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

The four sides of a quadrilateral are given by equ. $(xy+12-4x-4y{ ) }^{ 2 }=(2x-2y{ ) }^{ 2 }$. The equation of a line with slope $\sqrt { 3 } $ which divides the area of the quadrilateral in two equal parts is 

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

The equation (xy + 12 - 4x - 4y)^2 = (2x - 2y)^2 factors into (xy + 12 - 4x - 4y - 2x + 2y)(xy + 12 - 4x - 4y + 2x - 2y) = 0. This simplifies to (x-2)(y-6) = 0 and (x-6)(y-2) = 0, representing a rectangle with vertices (2,2), (6,2), (6,6), and (2,6). The center of this rectangle is (4,4), and any line passing through the center bisects the area.

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

If the equation  $2 x ^ { 2 } + 3 x y + b y ^ { 2 } - 11 x + 13 y + c = 0$  represents two perpendicular straight lines, then

  1. $b = - 2$
  2. $b = 2$
  3. $c = - 2$
  4. $c = 2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a general equation ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 to represent perpendicular lines, the sum of the coefficients of x^2 and y^2 must be zero. Thus, a + b = 0. Given a = 2, we have 2 + b = 0, so b = -2.

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

The product of the perpendiculars from origin to the pair of lines ${ ax }^{ 2 }+2hxy+{ by }^{ 2 }+2gx+2fy+c=0$ is


  1. $\frac { \left| c \right| }{ \sqrt { \left( { a+b } \right) ^{ 2 } } +{ 4h }^{ 2 } } $
  2. $\frac { \left| c \right| }{ \sqrt { \left( a+b \right) ^{ 2 }-{ 4h }^{ 2 } } } $
  3. $\frac { \left| c \right| }{ \sqrt { \left( a-b \right) ^{ 2 }-{ 4h }^{ 2 } } } $
  4. $\frac { \left| c \right| }{ \sqrt { \left( a-b \right) ^{ 2 }-{ 4h }^{ 2 } } } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice maths two dimensional analytical geometry pair of straight lines distances and midpoints distance formula in 2d

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

  1. $2fgh= bg^{2}+ch^{2}$
  2. $bg^{2}\neq ch^{2}$
  3. $abc= 2fgh$

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

As $\displaystyle s=ax^{2}+2hxy +by^{2}+2gx +2fy+c=0 $ represent a pair of line $\displaystyle \therefore \begin{vmatrix}a &h  &g \h  &b  &f \g  &f  &c \end{vmatrix}=0$
or $\displaystyle abc+2fgh-af^{2}-bg^{2}-ch^{2}=0....(1)$ Now say point ofintersection onY axis be $\displaystyle (0,y _{1} $ and  point of intersection of pair of line be obtained by solving the equations $\displaystyle \frac{\partial s}{\partial x}=0=\frac{\partial s}{\partial y}$ $\displaystyle \therefore \frac{\partial s}{\partial x}=0\Rightarrow ax+by+g=0$ $\displaystyle \begin{matrix}\Rightarrow  \ \Rightarrow  \end{matrix} \left{\begin{matrix}hy _{1}+g=0 \by _{1}+f=0 \end{matrix}\right.>  ()$ and $\displaystyle \frac{\partial s}{\partial y}=0\Rightarrow bx+by+f=0$  On compairing the equation given in () we get $\displaystyle bg=fh $ and  $\displaystyle bg^{2}=fgh ....(2) $ Again $\displaystyle ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ meet at y-axis $\displaystyle \therefore x=0$ $\displaystyle \Rightarrow by^{2}+2fy+c=0$ whose roots must be equal $\displaystyle \Rightarrow by^{2}+2fy+c=0$ whose roots must be equal $\displaystyle \therefore f^{2}=bc af^{2}=abc ......(3)$ Now using (2) and (3) in equation (I) we have $\displaystyle abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$ $\displaystyle \Rightarrow (abc-af^{2})+(fgh-bg^{2})+fgh-ch^{2}=0$ $\displaystyle \Rightarrow 0+0+fgh-ch^{2}=0 \therefore ch^{2}=fgh .....(4) $ Now adding (2) and (4) $\displaystyle 2fgh=ch^{2}+bg^{2}$

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

A line is at distance of $4$ units from origin and having both intercepts positive. If the perpendicular from the origin to this line makes an angle of ${60}^{o}$ with the line $x+y=0$ Then the equation of the line is

  1. $\left( \sqrt { 3 } +1 \right) x+\left( \sqrt { 3 } +2 \right) y=y=8\sqrt { 2 } $
  2. $\left( \sqrt { 3 } -1 \right) x+\left( \sqrt { 3 } +1 \right) y=y=8\sqrt { 2 } $
  3. $\left( \sqrt { 3 } +1 \right) x-\left( \sqrt { 3 } +1 \right) y=8\sqrt { 2 } $
  4. $\left( \sqrt { 3 } +2 \right) x+\left( \sqrt { 3 } +1 \right) y=8\sqrt { 2 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The line is at distance 4 from the origin. Using the normal form x cos(theta) + y sin(theta) = 4, and the condition that the normal makes 60 degrees with x+y=0 (which has a normal vector (1,1) at 45 degrees), the angle of the normal is 45 +/- 60 degrees. Calculating the intercepts and checking the positive intercept condition leads to the correct equation.

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

If one of the lines given by $6x^{2}-xy+4cy^{2}= 0$ is $3x+4y= 0$, then $c$ equals

  1. 3

  2. -1

  3. 1

  4. -3
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equation $\displaystyle ax^{2}+2hxy+by^{2}=0$

$\displaystyle  =(y-m _{1}x)(y-m _{2}x) $

$\displaystyle  \Rightarrow m _{1}+m _{2}=-\frac{2h}{b}=\frac{1}{4c }$.....(1) 

$\displaystyle  \Rightarrow m _{ _1}m _{2} =\frac{3}{2c} $
and $\displaystyle  3x+4y=0\Rightarrow m _{1} =-\dfrac {3}{4}$
 $\displaystyle \therefore m _{2} =-\frac{2}{c} $

 Now by $\displaystyle (1)$ we have $\displaystyle-\left(\frac{3}{4}+\frac{2}{c}\right) =\frac{1}{4c} $

$\displaystyle  \Rightarrow -\frac{3}{4}=\frac{1}{4c}+\frac{2}{c},\frac{3}{4}=\frac{1}{4c}+\frac{8}{4c}$ $\displaystyle  -\frac{3}{4}=\frac{9}{4c}$

$ \therefore c=-3$

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

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 $2x+7y-1=0$)

  1. $9x+12y+3=0$
  2. $9x+12y-3=0$
  3. $9x+12y-2=0$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Find points A and B by intersecting the line with L1 and L2. Use the section formula to find point P on the line 2x+7y-1=0 such that AP:PB=2:1. The required line is parallel to L1 and L2 (3x+4y+c=0) and passes through P.