Tag: transformation and symmetry in geometrical shapes

Questions Related to transformation and symmetry in geometrical shapes

Let $0<\alpha< \dfrac{\pi}{4}$ be a fixed angle. If $\mathrm{P}=(\cos\theta,\sin\theta)$ and $\mathrm{Q}=(\cos(\alpha-\theta),\sin(\alpha-\theta))$ then $\mathrm{Q}$ is obtained from $\mathrm{P}$ by :

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

  1. clockwise rotation around the origin through an angle $\alpha$

  2. anticlockwise rotation around the origin through an angle $\alpha$

  3. reflection in the line through origin with slope $\tan\alpha$

  4. reflection in the line through origin with slope $\displaystyle \tan\frac{\alpha}{2}$

Show answer
Correct Option: D
Explanation:
$P =\left(\cos \theta, \sin \theta\right)$ ; $Q=\left(\cos \left(\alpha -\theta \right),\sin \left(\alpha -\theta \right)\right)$

Angle between $P$ and $Q$ is $\tan^{-1}$  $\left(\dfrac{\sin\left(\alpha -\theta \right)-\sin\alpha }{\cos\left(\alpha -\theta \right)-\cos\alpha }  \right)$

$=\tan^{-1}\left( \tan \alpha  \right)=\alpha $

$\therefore$ mid point of $P$ and $Q$ is $ \left(\dfrac{\cos\theta + \cos(\alpha -\theta)}{2}, \dfrac{\sin\theta +\sin( \alpha -\theta)}{2}\right)$

$= \left(  \cos\dfrac{\alpha }{2} \cos\left(\dfrac{\theta-\alpha }{2}\right), \sin\dfrac{\alpha }{2}\cos\left(\dfrac{\theta-\alpha }{2}\right) \right)$

$= \cos\left(\dfrac {\theta -\alpha }{2}\right)\left( \cos\dfrac{\alpha }{2},\sin\dfrac{\alpha }{2} \right)$

Which is a point on line with slope $y=\tan\dfrac{\alpha }{2}$

$\therefore$ $Q$ is obtained by reflection of origin with slope $\tan$ $\dfrac{\alpha }{2}$.
Hence, option 'D' is correct.


The point $(4, 1)$ undergoes the following three transformations successively
i) Reflection about the line $\mathrm{y}=\mathrm{x}$
ii) Transformation through a distance of $2$ units along the $+\mathrm{v}\mathrm{e}$ direction of the x-axis
iii) Rotation through an angle $\displaystyle \frac{\pi}{4}$ about the origin in the anticlockwise direction. The final position of the point is given by the co-ordinates 

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

  1. $\left(\displaystyle \frac{-1}{\sqrt{2}}\frac{7}{\sqrt{2}}\right)$

  2. $(-2,7\sqrt{2})$

  3. $\left(\displaystyle \frac{7}{\sqrt{2}}\frac{1}{\sqrt{2}}\right)$

  4. $(7, 1)$

Show answer
Correct Option: C
Explanation:

Under the transformation reflection about $y = x$ the new point is $(1,4)$
Under the transformation of shifting it $+2$ units along x-axis the new point is $(1+2,4) = (3,4)$
Under the transformation of rotation of $\dfrac{\pi}{4}$
$y= -X \sin  \theta +y\cos\ \theta$

$X=\dfrac{3}{\sqrt{2}}+\dfrac{4}{\sqrt{2}}=\dfrac{7}{\sqrt{2}}$

$y=\dfrac{3}{\sqrt{2}}+\dfrac{4}{\sqrt{2}}=\dfrac{1}{\sqrt{2}}$

The image of the origin with reference to the line $4x + 3y - 25 = 0$, is

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

  1. $(-8, 6)$

  2. $(8, 6)$

  3. $(-3, 4)$

  4. $(8, -6)$

Show answer
Correct Option: B
Explanation:

Let the image or (reflection) of the origin with reference to the line $4x + 3y - 25 = 0$ is $(h, k)$.
$\therefore \dfrac {h - 0}{4} = \dfrac {k - 0}{3} = \dfrac {-2(0 + 0 - 25)}{16 + 9} = \dfrac {50}{25} = 2$
$\therefore \dfrac {h}{4} = 2\Rightarrow h = 8$
and $\dfrac {k}{3} = 2\Rightarrow k = 6$
$\therefore$ The required point is $(8, 6)$.

A light ray gets reflected from the $ x= -2 $ .If the reflected ray touches the circle $ x^{2}+y^{2}=4 $ and point of incident is $(-2,-4)$,then equation of incident ray is

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

  1. $ 4y+3x+22=0 $

  2. $ 3y+4x+20=0 $

  3. $ 4y+2x+20=0 $

  4. $ y+x+6=0 $

Show answer
Correct Option: A

The image of the point $(3, 8)$ with respect to the line $x + 3y = 7$ is

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

  1. $(-1, -4)$

  2. $(-1, 4)$

  3. $(1, -4)$

  4. $(1, 4)$

Show answer
Correct Option: A
Explanation:

Image of any Point $(\alpha , \beta)$ with respect to line $ax+by +c = 0$ is given by equation,


$\Rightarrow \dfrac{x -\alpha}{a} = \dfrac{y - \beta}{b} = \dfrac{-2(a\alpha + b \beta + c )}{a^2 + b^2}$  ....$(1)$

Now the given point is $(3,8)$ and the given line is $x + 3y - 7 = 0$

Putting the values in equation  $(1)$, we get,

$\Rightarrow \dfrac{x -3}{1} = \dfrac{y - 8} {3} = \dfrac{ -2(1.3 + 3.8 -7 )}{1^2 + 3^2} = -4$

Hence $\Rightarrow x= -1$ and $y = -4$

Correct option is $A$.

 The point $(4, 1)$ undergoes the following three transformations successively
(a) Reflection about the line $y = x$

(b) Transformation through a distance $2$ units along the positive direction of the x-axis.

(c) Rotation through an angle $p/4$ about the origin in the anti clockwise direction.

The final position of the point is given by the co-ordinates

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

  1. $\left(\dfrac{4}{\sqrt{2}} , \dfrac{1}{\sqrt{2}}\right)$

  2. $\left(-\dfrac{1}{\sqrt 2} , \dfrac{7}{\sqrt 2}\right)$

  3. $\left(\dfrac{1}{\sqrt{2}} , \dfrac{7}{\sqrt{2}}\right)$

  4. $\left(-\dfrac{3}{\sqrt{2}} , \dfrac{4}{\sqrt{2}}\right)$
Show answer
Correct Option: B
Explanation:

Given Point $P(4,1)$


$(a)$ Reflection of given point about the line $y=x$
In Point P $x=4$ Hence about the line y-coordinate be $y=x\Rightarrow y=4$
Point P become $P(1,4)$

$(b)$ Transformation of Point P through distance $2$ units along the positive direction of the x-axis
$P(1+2,4)\equiv P(3,4)$

$(c)$Rotation of Point P through angle $\dfrac{\pi}{4}$ about origin in anti clockwise direction
$P\left (  r\cos\left ( \alpha+\dfrac{\pi}{4} \right ),r\sin\left ( \alpha+\dfrac{\pi}{4} \right )\right )$

Given point $P(3,4)$
$r=\sqrt{3^2+4^2}=5$
$\tan\alpha=\dfrac{4}{3}$
Hence In right angle triangle here hypotenuse be $5$
$\cos\alpha=\dfrac{3}{5}$ and $\sin\alpha=\dfrac{4}{5}$

$\cos\left ( \alpha+\dfrac{\pi}{4} \right )=\cos\alpha\cos\dfrac{\pi}{4}-\sin\alpha\sin\dfrac{\pi}{4}$

$\cos\left ( \alpha+\dfrac{\pi}{4} \right )=\dfrac{3}{5}\times \dfrac{1}{\sqrt{2}}-\dfrac{4}{5}\times \dfrac{1}{\sqrt{2}}=-\dfrac{1}{5\sqrt{2}}$

$r\cos\left ( \alpha+\dfrac{\pi}{4}\right )=5\times \dfrac{-1}{5\sqrt{2}}=-\dfrac{1}{\sqrt{2}}$

$\sin\left ( \alpha+\dfrac{\pi}{4} \right )=\sin\alpha\cos\dfrac{\pi}{4}+\cos\alpha\sin\dfrac{\pi}{4}$

$\sin\left ( \alpha+\dfrac{\pi}{4} \right )=\dfrac{4}{5}\times \dfrac{1}{\sqrt{2}}+\dfrac{3}{5}\times \dfrac{1}{\sqrt{2}}=\dfrac{7}{5\sqrt{2}}$

$r\sin\left ( \alpha+\dfrac{\pi}{4}\right )=5\times \dfrac{7}{5\sqrt{2}}=\dfrac{7}{\sqrt{2}}$


Point be 

$P\left (  -\dfrac{1}{\sqrt{2}},\dfrac{7}{\sqrt{2}}\right )$


Image of the point $\left( -8,12 \right) $ with respect to the line mirror $4x+7y+13=0$ is 

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

  1. $\left( 16,2 \right) $

  2. $\left( -16,-2 \right) $

  3. $\left( -12,5 \right) $

  4. $\left( 12,-5 \right) $

Show answer
Correct Option: B
Explanation:

Here, $(x _1,y _1)$ is $(-8,12)$


Let $(\alpha,\beta)$ be image of the point.


$a=4,\,b=7$ and $c=13$

$\Rightarrow$  $\dfrac{\alpha+8}{4}=\dfrac{\beta-12}{7}=\dfrac{-2[4(-8)+7(12)+13]}{(4)^2+(7)^2}$

$\Rightarrow$  $\dfrac{\alpha+8}{4}=\dfrac{\beta-12}{7}=\dfrac{-2(-32+84+13)}{16+49}$

$\Rightarrow$  $\dfrac{\alpha+8}{4}=\dfrac{\beta-12}{7}=\dfrac{-2(65)}{65}$

$\Rightarrow$  $\dfrac{\alpha+8}{4}=\dfrac{\beta-12}{7}=-2$

$\Rightarrow$ $\dfrac{\alpha+8}{4}=-2$  and  $\dfrac{\beta-12}{7}=-2$

$\Rightarrow$  $\alpha+8=-8$   and  $\beta-12=-14$

$\Rightarrow$  $\alpha=-16$ and $\beta=-2$

So, the image of the point is $(-16,-2).$

Equation of line equidistant from lines $2x + 3y = 5$ and $4x + 6y = 11$ is

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

  1. $(x + y - 3)(x - y - 1) = 0$

  2. $8x + 12y = 1$

  3. $x - y = 2$

  4. None

Show answer
Correct Option: B

A ray of light along $x+\sqrt{3}y=\sqrt{3}$ get reflected upon reaching x-axis, the equation of the reflected ray is?

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

  1. $y=x+\sqrt{3}$

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

  3. $y=\sqrt{3}x-\sqrt{3}$

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

Show answer
Correct Option: B

What is the reflection of the point $(6,-1)$ in the line $y=2$?

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

  1. $(-2,-1)$

  2. $(-6,5)$

  3. $(6,5)$

  4. $(2,1)$

Show answer
Correct Option: C