Tag: two dimensional analytical geometry

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Questions Related to two dimensional analytical geometry

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

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

  1. $p+q=9$
  2. $p^{2}+q^{2}=0$
  3. $3p+2q=22$
  4. $4p+3q=31$
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
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
$p^{2}+q^{2}=(p+q)^{2}-2pq$
$=81-2(20)$
$=81-40$
$=41$

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

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

  1. $\displaystyle h=6, f=2 $
  2. $\displaystyle h=-6, f=-2 $
  3. $\displaystyle h=-6, f=2 $
  4. $\displaystyle h=6, f=-2 $
Reveal answer Fill a bubble to check yourself
A,B Correct answer
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 $\displaystyle \Rightarrow f=-2 \ \ \ \ \ if \ \ \ \ \ \ (h=-6)$

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

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

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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
$\left( x+y-3 \right) \left( 2x-y \right) =0$

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

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

  1. $\displaystyle \frac {1}{\sqrt 5}$
  2. $\sqrt 5$
  3. $2\sqrt 5$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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 } $

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

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

  1. $\displaystyle \sqrt{2}$
  2. $\displaystyle -\sqrt{2}$
  3. $\displaystyle \frac{1}{\sqrt{3}}$
  4. $\displaystyle -\frac{1}{\sqrt{2}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 
$m^2-9m^2+4=0$
$8m^2=4$
$m^2=\dfrac{1}{2}$
$m=\pm\dfrac{1}{\sqrt{2}}$
Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

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 $\displaystyle \frac{l^{2}+m^{2}}{n^{2}}=$

  1. $a + b$
  2. $(\mathrm{a}+\mathrm{b})^{2}$
  3. $\mathrm{a}^{2}+\mathrm{b}^{2}$
  4. $\mathrm{a}^{2}-\mathrm{b}^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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}}}$

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

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 

  1. $P _{3} < P _{2} < P _{1}$
  2. $P _{3} < P _{1} < P _{2}$
  3. $P _{2} < P _{1} < P _{3}$
  4. $P _{1} < P _{2} < P _{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 
$P _{3}<P _{2}<P _{1}$

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

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.

  1. 2k
  2. $ \frac{k}{2} $

  3. k

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the line be x cos(alpha) + y sin(alpha) = p. The sum of perpendiculars from (x1, y1) and (x2, y2) is |x1 cos(alpha) + y1 sin(alpha) - p| + |x2 cos(alpha) + y2 sin(alpha) - p| = 2k. This is a standard locus problem where the line touches a circle.

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

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

  1. Right-angled

  2. obtuse-angled isosceles

  3. acute-angled isosceles

  4. none
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

If the pair of lines $a{ x }^{ 2 }+2hxy+b{ y }^{ 2 }+2gx+2fy+c=0$ intersect on the y axis then

  1. $2fgh=b{ g }^{ 2 }+c{ h }^{ 2 }$
  2. $b{ g }^{ 2 }\neq c{ h }^{ 2 }$
  3. $abc=2fgh$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given pairs of lines

$S=ax^2+2hxy+by^2+2gx+2fy+c=0------(1)$ intersects at y-axis $x=0$ 

Let coordinate of point of intersection is $(0,y)$
To find the point of intersection: 
$\dfrac{d{S}}{d{x}}=0$

$2ax+2hy+2g=0$

Here point of intersection is $(0,y)$ SO above eq passes through $(0,y)$
$0+2hy=-2g$

$y=-\dfrac{g}{h}$
So Intersection point $\left (0,-\dfrac{g}{h} \right)$

Given pair passes through $\left (0,-\dfrac{g}{h} \right)$

$b\left( -\dfrac{g}{h}\right)^2+2f\left ( -\dfrac{g}{h} \right )+c=0$

$bg^2-2fgh+ch^2=0$

$2fgh=bg^2+ch^2$