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Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between 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
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The joint equation of the pair of lines is ax^2 + 2hxy + by^2 = 0, and the equation of their angle bisectors is (x^2 - y^2)/(a - b) = xy/h. Comparing coefficients or using standard transformations for the pair of bisector lines given by x^2 - 2pxy - y^2 = 0 yielding bisectors x^2 - 2qxy - y^2 = 0 leads to the condition pq + 1 = 0.

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 $\left (\dfrac{11-\sqrt{377}}{16}  \right )$
Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

$P: 2x^{2}-axy+6y^{2}=0$
$Q: 3x^{2}-8xy+4y^{2}=0$
If the bisectors of the angles between the lines represented by $P$ and $Q$ are same, the value of $a$ is

  1. $8$
  2. $\dfrac{16}{3}$
  3. $32$
  4. $-16$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Equation of angle bisector of $2x^{ 2 }-axy+6y^{ 2 }=0$ is

$\cfrac { { x }^{ 2 }-{ y }^{ 2 } }{ 2-6 } =\cfrac { xy }{ -a/2 } \Rightarrow a{ x }^{ 2 }-a{ y }^{ 2 }-8xy=0$   ----------(1)

And equation of angle bisector of $3x^{ 2 }-8xy+4y^{ 2 }=0$ is

$\cfrac { { x }^{ 2 }-{ y }^{ 2 } }{ 3-4 } =\cfrac { xy }{ -4 } \Rightarrow 4{ x }^{ 2 }-4{ y }^{ 2 }-xy=0$  -----------(2)

From (1) and (2), we get $a=32$

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

If the lines represented by $x^2-2pxy-y^2=0$ are rotated about the origin through an angle $\theta,$ one in clockwise direction and other in anti-clockwise direction, then the equation of the bisector of the angle between the lines in the new positions is

  1. $px^2+2xy-py^2=0$
  2. $px^2+2xy+py^2=0$
  3. $x^2-2pxy-y^2=0$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given eq 
$x^2-2pxy-y^2=0$
comparing above eq with general form of eq $ax^2+2hxy+by^2=0$
$a=1,h=-p,b=-1$
Now the line is rotated one in clockwise and other is anticlockwise so the both eq are replaced by each other and form the same eq as it was so here we finding eq of angle bisector by formula
$\dfrac{x^2-y^2}{a-b}=\dfrac{xy}{h}$
$\dfrac{x^2-y^2}{1+1}=\dfrac{xy}{-p}$
$\dfrac{x^2-y^2}{2}=\dfrac{xy}{-p}$
$(-p)(x^2-y^2)=2xy$
$-px^2+py^2=2xy$
$px^2+2xy-py^2=0$