Tag: transformation and symmetry in geometrical shapes

Questions Related to transformation and symmetry in geometrical shapes

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 $(0,0)$

If $\displaystyle \left ( -2, 6 \right )$ is the image of the point $\displaystyle \left ( 4,2 \right )$ with respect to the line $\displaystyle L=0$, then $\displaystyle L=$

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

  1. $\displaystyle 6x-4y-7=0$

  2. $\displaystyle 2x-3y-5=0$

  3. $\displaystyle 3x-2y+5=0$

  4. $\displaystyle 3x-2y+10=0$

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Correct Option: C
Explanation:

Slope of line joining the image $Q(-2,6)$ and the point $P(4,2)$ is $\displaystyle -\frac{2}{3}$

So, the slope of mirror $L$ is $\displaystyle \frac{3}{2}$

Mid-point of $PQ$ is $(1,4)$

Since, the image and point are equidistant from mirror. So, this point $(1,4)$ lies on the mirror.

So, the equation of mirror is
$y-4=\displaystyle \frac{3}{2} (x-1)$

$\Rightarrow 3x-2y+5=0$

The equation of image of pair of lines $y=|x-1|$ with respect to y-axis is 

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

  1. ${x^2} - {y^2} - 2x + 1 = 0$

  2. ${x^2} - {y^2} - 4x + 4 = 0$

  3. $4{x^2} - 4x - {y^2} + 1 = 0$

  4. ${x^2} - {y^2} + 2x + 1 = 0$

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Correct Option: D
Explanation:

We have $y=|x-1|$


$\Rightarrow y^2=(x-1)^2$

Change $x$ by $-x$, then the required image is 

$y^2=(-x-1)^2$

$\Rightarrow y^2=x^2+2x+1$

$\Rightarrow x^2-y^2+2x+1=0$