Tag: reflection w.r.t a line

Questions Related to reflection w.r.t a line

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 $

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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. $(1, 4)$

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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. $\left(-\dfrac{3}{\sqrt{2}} , \dfrac{4}{\sqrt{2}}\right)$
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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) $

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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

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

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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. $(2,1)$

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Correct Option: C

The image of (2, -3) in the y - axis is

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

  1. (2, 3)

  2. (-2, 3)

  3. (-2, -3)

  4. (2, -3)

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Correct Option: B

If ${ P } _{ 1 }\left( \dfrac { 1 }{ 5 } ,\alpha  \right)$ and ${P } _{ 2 }\left( \beta ,\dfrac { 18 }{ 5 }  \right)$ be the images of point $P\left( 1,\gamma  \right)$ about lines ${ L } _{ 1 }:2x-y=\lambda$ and ${ L } _{ 2 }:2y+x=4$ respectively, then the value of $\alpha$is-

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

  1. $-\dfrac { 3 }{ 5 }$

  2. $\dfrac { 2 }{ 5 }$

  3. $\dfrac { 7 }{ 5 }$

  4. $-\dfrac { 8 }{ 5 }$

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Correct Option: A

The image of $P(a, b)$ in the line $y= -x$ is $Q$ and the image of $Q$ in the line $y=x$ is $R$. Then the midpoint of $PR$ is

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

  1. $(a+b, b+a)$

  2. $\left(\dfrac{a+b}{2}, \dfrac{b+a}{2}\right)$

  3. $(a-b, b-a)$

  4. $(0, 0)$

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Correct Option: D
Explanation:

Consider point P and Q
The line which is perpendicular to $y=-x$ and passes through $(a,b)$ will be
$\dfrac{y-b}{x-a}=1$
$y=x-(a-b)$
Now the point where it intersects $y=-x$ is
$-x=x-(a-b)$
$2x=(a-b)$
$x=\dfrac{a-b}{2}$ and hence $y=\dfrac{b-a}{2}$
This will be the mid point of PQ since Q is the image of point P on the line $y=-x$
Therefore $Q=(-b,-a)$
Similarly $R=(-a,-b)$
Hence The midpoint of PR will be $(0,0)$