Tag: transformation and symmetry in geometrical shapes

Quizzes (2) Questions (65)

Questions Related to transformation and symmetry in geometrical shapes

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

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

  1. $7y+x-6=0$
  2. $7y-x-6=0$
  3. $x+7y=8$
  4. $7y-x+6=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.
Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

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 :-

  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) $
Reveal answer Fill a bubble to check yourself
B Correct answer
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) $

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

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:

  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. $(-2-\sqrt{2},-1-2\sqrt{2})$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. $6$
  2. $3$
  3. $8$
  4. $7$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $



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

${A}$ 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 $\tan\alpha=$ 

  1. $\displaystyle \frac{{a}+b}{{a}-b}$
  2. $\displaystyle \frac{{a}-b}{{a}+b}$
  3. $\dfrac{b}{a}$
  4. $\displaystyle \frac{{a}}{b}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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}$
Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

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 

  1. $\dfrac{\pi}{6}$
  2. $\dfrac{\pi}{4}$
  3. $\dfrac{\pi}{3}$
  4. $\dfrac{\pi}{12}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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}$
Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

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

  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. $\displaystyle \dfrac{-5}{3}x$
Reveal answer Fill a bubble to check yourself
B Correct answer
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} $

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

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 

  1. $15^{\mathrm{o}}$
  2. $30^{\mathrm{o}}$
  3. $45^{0}$
  4. $60^{\mathrm{o}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $= 30^{\circ}$
Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

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.

  1. $2$
  2. $5$
  3. $6$
  4. $7$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.