Tag: distances and midpoints

Questions Related to distances and midpoints

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

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

Show answer
Correct Option: C
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
$\cot A+\tan A=a+2$
$\dfrac {1}{\sin A \cos A}=a+2$

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

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

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

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

  2. $y=2\sqrt{3}+1$

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

  4. $x=2\sqrt{3}$

Show answer
Correct Option: A,C
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 $x+2\sqrt{3}=1$.

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