Tag: distances and midpoints

Questions Related to distances and midpoints

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. $None\ of\ these$

Show answer
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
$p^{2}+q^{2}=(p+q)^{2}-2pq$
$=81-2(20)$
$=81-40$
$=41$

$\displaystyle 9x^{2}+2hxy+4y^{2}+6x+2fy-3=0$ represents two parallel lines if

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

  1. $\displaystyle h=6, f=2 $

  2. $\displaystyle h=-6, f=-2 $

  3. $\displaystyle h=-6, f=2 $

  4. $\displaystyle h=6, f=-2 $

Show answer
Correct Option: A,B
Explanation:

Since the given equation represents a pair of parallel lines, we have
$\displaystyle h^{2}=9 \times 4\Rightarrow h= \pm 6$
and $\displaystyle \begin{vmatrix}
9 & h & 3\ 
 h& 4 & f\ 
 3& f & -3
\end{vmatrix}=0$
$\displaystyle \Rightarrow 9\left ( -12-f^{2} \right )-h\left ( -3h-3f \right )+3\left ( hf-12 \right )=0$
$\displaystyle \Rightarrow 3h^{2}+6hf-9f^{2}-144=0$
$\displaystyle \Rightarrow 108 \pm 36f-9f^{2}-144=0 \ \ \ \left ( \because h= \pm 6 \right )$
$\displaystyle \Rightarrow 9f^{2} \mp 36f+36=0 \ \ \ \ \  (if \ \  h= \pm 6)$
$\displaystyle \Rightarrow f=2 \ \ \ \ if \ \ \ \ \ ( h=6)$
and $\displaystyle \Rightarrow f=-2 \ \ \ \ \ if \ \ \ \ \ \ (h=-6)$

Joint equation of a pair of lines passing through the point of intersection of the lines $x^{2}+xy-2y^{2}-4x+7y-5=0$ and perpendicular to these lines is

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

  1. $2x^{2}-xy+y^{2}-4x+7y-5=0$

  2. $4x^{2}-7xy-2y^{2}+6x+15y-18=0$

  3. $2x^{2}+xy-y^{2}+2x+y+5=0$

  4. none of these

Show answer
Correct Option: D
Explanation:

$x^{ 2 }+xy-2y^{ 2 }-4x+7y-5=0\ \Rightarrow \left( x-y+1 \right) \left( x+2y-5 \right) =0$
Intersection point is $\left( 1,2 \right) $
Equation of line passing through $\left( 1,2 \right) $ and perpendicular to $x-y+1=0$ is
$x+y-3=0$
And equation of line passing through $\left( 1,2 \right) $ and perpendicular to $x+2y-5=0$ is
$2x-y=0$
Hence their joint equation is
$\left( x+y-3 \right) \left( 2x-y \right) =0$

Distance between two lines respresented by the line pair, $x^2 -4xy + 4y^2 + x -2y -6 = 0$ is

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

  1. $\displaystyle \frac {1}{\sqrt 5}$

  2. $\sqrt 5$

  3. $2\sqrt 5$

  4. none of these

Show answer
Correct Option: B
Explanation:

As ${ h }^{ 2 }=4=1\times 4=ab$ and $\displaystyle b{ g }^{ 2 }=4\times \left( \frac { 1 }{ 4 }  \right) =1\times \left( 1 \right) =a{ f }^{ 2 }$
Therefore the lines in ${ x }^{ 2 }-4xy+4{ y }^{ 2 }+x-2y-6=0$ are parallel
and the distance between the two lines
$\displaystyle =\frac { 2\sqrt { { g }^{ 2 }-ac }  }{ \sqrt { a\left( a+b \right)  }  } =\frac { 2\sqrt { \dfrac { 1 }{ 4 } -1\times \left( -6 \right)  }  }{ \sqrt { 1\left( 1+4 \right)  }  } =\frac { \sqrt { 25 }  }{ \sqrt { 5 }  } =\sqrt { 5 } $

A line passes through (3, 0) The slope of the line for which its intercept between y = x - 2 and y = -x + 2 subtends a right angle at the origin may be

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

  1. $\displaystyle \sqrt{2}$

  2. $\displaystyle -\sqrt{2}$

  3. $\displaystyle \frac{1}{\sqrt{3}}$

  4. $\displaystyle -\frac{1}{\sqrt{2}}$

Show answer
Correct Option: D
Explanation:
Given line 
$y=x-2\Rightarrow x-y-2=0----(1)$
$y=-x+2\Rightarrow x+y-2=0----(2)$
On multiplying eq (1) and (2)
$(x-y-2)(x+y-2)=0$
$x^2+4-4x-y^2=0$
$x^2-y^2-4x+4=0---(3)$
Equation of line from point $(3,0)$ with slope m 
$y=mx-3m$
$1=\dfrac{mx-y}{3m}$
From eq (3)
$x^2-y^2-4x\left ( \dfrac{mx-y}{3m} \right )+4\left ( \dfrac{mx-y}{3m} \right )^2=0$

$x^2-y^2-\left ( \dfrac{4mx^2-4xy}{3m} \right )+4\left ( \dfrac{m^2x^2+y^2-2mxy}{9m^2} \right )=0$

$9m^2x^2-9m^2y^2-12m^2x^2+12mxy+4m^2x^2+4y^2-8mxy=0$

$(m^2)x^2+(-9m^2+4)y^2+4mxy=0$

Since line subtends right angle 
$m^2-9m^2+4=0$
$8m^2=4$
$m^2=\dfrac{1}{2}$
$m=\pm\dfrac{1}{\sqrt{2}}$

The line $\mathrm{l}\mathrm{x}+\mathrm{m}\mathrm{y}+\mathrm{n}=0$ intersects the curve $\mathrm{a}\mathrm{x}^{2}+2\mathrm{h}\mathrm{x}\mathrm{y}+\mathrm{b}\mathrm{y}^{2}=1$ at $\mathrm{P}$ and $\mathrm{Q}$. The circle with $\mathrm{P}\mathrm{Q}$ as diameter passes through the origin then $\displaystyle \frac{l^{2}+m^{2}}{n^{2}}=$

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

  1. $a + b$

  2. $(\mathrm{a}+\mathrm{b})^{2}$

  3. $\mathrm{a}^{2}+\mathrm{b}^{2}$

  4. $\mathrm{a}^{2}-\mathrm{b}^{2}$

Show answer
Correct Option: A
Explanation:

$lx+my+x=0$
$\displaystyle y=\frac{-x-lx}{m}$ ---1
$\therefore $$ax^{2}+2hxy+by^{2}=1$
Put $\displaystyle y=\frac{-x-lx}{m}$ in the above equation
$\displaystyle ax^{2}+2hx(\frac{-h-lx}{m})+b(\frac{h+lx}{m})^{2}=1$
$\displaystyle ax^{2}-\frac{2xhx}{m}-\frac{2hlx^{2}}{m}+\frac{bx^{2}}{m^{2}}+\frac{bl^{2}x^{2}}{m^{2}}+\frac{2bxlx}{m}=1$
$\displaystyle (a-\frac{2hl}{m}+\frac{bl^{2}}{m^{2}})x^{2}+(\frac{2bxl}{m}-\frac{2xh}{m})x+\frac{bx^{2}}{m^{2}}-1=0$
$\displaystyle \therefore $$x _{2}+x _{2}=\displaystyle \dfrac{\dfrac{2xh-2bxl}{m}}{a-\dfrac{2hl}{m}+\dfrac{bl^{2}}{m^{2}}}$

Let $P _{1},\ P _{2},\ P _{3}$ be the perpendicular distances between pair of parallel lines represented by $x^{2}-3x-4=0$, $y^{2}-5y+6=0$, $4x^{2}+20xy+25y^{2}=0$ respectively then 

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

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

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

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

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

Show answer
Correct Option: A
Explanation:

$x^{2}-3x-4=0$
$(x-4)(x+1)=0$

$x=4$ and $x=-1$
Hence the perpendicular distance between these two lines 
$P _{1}=4-(-1)=5$.

$y^{2}-5y+6=0$
$(y-2)(y-3)=0$
$y=2$ and $y=3$
Hence perpendicular distance between these lines is 
$P _{2}=3-2=1$
Thus $P _{2}<P _{1}$ 

$4x^{2}+20xy+25y^{2}=0$
$x=\dfrac{-20y\pm\sqrt{400y^{2}-400y^{2}}}{8}$

$x=\dfrac{-20y}{8}$
Or 
$8x+20y=0$
$2x+5y=0$
Since we get a single line 
$P _{3}=0$
Therefore 
$P _{3}<P _{2}<P _{1}$

A straight lines moves such that the algebraic sum of the perpendicular drawn to it from two fixed points is equal to 2k than, the straight line always touches a fixed circle of radius.

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

  1. 2k

  2. $ \frac{k}{2} $

  3. k

  4. None of these

Show answer
Correct Option: C

Lines $x+y=4$, $3x+y=4$, $x+3y=4$ from a triangle which is

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

  1. Right-angled

  2. obtuse-angled isosceles

  3. acute-angled isosceles

  4. none

Show answer
Correct Option: C
Explanation:

Given

$x+y=4$........(1)
$3x+y=4$.......(2)
$x+3y=4$........(3)
solving (1) and (2)
$3x+y=4\ x+y=4\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ 2x=0$
$\therefore x=0\quad y=4\quad A=(0, 4)$
solving (1) and (3)
$x+3y=4\ x+y=4\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ 2y=0$
$\therefore y=0\quad x=4\quad B=(4, 0)$
solving (2) and (3)
$3x+y=4\ 3x+9y=12\ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ -8y=-8$
$\therefore y=1\quad x=1\quad C=(1, 1)$
Distance between $A$ & $B$
$=\sqrt{(0-4)^2+(4-0)^2}$
$=\sqrt{4^2+4^2}$
$=\sqrt{32}=4\sqrt 2$
= Distance between $A$ & $C$
$\sqrt{(0-1)^2+(4-1)^2}$
$=\sqrt{1^2+3^2}$
$=\sqrt {10}$
Distance between $B$ & $C$
$\sqrt{(4-1)^2+(0-1)^2}$
$=\sqrt{10}$
As $AC=BC$
So, $\triangle ABC$, is isosceles triangle
using pythagoras theorem,
$AC^2+BC^2<AB^2$
as $10+10=20 < (4\sqrt 2)^2$
$20<32$
So $\triangle ABC$ is acute $\angle$  ed isosceles triangle.