Tag: position and movement

Questions Related to position and movement

The line $3x-4y+7=0$ is rotated through an angle $\dfrac {\pi}{4}$ in clockwise direction about the point $\left (1,1\right)$. The equation of the line in its new position is

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $7y+x-6=0$

  2. $7y-x-6=0$

  3. $x+7y=8$

  4. $7y-x+6=0$

Show answer
Correct Option: A
Explanation:

Given slope $=\dfrac{3}{4} < 1$
$< 45^{o}$
$\therefore $ after rational clock slope because negative
$\dfrac{m _1-m _{2}}{1+m _{1},m _{2}}=\tan 45$
$m _{1}=\dfrac{3}{4} \,\,\,m _{2}=$ new slope
$\left| \dfrac{\dfrac{3}{4}-m}{1+\dfrac{3}{4} m} \right|=1$
$\dfrac{3}{4}-m= \pm \left( 1+\dfrac{3}{4}m \right)$
$\dfrac{3}{4}-m=1+ \dfrac{3}{4} m$
$\dfrac{7}{4}m=\dfrac{-1}{4}$
$m=\dfrac{-1}{7}$ appeared 
$\therefore$ satisfy slope and pt in option to save time.

Let $\displaystyle A\equiv \left( 2,0 \right) $ and $\displaystyle B\equiv \left( 3,1 \right) $. The line $\displaystyle AB$ turns about $\displaystyle A$ through an angle $\displaystyle \frac { \pi  }{ 12 } $ in the clockwise sense, and the new position of $\displaystyle B$ is $\displaystyle B'$. Then $\displaystyle B'$ has the co-ordinates :-

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $\displaystyle \left( \frac { 2\sqrt { 2 } -\sqrt { 3 } }{ \sqrt { 2 } } ,\frac { 1 }{ \sqrt { 2 } } \right) $

  2. $\displaystyle \left( \frac { 2\sqrt { 2 } +\sqrt { 3 } }{ \sqrt { 2 } } ,\frac { 1 }{ \sqrt { 2 } } \right) $

  3. $\displaystyle \left( \frac { \sqrt { 3 } -2\sqrt { 2 } }{ \sqrt { 2 } } ,\frac { 1 }{ \sqrt { 2 } } \right) $

  4. $\displaystyle \left( \frac { \sqrt { 3 } -2\sqrt { 2 } }{ 2 } ,\frac { 1 }{ \sqrt { 2 } } \right) $

Show answer
Correct Option: B
Explanation:

Slope of the line $\displaystyle AB=\frac { 0-1 }{ 2-3 } =1$
$\therefore \angle BAX={ 45 }^{ o }$
Given $\angle B'AB={ 15 }^{ o }\Rightarrow \angle B'AX={ 30 }^{ o }$
Therefore slope of the line $\displaystyle AB'=\tan { { 30 }^{ o } } =\frac { 1 }{ \sqrt { 3 }  } $
Now line $AB'$ makes an angle of ${ 30 }^{ o }$ with positive direction of $x$-axis and 
$AB'=AB=\sqrt { { \left( 3-2 \right)  }^{ 2 }+{ \left( 1-0 \right)  }^{ 2 } } =\sqrt { 2 } $
Therefore coordinates are $\displaystyle \left( 2+\sqrt { 2 } \cos { { 30 }^{ o } } ,0+\sqrt { 2 } \sin { { 30 }^{ o } }  \right) =\left( \frac { 2\sqrt { 2 } +\sqrt { 3 }  }{ \sqrt { 2 }  } ,\frac { 1 }{ \sqrt { 2 }  }  \right) $

Without changing the direction of coordinates axes,  origin is transferred to $(\alpha ,\ \beta)$ so that linear term in the equation $x^{2}+y^{2}+2x-4y+6=0$ are eliminated the point $(\alpha ,\ \beta)$ is   

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $(-1,\ 2)$

  2. $(1,\ -2)$

  3. $(1,\ 2)$

  4. $(-1,\ -2)$

Show answer
Correct Option: B

The point $\mathrm{A}(2,1)$ is translated parallel to the line $x-y=3$ by a distance $4$ units. If the new position $A'$ is in third quadrant, then the coordinates of $A'$ are:

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $(2+2\sqrt{2},2+2\sqrt{2})$

  2. $(-2+\sqrt{2},-1-2\sqrt{2})$

  3. $(2-2\sqrt{2},1-2\sqrt{2})$

  4. $(-2-\sqrt{2},-1-2\sqrt{2})$

Show answer
Correct Option: C
Explanation:

$ y = x-3$ $\Rightarrow m=1$ ; $tan \theta =1$
By parametrization we have $x=2 \pm r cos\theta $
$y=1 \pm r sin\theta $

$x =2 \pm 4\times \dfrac{1}{\sqrt{2}} ;\ y=1 \pm 4\times \dfrac{1}{\sqrt{2}}$.....(consider -ive sign for third quadrant)
$x=2-2\sqrt{2};\ y=1-2\sqrt{2}$ since they are in third quadrant.

If the points $(5, 5), (7, 7)$ and $(a, 8)$ are collinear then the value of a is

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $6$

  2. $3$

  3. $8$

  4. $7$

Show answer
Correct Option: C
Explanation:

When three points are collinear, Slope of line joining any two points is same as the slope of line joining any other two points


Slope of line joining two points $ ({x} _{1}, {y} _{1}) $ and $ ({x} _{2}, {y} _{2}) $ is $\dfrac { {y} _{2} - {y} _{1}}{ {x} _{2} - {x} _{1}} $

So, Slope of line joining $ (5,5) ;  (7,7) $ is $ \dfrac {7-5}{7-5} = \dfrac {2}{2} = 1 $

And Slope of line joining $ (a,8) ;  (7,7) $ is $ \dfrac {7-a}{7-8} = a - 7 $

As they are collinear $ a - 7 = 1 => a = 8 $



${A}$ line has intercepts $ a$ and ${b}$ on the co ordinate axes. When the axes are rotated through an angle $\alpha$, keeping the origin fixed, the line makes equal intercepts on the coordinate axes, then $\tan\alpha=$ 

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $\displaystyle \frac{{a}+b}{{a}-b}$

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

  3. $\dfrac{b}{a}$

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

Show answer
Correct Option: B
Explanation:
Let the equation of line be $ \displaystyle \frac{x}{a}+\frac{y}{b}=1.$

When axes are rotated through an angle $ \alpha$, the new coordinates $XY$ are related to old coordinates $xy$ as follows:

$x=X \cos\alpha -Y\sin \alpha $

$y=X \sin\alpha +Y\ \cos \alpha $

Substituting these values in the equation of line, we get
$ \displaystyle \frac { X\cos { \alpha  } -Y\sin { \alpha  }  }{ a } +\frac { X\sin { \alpha  } +Y\cos { \alpha  }  }{ b } =1\\ \displaystyle \Rightarrow X\left( \frac { \cos { \alpha  }  }{ a } +\frac { \sin { \alpha  }  }{ b }  \right) +Y\left( \frac { \cos { \alpha  }  }{ b } -\frac { \sin { \alpha  }  }{ a }  \right) =1$

As it makes equal intercepts in the new coordinate system, we get

$ \displaystyle \Rightarrow \frac{\cos \alpha }{a}+\frac{\sin \alpha }{b}=\frac{\cos \alpha }{b}-\frac{\sin \alpha }{a}$

$\Rightarrow \cos \alpha \left ( b-a \right )=-\sin \alpha \left ( a+b \right )$

$ \displaystyle \Rightarrow \tan  \alpha =\dfrac{a-b}{a+b}$

The angle of rotation of the axes so that the equation $\sqrt{3}\mathrm{x}-\mathrm{y}+5=0$ may be reduced to the form $\mathrm{Y}=\mathrm{k}$, where $\mathrm{k}$ is a constant is 

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $\dfrac{\pi}{6}$

  2. $\dfrac{\pi}{4}$

  3. $\dfrac{\pi}{3}$

  4. $\dfrac{\pi}{12}$

Show answer
Correct Option: C
Explanation:
Let axis be rotated through an angle $\theta $ then 
$ x= x^{1} \cos \theta - y^{1} \sin\theta $
$ y = x^{1} \sin \theta + y^{1} \cos\theta $
$ \sqrt{3}\times x -y +5 = 0$
$ \sqrt{3} (x^{1} \cos \theta - y^{1} \sin\theta) - (x^{1} \sin \theta + y^{1} \cos\theta) +5 = 0$
$x^{1} (\sqrt{3} \cos \theta - \sin\theta ) = 0$
$ \Rightarrow \tan\theta  = \sqrt{3}$
$ \theta = 60^{\circ} = \dfrac{\pi }{3}$

Find the equation of a line whose inclination is $\displaystyle 30^{\circ}$ and making an intercept of -3/5 on the y-axis

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $ y = \dfrac {5}{\sqrt{6}}x  +\dfrac {2}{5} $

  2. $ y = \dfrac {1}{\sqrt{3}}x  -\dfrac {3}{5} $

  3. $ y = \dfrac {3}{\sqrt{7}}x  -\dfrac {1}{3} $

  4. $\displaystyle \dfrac{-5}{3}x$

Show answer
Correct Option: B
Explanation:

The equation of any straight line can be written as $ y = mx + c $, where $m$ is its slope and $c$ is its y - intercept.

As inclination is $ 30^o $, slope of the line $ =  tan (30 ^o) = \dfrac {1}{\sqrt{3}} $

So equation of line is $ y = \dfrac {1}{\sqrt{3}}x  -\dfrac {3}{5} $

lf the equation $4\mathrm{x}^{2}+2\sqrt{3}\mathrm{x}\mathrm{y}+2\mathrm{y}^{2}-1=0$ becomes $5\mathrm{X}^{2}+\mathrm{Y}^{2}=1$, when the axes are rotated through an angle $\theta$, then $\theta$ is 

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $15^{\mathrm{o}}$

  2. $30^{\mathrm{o}}$

  3. $45^{0}$

  4. $60^{\mathrm{o}}$

Show answer
Correct Option: B
Explanation:
By  rotation  of  axes  through  $\theta $,  co-ordinates  become
$x = x^{1}  \cos\theta  - y^{1}  \sin\theta $
$y = x^{1}  \sin\theta  + y^{1}  \cos\theta $
$4x^{2} + 2\sqrt{3}xy + 2y^{2} - 1=0$
$\Rightarrow 4(x^{1}\cos\theta - y^{1} \sin\theta )^{2} + 2\sqrt{3} (x^{1} \cos\theta  -y^{1} \sin\theta )  (x^{1} \sin\theta + y^{1} \cos\theta )  +2 (x^{1}\sin\theta +y^{1} \cos\theta )^{2} -1 =0$
coeff  of $xy=0$
$\Rightarrow 4(-\sin2\theta )+2\sqrt{3}   \cos20  +  2 \sin2\theta  = 0$
$2\sqrt{3}   \cos2\theta  = 2\sin2\theta $
$\tan2\theta = \sqrt{3}$
$2\theta = \dfrac{\pi }{3}$
$\theta  = \dfrac{\pi }{6} = 30^{\circ}$
$\therefore $ angle  to  be  rotated $= 30^{\circ}$

lf the distance between two given points is $2$ units and the points are transferred by shifting the origin to $(2, 2)$, then the distance between the points in their new position is.

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $2$

  2. $5$

  3. $6$

  4. $7$

Show answer
Correct Option: A
Explanation:

Shifting the origin to $(2 , 2)$ transforms the coordinates to
$(x , y)$ to $(x - 2, y - 2)$.

$\therefore$ distance between two points is

$\sqrt{(x _{1}-x _{2})^{2}+(y _{1}-y _{2})^{2}}$

$=\sqrt{((x _{1}-2)-(x _{2}-2))^{2}+((y _{1}-2)-(y _{2}-2))}^{2}$

$=\sqrt{(x _{1}-x _{2})^{2}+(y _{1}-y _{2})}^{2}$

$\therefore$ distance is unaltered $ = 2$ units.