Tag: maths

Questions Related to maths

If $4xy+2x+2fy+3=0$ represents a pair of lines then $f=$

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

  1. $2$

  2. $3$

  3. $4$

  4. $5$

Show answer
Correct Option: B
Explanation:
By factorising,
$4xy+2x+2fy+3=0$
$2x\left(2y+1\right)+3\left(\dfrac{2fy}{3}+1\right)=0$
$\Rightarrow \dfrac{2f}{3}=2$
$\Rightarrow f=3$
$\therefore$ the value of $f=3$

If the two pair of lines $x^2-2mxy-y^2=0$ and $x^2-2nxy-y^2=0$ are such that one of them represents the bisectors of the angles between the other, then 

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

  1. $mn+1=0$

  2. $mn-1=0$

  3. $1/m+1/n=0$

  4. $1/m -1/n=0$

Show answer
Correct Option: A

If the equation of the pair of straight lines passing through the point $(1, 1),$ one making an angle $\theta$ with the positive direction of x-axis and the other making the same angle with the positive direction of y-axis is $x^{2}- (a + 2)xy + y^{2} + a(x + y -1) = 0, a \neq -2,$ then the value of $\sin 2\theta $ is

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

  1. $a -2$

  2. $a + 2$

  3. $\dfrac2{(a + 2)}$

  4. $ \dfrac2a$

Show answer
Correct Option: C
Explanation:

Equations of the given lines are $y -1 = \tan \theta (x -1) $ and $y -1 =\ cot \theta (x -1)$ 


so their joint equation is 

$[(y-1)-\tan \theta (x -1)][(y -1) -\cot \theta (x-1)] = 0$

$\Rightarrow (y -1)^{2} -(\tan \theta +\cot \theta) (x-l)(y -1) +(x-l)^{2}= 0$

$\Rightarrow x^{2} -(\tan \theta + cot \theta) xy + y^{2} + (\tan \theta+ \cot \theta -2) (x+y -1)=0$

Comparing with the given equation we get $\tan \theta + \cot \theta= a + 2$

$\displaystyle \Rightarrow \frac{1}{\sin \theta \cos\theta }= a + 2 \Rightarrow \sin 2\theta = \frac{2}{a+2}$

The absolute value of difference of the slope of the lines $\displaystyle x^{2}\left ( \sec ^{2}\theta -\sin ^{2}\theta  \right )-2xy\tan \theta +y^{2}\sin ^{2}\theta =0$ is

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

  1. $-2$

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

  3. $2$

  4. $1$

Show answer
Correct Option: C
Explanation:
Given pair of lines
$x^2(\sec^2\theta-\sin^2\theta)-2xy\tan\theta+y^2\sin^2\theta=0$

$y^2\sin^2\theta-2xy\tan\theta+x^2(\sec^2\theta-\sin^2\theta)=0$

$y=\dfrac{2x\tan\theta\pm\sqrt{4x^2\tan^2\theta-4\sin^2\theta x^2(\sec^2\theta-\sin^2\theta)}}{2\sin^2\theta}$

$y=\dfrac{2x\tan\theta\pm\sqrt{4x^2\tan^2\theta-4\tan^2\theta x^2+4x^2\sin^4\theta}}{2\sin^2\theta}$

$y=\dfrac{2x\tan\theta\pm\sqrt{4x^2\sin^4\theta}}{2\sin^2\theta}$

$y=\dfrac{2x\tan\theta\pm2x\sin^2\theta}{2\sin^2\theta}$

$y=\dfrac{(\tan\theta\pm\sin^2\theta)}{\sin^2\theta}x$

On comparing above equation with $y=mx+c$ we get
$m=\dfrac{(\tan\theta\pm\sin^2\theta)}{\sin^2\theta}$

Here $m _{1}=\dfrac{(\tan\theta+\sin^2\theta)}{\sin^2\theta}$ and $m _{2}=\dfrac{(\tan\theta-\sin^2\theta)}{\sin^2\theta}$

$m _{1}-m _{2}=\dfrac{(\tan\theta+\sin^2\theta)}{\sin^2\theta}-\dfrac{(\tan\theta-\sin^2\theta)}{\sin^2\theta}$

$\Rightarrow m _{1}-m _{2}=\dfrac{\tan\theta+\sin^2\theta-\tan\theta+\sin^2\theta}{\sin^2\theta}$

$\Rightarrow m _{1}-m _{2}=\dfrac{2\sin^2\theta}{\sin^2\theta}$

$\Rightarrow m _{1}-m _{2}=2$

Two pair of straight lines have the equation $\displaystyle x^{2}+6xy+9y^{2}=0: : and: : ax^{2}+2bxy+cy^{2}=0 $. If one line among them is common, then the value of $9a - 6b + c$ is

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

  1. $1$

  2. $3$

  3. $0$

  4. $2$

Show answer
Correct Option: C
Explanation:
Given pairs 
$x^2+6xy+9y^2=0$

$(x+3y)^2=0\Rightarrow x=-3y$

On comparing $3y=-x$ with $y=mx+c$ we get slope 

$m=-\dfrac{1}{3}$
Second pair of line 

$cy^2+2bxy+ax^2=0$

$y=\dfrac{-2bx\pm\sqrt{4b^2x^2-4acx^2}}{2c}$

$y=\dfrac{-2bx\pm2x\sqrt{b^2-ac}}{2c}$


$y=\left (\dfrac{-b\pm \sqrt{b^2-ac}}{c}  \right )x$

On comparing above eq with $y=mx+c$ we get 

$m _{1}=\dfrac{-b+ \sqrt{b^2-ac}}{c}$ and $m _{2}=\dfrac{-b- \sqrt{b^2-ac}}{c}$

Here one line is common in both pairs so slope will be same 

$m _{1}=-\dfrac{1}{3}$

$\dfrac{-b+ \sqrt{b^2-ac}}{c}=-\dfrac{1}{3}$

$-3b+3 \sqrt{b^2-ac}=-c$

$3 \sqrt{b^2-ac}=-c+3b$

On squaring both sides 
$9b^2-9ac=c^2+9b^2-6bc$
$9ac+c^2-6bc=0$
$c(9a-6b+c)=0$
$c=0$ and $9a-6b+c=0$

If the equation $ax^{3}+3bx^{2}y+3cxy^{2}+dy^{3}=0$ $(a, b,c, d\neq 0)$ represents three coincident lines, then 

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

  1. $a=c$

  2. $b=d$

  3. $\displaystyle {\frac{a}{b}=\frac{b}{c}=\frac{c}{d}}$

  4. $ac=bd$

Show answer
Correct Option: C
Explanation:

$ax^3+3bx^2y+3cxy^2+dy^3=0$  ---(1)   represent three coincident line

Let $ y=mx$ is that line

$(y-mx)^3=0$

$y^3+3m^2xy-3y^2(mx)-m^3n^3=0$

$m^3x^3+3mny^2-3m^2xy-y^3=0$   ---(2)

Compare 1 & 2,

$\dfrac{a}{m^3}=\dfrac{b}{m}=\dfrac{c}{-m^2}=\dfrac{d}{-1}$

$\therefore \dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}$

lf the equation of the pair of straight lines passing through the point $(1,1 )$ , one making an angle ` $\theta$' with the postive direction of x-axis and the other making the same angle with the positive direction of y-axis is $x^{2}-(a+2)xy+y^{2}+a(x+y-1)=0$, $a\neq-2$, then the value of $\sin 2\theta$ is

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

  1. $a-2$

  2. $a+2$

  3. $\frac{\displaystyle 2}{\displaystyle a+2}$

  4. $\frac{\displaystyle 2}{\displaystyle a}$

Show answer
Correct Option: C
Explanation:

Equations of the given lines are 
$y-1=\tan { \theta  } \left( x-1 \right) $ and $y-1=\cot { \theta  } \left( x-1 \right) $
Their combined equation is 
$\left( y-1-\tan { \theta  } \left( x-1 \right)  \right) \left( y-1-\cot { \theta  } \left( x-1 \right)  \right) =0\ \Rightarrow { x }^{ 2 }-\left( \tan { \theta  } +\cot { \theta  }  \right) xy+{ y }^{ 2 }+\left( \tan { \theta  } +\cot { \theta  } -2 \right) \left( x+y-1 \right) =0$
Comparing this with given equation we get
$\tan { \theta  } +\cot { \theta  } =a+2\ \Rightarrow \cfrac { 1 }{ \sin { \theta  } \cos { \theta  }  } =a+2\ \Rightarrow \sin { 2\theta  } =\cfrac { 2 }{ a+2 } $

If $P _{1},\ P _{2},\ P _{3}$ be the product of perpendiculars from $(0,0)$ to $xy+x+y+1=0$, $x^{2}-y^{2}+2x+1=0$, $2x^{2}+3xy-2y^{2}+3x+y+1=0$ respectively then?

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

  1. $P _{1} < P _{2}< P _{3}$

  2. $P _{3} < P _{2}< P _{1}$

  3. $P _{2} < P _{3}< P _{1}$

  4. $P _{1} < P _{3}< P _{2}$

Show answer
Correct Option: B
Explanation:
Given 
$xy+x+y+1=0$
comparing above eq with $ax^2+2hxy+by^2+2gx+2fy+c=0$
we get $h=\dfrac{1}{2},g=\dfrac{1}{2},f=\dfrac{1}{2},c=1$
Product of perpendicular from origin is $\left | \dfrac{c}{\sqrt{(a-b)^2+4h^2}} \right |$ 
$P _{1}=\left | \dfrac{1}{\sqrt{4\left ( \dfrac{1}{2} \right )^2}} \right |$
$P _{1}=\dfrac{1}{\sqrt{4\left ( \dfrac{1}{4} \right )}} $
$P _{1}=\dfrac{1}{\sqrt{1}} $
$P _{1}=1$
Now given eq $x^2-y^2+2x+1=0$ 
comparing above eq with $ax^2+2hxy+by^2+2gx+2fy+c=0$
we get $a=1,b=-1,h=0,g=1,f=0,c=1$
Product of perpendicular from origin is $\left | \dfrac{c}{\sqrt{(a-b)^2+4h^2}} \right |$ 
$P _{2}=\left | \dfrac{1}{\sqrt{\left ( 1+1 \right )^2}} \right |$
$P _{2}=\dfrac{1}{\sqrt{4}} $
$P _{2}=\dfrac{1}{2} $

Now given eq $2x^2+3xy-2y^2+3x+y+1=0$ 
comparing above eq with $ax^2+2hxy+by^2+2gx+2fy+c=0$
we get $a=2,b=-2,h=\dfrac{3}{2},g=\dfrac{3}{2},f=\dfrac{1}{2},c=1$
Product of perpendicular from origin is $\left | \dfrac{c}{\sqrt{(a-b)^2+4h^2}} \right |$ 
$P _{3}=\left | \dfrac{1}{\sqrt{\left ( 2+2 \right )^2-4\left ( \dfrac{3}{2} \right )^2}} \right |$
$P _{3}=\left | \dfrac{1}{\sqrt{16-4\left ( \dfrac{9}{4} \right )}} \right |$
$P _{3}=\left | \dfrac{1}{\sqrt{16-9}} \right |$
$P _{3}=\dfrac{1}{\sqrt{7}} $

SO $P _{3}< P _{2}< P _{1}$

Assertion (A): The distance between the lines represented by $x^{2}+2\sqrt{2}xy+2y^{2}+4\sqrt{2}x+4y+1=0$ is 2 
Reason (R): Distance between the lines $ax+by+c=0$ and $ax+by+c _{1}=0$ is $\displaystyle \frac{|c-c _{1}|}{\sqrt{(a^{2}+b^{2})}}$ 

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

  1. Assertion is true.

  2. Reason is true.

  3. Both Assertion and Reason are true.

  4. Neither Assertion nor Reason are true.

Show answer
Correct Option: B
Explanation:

Reason is true 
Assertion
$x^{ 2 }+2\sqrt { 2 } xy+2y^{ 2 }+4\sqrt { 2 } x+4y+1=0$ represents a pair of lines when $\Delta =0$
But $\Delta =\begin{vmatrix} 1 & \sqrt { 2 }  & 2\sqrt { 2 }  \ \sqrt { 2 }  & 4 & 1 \ 2\sqrt { 2 }  & 1 & 1 \end{vmatrix}=-2$

lf the expression $3x^{2}+2pxy+2y^{2}+2ax-4y+1$ can be resolved into two linear factors, then $p$ must be a root of the equation

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

  1. $x^{2}+ax+6=0$

  2. $x^{2}+4ax+6=0$

  3. $x^{2}+4ax+2a^{2}+6=0$

  4. $x^{2}-4ax+6=0$

Show answer
Correct Option: C
Explanation:

If $3x^{2}+2pxy+2y^{2}+2ax-4y+1=0.$, then.
$\Delta =abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$
$=3(2)(1)+2(-2)(a)(p)-3(-2)^{2}-2(a)^{2}-1(p)^{2}=0$
$6-4ap-12-2a^{2}-p^{2}=0$
$p^{2}+4ap+2a^{2}+6=0$
$\Rightarrow p $ is a solution of $x^{2}+4ax+2a^{2}+6=0$