Tag: maths

Questions Related to maths

If the equation  $2 x ^ { 2 } + 3 x y + b y ^ { 2 } - 11 x + 13 y + c = 0$  represents two perpendicular straight lines, then

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $b = - 2$

  2. $b = 2$

  3. $c = - 2$

  4. $c = 2$

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Correct Option: A

The product of the perpendiculars from origin to the pair of lines ${ ax }^{ 2 }+2hxy+{ by }^{ 2 }+2gx+2fy+c=0$ is


maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $\frac { \left| c \right| }{ \sqrt { \left( { a+b } \right) ^{ 2 } } +{ 4h }^{ 2 } } $

  2. $\frac { \left| c \right| }{ \sqrt { \left( a+b \right) ^{ 2 }-{ 4h }^{ 2 } } } $

  3. $\frac { \left| c \right| }{ \sqrt { \left( a-b \right) ^{ 2 }-{ 4h }^{ 2 } } } $

  4. $\frac { \left| c \right| }{ \sqrt { \left( a-b \right) ^{ 2 }-{ 4h }^{ 2 } } } $

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Correct Option: A

If $6x^{2}-5xy+by^{2}+4x+7y+c=0$ represents a pair of perpendicular lines, then :

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $b=6,\ c=-2$

  2. $b=-6,\ c=2$

  3. $b=-6,\ c=-2$

  4. $b=-2,\ c=-6$

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Correct Option: A

If the lines $ x ^ { 2 } + ( 2 + k ) x y - 4 y ^ { 2 } = 0 $ are equally inclined to the coordinate axes, then  k =

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. -1

  2. -2

  3. -3

  4. -4

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Correct Option: A

If the pair of lines $ax^{2}+2hxy+by^{2}+2gx+2fy+c= 0$ intersect on $y$ axis then

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $2fgh= bg^{2}+ch^{2}$

  2. $bg^{2}\neq ch^{2}$

  3. $abc= 2fgh$

  4. None of these

Show answer
Correct Option: A
Explanation:

As $\displaystyle s=ax^{2}+2hxy +by^{2}+2gx +2fy+c=0 $ represent a pair of line $\displaystyle \therefore \begin{vmatrix}a &h  &g \h  &b  &f \g  &f  &c \end{vmatrix}=0$
or $\displaystyle abc+2fgh-af^{2}-bg^{2}-ch^{2}=0....(1)$ Now say point ofintersection onY axis be $\displaystyle (0,y _{1} $ and  point of intersection of pair of line be obtained by solving the equations $\displaystyle \frac{\partial s}{\partial x}=0=\frac{\partial s}{\partial y}$ $\displaystyle \therefore \frac{\partial s}{\partial x}=0\Rightarrow ax+by+g=0$ $\displaystyle \begin{matrix}\Rightarrow  \ \Rightarrow  \end{matrix} \left{\begin{matrix}hy _{1}+g=0 \by _{1}+f=0 \end{matrix}\right.>  ()$ and $\displaystyle \frac{\partial s}{\partial y}=0\Rightarrow bx+by+f=0$  On compairing the equation given in () we get $\displaystyle bg=fh $ and  $\displaystyle bg^{2}=fgh ....(2) $ Again $\displaystyle ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ meet at y-axis $\displaystyle \therefore x=0$ $\displaystyle \Rightarrow by^{2}+2fy+c=0$ whose roots must be equal $\displaystyle \Rightarrow by^{2}+2fy+c=0$ whose roots must be equal $\displaystyle \therefore f^{2}=bc af^{2}=abc ......(3)$ Now using (2) and (3) in equation (I) we have $\displaystyle abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$ $\displaystyle \Rightarrow (abc-af^{2})+(fgh-bg^{2})+fgh-ch^{2}=0$ $\displaystyle \Rightarrow 0+0+fgh-ch^{2}=0 \therefore ch^{2}=fgh .....(4) $ Now adding (2) and (4) $\displaystyle 2fgh=ch^{2}+bg^{2}$

A line is at distance of $4$ units from origin and having both intercepts positive. If the perpendicular from the origin to this line makes an angle of ${60}^{o}$ with the line $x+y=0$ Then the equation of the line is

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $\left( \sqrt { 3 } +1 \right) x+\left( \sqrt { 3 } +2 \right) y=y=8\sqrt { 2 } $

  2. $\left( \sqrt { 3 } -1 \right) x+\left( \sqrt { 3 } +1 \right) y=y=8\sqrt { 2 } $

  3. $\left( \sqrt { 3 } +1 \right) x-\left( \sqrt { 3 } +1 \right) y=8\sqrt { 2 } $

  4. $\left( \sqrt { 3 } +2 \right) x+\left( \sqrt { 3 } +1 \right) y=8\sqrt { 2 } $

Show answer
Correct Option: B

If two lines $\dfrac{x-1}{1}=\dfrac{y-2}{k}=\dfrac{z-3}{1}$ and $\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{k}$ intersect, then the value of k is?

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $4$

  2. $5$

  3. $3$

  4. $6$

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Correct Option: A

If one of the lines given by $6x^{2}-xy+4cy^{2}= 0$ is $3x+4y= 0$, then $c$ equals

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. 3

  2. -1

  3. 1

  4. -3

Show answer
Correct Option: D
Explanation:

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

$\displaystyle  =(y-m _{1}x)(y-m _{2}x) $

$\displaystyle  \Rightarrow m _{1}+m _{2}=-\frac{2h}{b}=\frac{1}{4c }$.....(1) 

$\displaystyle  \Rightarrow m _{ _1}m _{2} =\frac{3}{2c} $
and $\displaystyle  3x+4y=0\Rightarrow m _{1} =-\dfrac {3}{4}$
 $\displaystyle \therefore m _{2} =-\frac{2}{c} $

 Now by $\displaystyle (1)$ we have $\displaystyle-\left(\frac{3}{4}+\frac{2}{c}\right) =\frac{1}{4c} $

$\displaystyle  \Rightarrow -\frac{3}{4}=\frac{1}{4c}+\frac{2}{c},\frac{3}{4}=\frac{1}{4c}+\frac{8}{4c}$ $\displaystyle  -\frac{3}{4}=\frac{9}{4c}$

$ \therefore c=-3$

If line $2x+7y-1=0$ intersect the lines $L _1=3x+4y+1=0$ and $L _2=6x+8y-3=0$ in $A$ and $B$ respectively, then equation of a line parallel to $L _1$ and $L _2$ and passes through a point $P$ such that $AP : PB=2:1$ (internally) is ($P$ is on the line $2x+7y-1=0$)

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $9x+12y+3=0$

  2. $9x+12y-3=0$

  3. $9x+12y-2=0$

  4. $None\ of\ these$

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Correct Option: C

If the pair of lines represented by the equation $6x^{2}+17xy+12y^{2}+22x+31y+20=0$ be $2x+3y+p=0$ and $3x+4y+q=0$, then

maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

  1. $p+q=9$

  2. $p^{2}+q^{2}=0$

  3. $3p+2q=22$

  4. $4p+3q=31$

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

$(2x+3y+p)(3x+4y+q)=0$

$6x^{2}+8xy+2qx+9xy+12y^{2}+3qy+3px+4py+pq=0$

$6x^{2}+17xy+12y^{2}+x(2q+3p)+y(3q+4p)+pq=0$
$\rightarrow 6x^{2}+17xy+12y^{2}+22x+31y+20=0$

Hence comparison gives us
$pq=20$
$3p+2q=22$
$4p+3q=31$
$(4p+3q)-(3p+2q)=31-22$
$p+q=9$.
Therefore
$p^{2}+q^{2}=(p+q)^{2}-2pq$
$=81-2(20)$
$=81-40$
$=41$