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

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 

  1. $mn+1=0$
  2. $mn-1=0$
  3. $1/m+1/n=0$
  4. $1/m -1/n=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The angle bisectors of ax^2 + 2hxy + by^2 = 0 are given by (x^2-y^2)/(a-b) = xy/h. For x^2-2mxy-y^2=0 and x^2-2nxy-y^2=0, the bisectors of the first are (x^2-y^2)/(1-(-1)) = xy/(-m), which simplifies to x^2-y^2 = -2xy/m, or x^2 + (2/m)xy - y^2 = 0. Comparing this to the second equation, -2n = 2/m, so mn = -1, or mn+1=0.

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

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

  1. $a -2$
  2. $a + 2$
  3. $\dfrac2{(a + 2)}$
  4. $ \dfrac2a$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $\tan \theta + \cot \theta= a + 2$

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

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

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

  1. $-2$
  2. $\dfrac{1}{2}$
  3. $2$
  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $m _{2}=\dfrac{(\tan\theta-\sin^2\theta)}{\sin^2\theta}$

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

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

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

$\Rightarrow m _{1}-m _{2}=2$
Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

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

  1. $1$
  2. $3$
  3. $0$
  4. $2$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $9a-6b+c=0$
Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

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

  1. $a=c$
  2. $b=d$
  3. $\displaystyle {\frac{a}{b}=\frac{b}{c}=\frac{c}{d}}$
  4. $ac=bd$
Reveal answer Fill a bubble to check yourself
C Correct answer
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,

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

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

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

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

  1. $a-2$
  2. $a+2$
  3. $\frac{\displaystyle 2}{\displaystyle a+2}$
  4. $\frac{\displaystyle 2}{\displaystyle a}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 } $

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

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?

  1. $P _{1} < P _{2}< P _{3}$
  2. $P _{3} < P _{2}< P _{1}$
  3. $P _{2} < P _{3}< P _{1}$
  4. $P _{1} < P _{3}< P _{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $P _{3}< P _{2}< P _{1}$
Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

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 $\displaystyle \frac{|c-c _{1}|}{\sqrt{(a^{2}+b^{2})}}$ 

  1. Assertion is true.

  2. Reason is true.

  3. Both Assertion and Reason are true.

  4. Neither Assertion nor Reason are true.
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\Delta =\begin{vmatrix} 1 & \sqrt { 2 }  & 2\sqrt { 2 }  \ \sqrt { 2 }  & 4 & 1 \ 2\sqrt { 2 }  & 1 & 1 \end{vmatrix}=-2$

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

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

  1. $x^{2}+ax+6=0$
  2. $x^{2}+4ax+6=0$
  3. $x^{2}+4ax+2a^{2}+6=0$
  4. $x^{2}-4ax+6=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $x^{2}+4ax+2a^{2}+6=0$