Tag: bisector of angle between lines

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Questions Related to bisector of angle between lines

Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles 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. $-(\frac{\sqrt{5} -1}{2})$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The angle bisectors of the pair of lines ax^2 + 2hxy + by^2 = 0 are given by (x^2 - y^2)/(a - b) = xy/h. Substituting the coefficients from x^2 + 4xy - y^2 = 0, we find the slopes of the bisectors.

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

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. $3x+2y=4$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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. $p= q$
Reveal answer Fill a bubble to check yourself
C Correct answer
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


$\displaystyle \therefore \frac{1}{q}=\frac{-p}{1}\Rightarrow pq=-1$
Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $x-2y+3=0$
Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The lines are x + 2y - 11 = 0 and 3x + 6y - 5 = 0. These are parallel lines. The bisector of parallel lines is a line parallel to them, equidistant from both. Calculating the midpoint of the perpendicular distance gives the result.

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

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} $
Reveal answer Fill a bubble to check yourself
B Correct answer
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}} $

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

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

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

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

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

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. $-15$
Reveal answer Fill a bubble to check yourself
D Correct answer
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)
$\dfrac{\dfrac{x^2-y^2}{10}}{3(x^2-y^2)}=\dfrac{\dfrac{xy}{-q}}{2pxy}$

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

$pq=-15$ 
Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

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. $\displaystyle \frac{-3-2\sqrt{3}}{7}$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
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 $\left (\dfrac{11-\sqrt{377}}{16}  \right )$