxy+by^2=0$ is

$h(x^2-y^2)-(a-b)xy=0$......(1).

Since $y=mx $ is given to be the bisector of the pair of straight lines, then the line will satisfy the equation (1).

Then we get,
$h(1-m^2)-(a-b)m=0$

$hm^2+(a-b)m-h=0$.

So $m$ satisfies the equation $hx^2+(a-b)x-h=0$.

Joint equation of perpendicular lines passing through $(0,0)$ one of which is parallel to $6x-4y+3=0$ is

maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

  1. $6x^{2}-5xy-6y^{2}=0$

  2. $6x^{2}+5xy-6y^{2}=0$

  3. $5x^{2}+5xy-6y^{2}=0$

  4. $6x^{2}-5xy-5y^{2}=0$

Show answer
Correct Option: A