Tag: maths

Questions Related to maths

$\displaystyle ax^{2}+2hxy+by^{2}=0$ 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

The Joint equation of the bisectors of the angle between the lines represented by $\displaystyle ax^{2}+2hxy+by^{2}=0 $ is

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

  1. a pair of perpendicular lines

  2. a pair of parallel lines

  3. a pair of intersecting lines but not $\displaystyle \perp$er

  4. None of these

Show answer
Correct Option: A
Explanation:

Equation of bisectors of angle between the pair of straight line $\displaystyle ax^{2}+2hxy+by^{2}=0$ is given by $\displaystyle \frac{x^{2}-y^{2}}{a-b}=\frac{xy}{h}$


$\displaystyle\Rightarrow hx^{2}-\left ( a-b \right )xy-hy^{2}=0$

Here coefficient of $x^{2}+$ coefficient of $y^{2}=h+(-h)=0$

$\displaystyle \therefore $ above equation represent the pair of $\perp$er lines

Hence choice (a) is correct.

The equation $a^2 x^2 + 2h(a+b) xy + b^2 y^2 = 0$ and $ax^2 + 2hxy + by^2 = 0$ represent

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

  1. two pairs of perpendicular straight lines

  2. two pairs of parallel straight lines

  3. two pairs of straight lines which are equally inclined to each other

  4. None of these

Show answer
Correct Option: C
Explanation:

$ax^2+2hxy+by^2=0$
Equation of the angle bisectors is given by
$\dfrac{x^2-y^2}{a-b}=\dfrac{xy}{h}$ ...(i)

For $ a^2x^2+2h(a+b)xy+b^2y^2=0$
The equations of the angle bisector is given by
$\dfrac{x^2-y^2}{a^2-b^2}=\dfrac{xy}{h(a+b)}$
 $\dfrac{x^2-y^2}{a-b}=\dfrac{xy}{h}$ ...(ii)
Since (i) is equal to (ii), the above lines are equally inclined to each other.

If one of the lines of $my^2 + (1-m^2) xy - mx^2 = 0$ is a bisector of the angle between the lines $xy = 0$, then $m$ is

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

  1. $3$

  2. $2$

  3. $-\dfrac{1}{2}$

  4. $-1$

Show answer
Correct Option: D
Explanation:

$my^2+(1-m^2)xy-mx^2$
$(my+x)(y-mx)=0$
Therefore the lines are $y=mx$ and $y=\dfrac{-x}{m}$ ...(i)
The angle bisectors of the $xy=0$ is $y=x$ and $y=-x$
Comparing the slopes with the equations of the (i) we get
$m=\pm1$

If one of the lines of is $my^{2}+\left ( 1-m^{2} \right )xy-mx^{2}=0$ is a bisector of the angle between the lines $\displaystyle xy = 0,$ then $m$ is

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

  1. $\displaystyle1$

  2. $\displaystyle2$

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

  4. $\displaystyle-1$

Show answer
Correct Option: A,D
Explanation:

Given pair of lines is
$\displaystyle { my }^{ 2 }+\left( 1-{ m }^{ 2 } \right) xy-{ mx }^{ 2 }=0\Rightarrow m{ \left( \frac { y }{ x }  \right)  }^{ 2 }+{ \left( 1-m \right)  }^{ 2 }\frac { y }{ x }- m=0$    ...(1)

Lines $xy=0$ are $x=0$ and $y=0.$
i.e the axes bisector of angle between the axes are $y=x$ and $y=-x$ $\displaystyle \Rightarrow \frac { y }{ x } =1$ or $\displaystyle \frac { y }{ x } =-1$

If $\displaystyle \frac { y }{ x } =1$ is represented by (1), then
$m{ \left( 1 \right)  }^{ 2 }+\left( 1-{ m }^{ 2 } \right) \left( -1 \right) -m=0\Rightarrow 1-{ m }^{ 2 }=0\Rightarrow m=\pm 1$

If $\displaystyle \frac { y }{ x } =-1$ is represented by (1), then
$m{ \left( -1 \right)  }^{ 2 }+\left( 1-{ m }^{ 2 } \right) \left( -1 \right) -m=0\Rightarrow 1-{ m }^{ 2 }=0\Rightarrow m=\pm 1$
In both cases either $m=1$ or $m=-1.$

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. $pq=-1$

  2. $p=q$

  3. $p=-q$

  4. $pq=1$

Show answer
Correct Option: B

The pairs of straight lines $ax^{2}+2hxy-ay^{2}=0$ and $hx^{2}-2axy-hy^{2}=0$ are such that

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

  1. one pair bisects the angles between the other pair

  2. the lines of one pair are equally inclined to the lines of the other pair

  3. the lines of one pair are perpendicular to the `lines of the other pair

  4. none of these

Show answer
Correct Option: A,B
Explanation:

Given pairs of straight lines are $ax^{2}+2hxy-ay^{2}=0$ and $hx^{2}-2axy-hy^{2}=0$.

Pair of angular bisectors of $ax^{2}+2hxy-ay^{2}=0$ is $h(x^2-y^2)=(a-(-a))xy$

$\Rightarrow hx^2-2axy-hy^2$

$\therefore$ One pair bisects the angle between the other.

Clearly, if one pair bisects the angle between the other, the lines of one pair are equally inclined to the lines of the other pair.
Hence, option A and B.

If one of the lines of $my^2 + (1- m^2) xy - mx^2 = 0$ is a bisector of the angle between the lines $xy = 0$, then $m$ is

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

  1. $1$

  2. $2$

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

  4. $-1$

Show answer
Correct Option: A,D
Explanation:

Angle bisectors of $xy=0$ are $x+y=0 $  and $x=y$
$my^2+xy-m^2 xy -mx^2=0$
$\therefore y(my+x)-mx(my+x)=0$
$\therefore (y-mx)(my+x)=0$
Comparing $y-mx=0  $ and $  my+x=0  $  with  $y=x $ and  $ y=-x$,  we get $m=\pm 1$ 

The straight lines $7x^{2}+6xy+4y^{2}=0$ have the same pair of bisectors as those of the lines given by

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

  1. $49x^{2}+66xy+16y^{2}=0$

  2. $10x^{2}+6xy+7y^{2}=0$

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

  4. $4x^{2}-6xy+7y^{2}=0$

Show answer
Correct Option: A,B,C,D
Explanation:

For $7x^{ 2 }+6xy+4y^{ 2 }=0$ 
Equation of angle bisector is 
$\cfrac { { x }^{ 2 }-{ y }^{ 2 } }{ 7-4 } =\cfrac { xy }{ 3 } \Rightarrow { x }^{ 2 }-xy-{ y }^{ 2 }=0$
For Option A 
Equation of angle bisector of $49x^{ 2 }+66xy+16y^{ 2 }=0$ is
$\cfrac { { x }^{ 2 }-{ y }^{ 2 } }{ 33 } =\cfrac { xy }{ 33 } \Rightarrow { x }^{ 2 }-xy-{ y }^{ 2 }=0$
For Option B
Equation of angle bisector of $10x^{ 2 }+6xy+7y^{ 2 }=0$ is
$\cfrac { { x }^{ 2 }-{ y }^{ 2 } }{ 3 } =\cfrac { xy }{ 3 } \Rightarrow { x }^{ 2 }-xy-{ y }^{ 2 }=0$
For Option C
Equation of angle bisector of $5x^{ 2 }+6xy+2y^{ 2 }=0$ is
$\cfrac { { x }^{ 2 }-{ y }^{ 2 } }{ 3 } =\cfrac { xy }{ 3 } \Rightarrow { x }^{ 2 }-xy-{ y }^{ 2 }=0$
For Option D
Equation of angle bisector of $4x^{ 2 }-6xy+7y^{ 2 }=0$ is
$\cfrac { { x }^{ 2 }-{ y }^{ 2 } }{ -3 } =\cfrac { xy }{ -3 } \Rightarrow { x }^{ 2 }-xy-{ y }^{ 2 }=0$

$\displaystyle ax^{2}+2hxy+by^{2}=0$ 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 $

Show answer
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

Show answer
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 $h(x^2-y^2)=(a-b)xy$
$\Rightarrow \displaystyle\frac{-25}{2}(x^2-y^2)=(24-2)xy$
$\Rightarrow \displaystyle\frac{25}{2}*(x^2-y^2)=-22xy$

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

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