Tag: bisector of angle between lines

Questions Related to bisector of angle between lines

If the line y = mx is one of the bisector of the lines $x^2 + 4xy - y^2 = 0$, then the value of no ___________.

  1. $\frac{\sqrt{5} - 1}{2}$

  2. $\frac{\sqrt{5} + 1}{2}$

  3. $-(\frac{\sqrt{5} + 1}{2})$

  4. $-(\frac{\sqrt{5} -1}{2})$


Correct Option: A

The Straight lines represented by the equation $135{ x }^{ 2 }-136xy+33{ y }^{ 2 }=0$ are equally inclined to the line 

  1. $x-2y=7$

  2. $x+2y=7$

  3. $x-2y=4$

  4. $3x+2y=4$


Correct Option: B
Explanation:

Give pair of lines is $135{ x }^{ 2 }-136xy+33{ y }^{ 2 }=0$   ...(1)


The equation of bisector of angles between pair of lines (1) is

$\displaystyle \frac { { x }^{ 2 }-{ y }^{ 2 } }{ a-b } =\frac { xy }{ h } \Rightarrow \frac { { x }^{ 2 }-{ y }^{ 2 } }{ 135-33 } =\frac { xy }{ -68 } $

$\Rightarrow 2{ x }^{ 2 }+3xy-2{ y }^{ 2 }=0\Rightarrow \left( x+2y \right) \left( 2x-y \right) =0$

One of the bisectors is $x+2y=0$ which is parallel to the line $x+2y=7$.

Hence, the line $x+2y=7$ is equally inclined to the given lines.

If the bisectors of the lines $x^2 - 2pxy - y^2 = 0$ be $x^2 - 2qxy - y^2 = 0$. then

  1. 2p + q = 0

  2. 2p + 3q = 0

  3. pq = 1

  4. pq + 1 = 0


Correct Option: A

If the pair of straight lines $x^{2}-2pxy-y^{2}= 0$ and $x^{2}-2qxy-y^{2}= 0$ be such that each pair bisects the angle between the other pair, then

  1. $p= -q$

  2. $pq= 1$

  3. $pq= -1$

  4. $p= q$


Correct Option: C
Explanation:

Given equations are $\displaystyle  x^{2}-2qxy-y^{2}=0 ...(1) $ $\displaystyle  x^{2}-2pxy-y^{2}=0 ...(2) $ Joint equation of angle bisector of the line (i) and (ii) are same $\displaystyle \therefore qx^{2}+2xy-qy^{2}=0....(3) $. 


Now (2).and (3) are same, taking the ratio of their coefficients


$\displaystyle \therefore \frac{1}{q}=\frac{-p}{1}\Rightarrow pq=-1$

2x + y - 4 = 0 is a besector of angles between the lines a(x - 1) + b(y - 2) = 0, c(x - 1) + d(y - 2) = 0 the other angular bisector is _______________.

  1. x - 2y + 1 = 0

  2. x - 2y - 3 = 0

  3. x - 2y + 3 = 0

  4. x + 2y - 5 = 0


Correct Option: C
Explanation:
We have $a\left(x-1\right)+b\left(y-2\right)=0$       .....$(1)$
and $c\left(x-1\right)+d\left(y-2\right)=0$       .....$(2)$

Clearly $\left(1,2\right)$ lie on both the lines and hence $\left(1,2\right)$ is their point of intersection.

Both the bisectors will pass through $\left(1,2\right)$
One of the bisector is $2x+y-4=0$

Other bisector will be perpendicular to this bisector.
Hence its equation will be $x-2y=\lambda$

It passes through $\left(1,2\right)$
$\Rightarrow\,1-4=\lambda$
$\Rightarrow\,\lambda=-3$

Hence the equation is $x-2y=-3$ or $x-2y+3=0$

The equations of the bisectors of that angle between the lines $x+2y-11=0,:3x+6y-5=0$ which contains the point $\left(1,-3\right)$ is 

  1. $3x=19$

  2. $3y=7$

  3. $3x=19$ and $3y=7$

  4. None of these


Correct Option: A

The line $L$ has intercepts $a$ and $b$ on the co-ordinate axes keeping the origin fixed, the co-ordinate axes are related through a fixed angle. If the same line has intercepts c and d then

  1. $ \displaystyle \frac{1}{a^{2}}+\frac{1}{c^{2}}= \frac{1}{b^{2}+d^{2}} $

  2. $ \displaystyle \frac{1}{a^{2}}+\frac{1}{b^{2}}= \frac{1}{c^{2}}+\frac{1}{d^{2}} $

  3. $ \displaystyle a^{2}+c^{2}= b^{2}+d^{2} $

  4. $ \displaystyle a^{2}+b^{2}= c^{2}+d^{2} $


Correct Option: B
Explanation:

Suppose we state he coordinate axis in the anti-clockwise direction through an angle $\alpha$.
The equation of the line $\alpha$ with respect to old axes is $\displaystyle\frac{x}{a}+\frac{y}{b}=1$
In this equation replacing $x$ by $x\cos\alpha-y\sin\alpha$
The equation of the line with respect to new axes is
$\displaystyle\frac{x\cos\alpha-y\sin\alpha}{a}+\frac{x\sin\alpha+y\cos\alpha}{b}=1$
$\displaystyle\Rightarrow x\left( \frac { \cos { \alpha  }  }{ a } +\frac { \sin { \alpha  }  }{ b }  \right) +y\left( \frac { \cos { \alpha  }  }{ b } -\frac { \sin { \alpha  }  }{ a }  \right) =1$   ...(1)
The intercept mode by (1) on the co-ordinate axes are given as $c$ and $d$.
Therefore, $\displaystyle\frac{1}{c}=\frac { \cos { \alpha  }  }{ a } +\frac { \sin { \alpha  }  }{ b } $ and $\displaystyle\frac{1}{d}=\frac { \cos { \alpha  }  }{ b } -\frac { \sin { \alpha  }  }{ a }$
Squaring and adding, we get $ \displaystyle \frac{1}{c^{2}}+\frac{1}{d^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}} $

$P: x^{2}-y^{2}+2y-1=0$
$L: x+y=3$

Equation of the angle bisectors of the pairs of lines P is

  1. $xy-y=0$

  2. $xy-x=0$

  3. $xy=0$

  4. $xy+y=0$


Correct Option: B
Explanation:

$P:{ x }^{ 2 }-{ \left( y-1 \right)  }^{ 2 }=0$

$\Rightarrow x+y-1=0$ and $x-y+1=0$

Equation of the angle bisector is

$\cfrac { A _1x+B _1y+C _1 }{ \sqrt {A _1^2+B _1^2 }  } =\pm \cfrac { A _2x+B _2y+C _2 }{ \sqrt {A _2^2+B _2^2}  } $
$\cfrac { x+y-1 }{ \sqrt { 2 }  } =\pm \cfrac { x-y+1 }{ \sqrt { 2 }  } $

$\Rightarrow x=0$ or $y-1=0$

If pairs of lines $3x^{2}-2pxy-3y^{2}=0$ and $5x^{2}-2qxy-5y^{2}=0$ are such that each pair bisects the angle between the other pair, then $pq$ is equal to

  1. $-1$

  2. $-3$

  3. $-5$

  4. $-15$


Correct Option: D
Explanation:
Given pairs 
$3x^2-2pxy-3y^2=0$----(1)

$5x^2-2qxy-5y^2=0$----(2)

Equation (1) can be written as 
$3(x^2-y^2)=2pxy$----(3)

Eq of angle bisector of eq (2)
$\dfrac{x^2-y^2}{5+5}=\dfrac{xy}{-q}$

$\dfrac{x^2-y^2}{10}=\dfrac{xy}{-q}$----(4)

Now as the given in question the eq of angle bisector of one pair bisects the other pair

So dividing eq (4) by (3)
$\dfrac{\dfrac{x^2-y^2}{10}}{3(x^2-y^2)}=\dfrac{\dfrac{xy}{-q}}{2pxy}$

$\dfrac{1}{30}=\dfrac{1}{-2pq}$

$pq=-15$ 

Slope of a bisector of the angle between the lines $4x^{2}-16xy-7y^{2}=0$ is

  1. $\displaystyle \frac{11+\sqrt{377}}{16}$

  2. $\displaystyle \frac{11-\sqrt{377}}{16}$

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

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


Correct Option: A,B
Explanation:
Given pair 
$4x^2-16xy-7y^2=0$
On comparing given eq with $ax^2+2hxy+by^2=0$
$a=4,b=-7,h=-8$
Eq of pair of Angle bisector 
$\dfrac{x^2-y^2}{a-b}=\dfrac{xy}{h}$
$\dfrac{x^2-y^2}{11}=\dfrac{xy}{-8}$
$-8x^2+8y^2=11xy$
$8y^2-11xy-8x^2=0$
$y=\dfrac{-(-11x)\pm\sqrt{121x^2+256x^2}}{16}$

$y=\dfrac{11x\pm\sqrt{377x^2}}{16}$

$y=\dfrac{11x\pm\sqrt{377}x}{16}$

$y=\left (\dfrac{11\pm\sqrt{377}}{16}  \right )x$
Comparing above eq with $y=mx+c$
$m=\left (\dfrac{11+\sqrt{377}}{16}  \right )$ and $\left (\dfrac{11-\sqrt{377}}{16}  \right )$