Tag: reflection w.r.t a line

Questions Related to reflection w.r.t a line

The point $A(4, 1)$ undergoes following transformations successively:
(i) reflection about line $y=x$
(ii) translation through a distance of $3$ units in the positive direction of x-axis.
(iii) rotation through an angle $105^o$ in anti-clockwise direction about origin O.
Then the final position of point A is?

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

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

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

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

  4. $(-2\sqrt{6}, 2\sqrt{2})$

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Correct Option: A

The reflection of the point $(4, -13)$ in the line $5x+y+6=0$ is

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

  1. $(-1, -14)$

  2. $(3,4)$

  3. $(1,2)$

  4. $(-4, 13)$

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Correct Option: A

The image of the pair of lines represented by $\displaystyle 3x^{2}+4xy+5y^{2}=0 $ in the line mirror $x = 0$ is

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

  1. $\displaystyle 3x^{2}-4xy+5y^{2}=0 $

  2. $\displaystyle 3x^{2}-4xy-5y^{2}=0 $

  3. $\displaystyle 5y^{2}-4xy-3x^{2}=0 $

  4. none of these

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Correct Option: A
Explanation:

Here the mirror image of line $3x^2+4xy+5y^2=0$ in mirror $x=0$

i.e about $Y-axis $
So the x-coordinates About $Y-axis $ becomes negative and the y-coordiantes remains same Hence putting $x=-x$ in given eq 
$3(-x)^2-4(-x)y+5y^2=0$
$3x^2-4xy+5y^2=0$

The point A(4, 1) undergoes following transformations successively
(i) reflection about line y = x
(ii) translation through a distance of 2 units in the positive direction of x axis
(iii) rotation through an angle $\displaystyle \pi/4 $ in anti clockwise direction about origin O
Then the final position of point A is

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

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

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

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

  4. none of these

Show answer
Correct Option: C
Explanation:
Given Point $A(4,1)$
$(i) $ reflection about $y=x$
Hence putting coordinates of point A in given eq $y=4,x=1$
The point becomes $A(1,4)$

$(ii)$ translation through distance of 2 units in positive X axis 
$A(1+2,4)\Rightarrow A(3,4)$

$(iii)$ rotation of point through and angle $\dfrac{\pi}{4}$ in anticlockwise about origin O
After rotation Point will be $A(r\cos(\alpha+\dfrac{pi}{4}),r\sin(\alpha+\dfrac{pi}{4}))$
Converting into polar form 
$r=\sqrt{3^2+4^2}=5$
$\tan\alpha=\dfrac{4}{3}$
Hence $\cos\alpha=\dfrac{3}{5}$
$\sin\alpha=\dfrac{4}{5}$
$\cos(\alpha+\dfrac{\pi}{4})=\cos\alpha\cos\dfrac{\pi}{4}-\sin\alpha\sin\dfrac{\pi}{4}$
$\cos(\alpha+\dfrac{\pi}{4})=\dfrac{3}{5}\dfrac{1}{\sqrt{2}}-\dfrac{4}{5}\dfrac{1}{\sqrt{2}}$
$\cos(\alpha+\dfrac{\pi}{4})=\dfrac{-1}{5\sqrt{2}}$

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

The co-ordinates of the point of reflection of the origin $(0, 0)$ in the line $4x -2y - 5 = 0$ is

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

  1. $(-1, 2)$

  2. $(2, -1)$

  3. $\displaystyle \left (\frac {4}{5}, -\frac {2}{5}\right )$

  4. $(2, 5)$

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Correct Option: B
Explanation:

Let $(h,k)$ be the point of reflection.
Line joining $(0,0)$ and $(h,k)$ is perpendicular to $4x-2y-5=0$
Product of slopes of these lines is $-1$
$\dfrac{k}h*2=-1=>k=-\dfrac{h}2$
Midpoint of $(0,0)$ and $(h,k)$ i.e, $(\dfrac{h}{2},\dfrac{k}{2})$ lies on $4x-2y-5=0$
Therefore $4\dfrac{h}2-2\dfrac{k}2-5=0=>2h-k-5=0$ but $k=-\dfrac{h}2$
$2h+\dfrac{h}2-5=0=>h=2 $ and $k=-1$
Point of reflection is $(2,-1)$.

The equation of the image of the circle $\displaystyle x^{2}+y^{2}+16x-24y+183=0 $ along 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. $\displaystyle x^{2}+y^{2}+32x-4y+235=0 $

  2. $\displaystyle x^{2}+y^{2}+32x+4y-235=0 $

  3. $\displaystyle x^{2}+y^{2}+32x-4y-235=0 $

  4. $\displaystyle x^{2}+y^{2}+32x+4y+235=0 $

Show answer
Correct Option: D
Explanation:

The given equation of the circle is $x^2 + y^2 + 16x - 24y + 183 = 0$, which can be written as,


$\Rightarrow (x+8)^2 + (y-12)^2 = (5)^2$

Hence we can see the center of the circle , let's say $O(- 8, 12)$ and radius of the circle is $r = 5$

If we mirror the image of the given circle by the line $4x +7y +13 =0$, the radius of the circle won't change. But the position of center will get change. Let's assume the new center will be $O'( \alpha, \beta)$

By using below equation to find $O'(\alpha, \beta)$, 

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

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

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

Hence new center $O'$ is $O'(-16,-2)$

The equation of the image of the circle through the mirror will be,

$\Rightarrow (x+16)^2 + (y +2)^2 = (5)^2$

$\Rightarrow x^2 +y^2 + 32x +4y + 235$

So correct option is $D$

The image of the pair of lines represented by $\displaystyle  3x^{2}+4xy+5y^{2}=0 $ in the line mirror x = 0 is

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

  1. $\displaystyle 3x^{2}-4xy+5y^{2}=0 $

  2. $\displaystyle 3x^{2}-4xy-5y^{2}=0 $

  3. $\displaystyle 5y^{2}-4xy-3x^{2}=0 $

  4. none of these

Show answer
Correct Option: A
Explanation:
Given pair
$3x^2+4xy+5y^2=0$
$x=0$ is the $Y-axis $ hence the $x$-coordinates will become $-x$ and $y$-coordinates remains same 
Hence 
$3(-x)^2+4(-x)y+5y^2=0$
$3x^2-4xy+5y^2=0$

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 $(8, 6)$.