Tag: maths

Questions Related to maths

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

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

  1. 2p + q = 0

  2. 2p + 3q = 0

  3. pq = 1

  4. pq + 1 = 0

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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

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

  1. $p= -q$

  2. $pq= 1$

  3. $pq= -1$

  4. $p= q$

Show answer
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 _______________.

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

  1. x - 2y + 1 = 0

  2. x - 2y - 3 = 0

  3. x - 2y + 3 = 0

  4. x + 2y - 5 = 0

Show answer
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 

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

  1. $3x=19$

  2. $3y=7$

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

  4. None of these

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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

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

  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} $

Show answer
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

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

  1. $xy-y=0$

  2. $xy-x=0$

  3. $xy=0$

  4. $xy+y=0$

Show answer
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

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

  1. $-1$

  2. $-3$

  3. $-5$

  4. $-15$

Show answer
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

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

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

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

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

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

Show answer
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 $\left (\dfrac{11-\sqrt{377}}{16}  \right )$

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

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

  1. $8$

  2. $\dfrac{16}{3}$

  3. $32$

  4. $-16$

Show answer
Correct Option: C
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 $a=32$

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

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

  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

Show answer
Correct Option: A
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
$\dfrac{x^2-y^2}{a-b}=\dfrac{xy}{h}$
$\dfrac{x^2-y^2}{1+1}=\dfrac{xy}{-p}$
$\dfrac{x^2-y^2}{2}=\dfrac{xy}{-p}$
$(-p)(x^2-y^2)=2xy$
$-px^2+py^2=2xy$
$px^2+2xy-py^2=0$