Tag: distance formula in 2d

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

The lines will be
$y-1=\tan A(x-1)$
and $y-1=\cot A(x-1)$
Therefore their joint equation will be
$(y-1-\cot A(x-1))(y-1-\tan A(x-1))=0$
$(y-1)^{2}-(\cot A+ \tan A)(x-1)(y-1)+(x-1)^{2}=0$
$y^2-2y+1-(\cot A+\tan A)(xy-x-y+1)+(x^2-2x+1)=0$
$x^2+y^2-(\cot A+\tan A)(xy)+((\cot A+\tan A)-2)(x+y-1)=0$
Comparing coefficients we get
$\cot A+\tan A=a+2$
$\dfrac {1}{\sin A \cos A}=a+2$

$2\sin A\cos A=\dfrac{2}{a+2}$
$=\sin 2A$
$=\sin 2\theta$

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

The combined equation of two sides of an equilateral tringle is $x^{2}-3y^{2}-2x+1=0$. If the length of a side of the triangle is $4$ then the equation of the third side is

  1. $x=2\sqrt{3}+1$
  2. $y=2\sqrt{3}+1$
  3. $x+2\sqrt{3}=1$
  4. $x=2\sqrt{3}$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

$x^{2}-3y^{2}-2x+1=0$

$(x-1)^{2}=3y^{2}$

$x-1=\pm\sqrt{3}y$

Hence the equation of the sides are 

$x-\sqrt{3}y=1$ and $x+\sqrt{3}y=1$

They intersect at $(1,0)$. Hence one of the vertex will be $(1,0)$.

Now we can clearly observe that the equation of the third side will be perpendicular to x-axis and parallel to y-axis since

$x-\sqrt{3}y=1$ and $x+\sqrt{3}y=1$ are equally inclined to positive x axis- one in clockwise sense and another in anticlockwise sense, and both have the same x-intercept while equal and opposite y intercept. In other words we can imagine $x-\sqrt{3}y=1$ as the image of the line $x+\sqrt{3}y=1$ with respect to x axis.

Hence the third line will be of the form $x=c$.

Now distance of the vertex $(1,0)$ from the above line will be 

$=asin60^{0}$

$=4sin60^{0}$

$=2\sqrt{3}$.

$=\dfrac{|1-c|}{1}$

Or 

$|1-c|=2\sqrt{3}$

Hence

$c-1=2\sqrt{3}$

$c=2\sqrt{3}+1$ and $c=-2\sqrt{3}+1$

Hence the corresponding equations are 

$x=2\sqrt{3}+1$ and $x+2\sqrt{3}=1$.

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