Tag: reflection w.r.t a line

Questions Related to reflection w.r.t a line

State whether true/false:
The size of the figure after reflection changes but the lines and angles remains the same.

maths does it look the same? reflection on coordinate axis reflection w.r.t a line reflection

  1. True

  2. False

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Correct Option: B
Explanation:

The size of the figure does not changes after reflection. So the statement is false.

A circle with centre $(1,1)$ intersects X axis at $(1,0)$ and  Y axis at $(0,1)$. Find the centre of the circle when reflected through X axis.

maths does it look the same? reflection on coordinate axis reflection w.r.t a line reflection

  1. $(1,-1)$

  2. $(1,1)$

  3. $(-1,-1)$

  4. None of the above

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Correct Option: A
Explanation:
The graph is symmetric(reflection) w.r.t to the $x$-axis i.e. $y=0$ , If $(x,y)$ is a point on the graph ,then $(x,-y)$ is also a point on the graph .

Given center at $(1,1)$
using above concept reflection of center about $X$-axis lies at  $(1,-1)$

State whether true/false:
A shape is symmetrical if both sides of it are the same when a mirror line is drawn.

maths does it look the same? reflection on coordinate axis reflection w.r.t a line reflection

  1. True

  2. False

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Correct Option: A
Explanation:

The line that divides a figure into identical halves is called the line of symmetry or the axis of symmetry. The line of symmetry is also called as mirror line because it produces two reflections of an image that coincide.

A circle with centre $(0,0)$ and of radius $2$ cm is plotted in the XY plane. Take the part of circle in first quadrant and find its reflection through X axis.

maths does it look the same? reflection on coordinate axis reflection w.r.t a line reflection

  1. The reflected image will form in $3rd$ quadrant.

  2. The reflected image will form in $1st$ quadrant.

  3. The reflected image will form in $2nd$ quadrant.

  4. The reflected image will form in $4rth$ quadrant.

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Correct Option: D
Explanation:
The graph is symmetric(reflection) w.r.t to the $x$-axis i.e. $y=0$ , If $(x,y)$ is a point on the graph ,then $(x,-y)$ is also a point on the graph .

in first quadrant both $x$ and $y$ are positive ,
using above concept ,all the point of circle which lie on the first quadrant has a reflection about $X$-axis at $(x,-y)$ .

And we know points $(x,-y)$ always lie in $4^{th}$ quadrant .

A circle with centre $(0,0)$ and of radius $2$ cm is plotted in the XY plane. Take the part of circle in first quadrant and find its reflection through Y axis.

maths does it look the same? reflection on coordinate axis reflection w.r.t a line reflection

  1. The reflected image would get formed in $1st$ quadrant.

  2. The reflected image would get formed in $2nd$ quadrant.

  3. The reflected image would get formed in $4rth$ quadrant.

  4. None of these

Show answer
Correct Option: B
Explanation:
The graph is symmetric (reflection) w.r.t to the $y$-axis i.e. $x=0$ , If $(x,y)$ is a point on the graph ,then $(-x,y)$ is also a point on the graph .

As we know points $(x,y)$ lie at $1^{st}$ quadrant --------   (both $x$ and $y$ are positive ).
Using above concept , all the reflected points of first quadrant about $Y$-axis lie at $(-x,y)$.
And we know points $(-x,y)$ lie in $2^{nd}$ quadrant .

 Line symmetry and mirror reflection are naturally related to each other.

maths does it look the same? reflection on coordinate axis reflection w.r.t a line reflection

  1. True

  2. False

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Correct Option: A
Explanation:

A figure may have both horizontal and vertical lines of reflection.An object and its images are always at the same distance from the surface of a mirror, which is called the mirror line. Line symmetry and mirror reflection are naturally related and linked to each other. So, the option is true.

The distance of the object from the mirror line is ____ the distance of the image from the mirror line.

maths does it look the same? reflection on coordinate axis reflection w.r.t a line reflection

  1. same as

  2. greater than

  3. less than

  4. None of the above

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Correct Option: A
Explanation:

An objects and its image are always at the same distance from the surface of mirror,  which is called the mirror line. Therefore the distance of the object from the mirror line is same as the distance of the image from the mirror line

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

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

  1. $7y+x-6=0$

  2. $7y-x-6=0$

  3. $x+7y=8$

  4. $7y-x+6=0$

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Correct Option: A
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.

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 :-

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

  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) $

Show answer
Correct Option: B
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) $

Without changing the direction of coordinates axes,  origin is transferred to $(\alpha ,\ \beta)$ so that linear term in the equation $x^{2}+y^{2}+2x-4y+6=0$ are eliminated the point $(\alpha ,\ \beta)$ is   

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

  1. $(-1,\ 2)$

  2. $(1,\ -2)$

  3. $(1,\ 2)$

  4. $(-1,\ -2)$

Show answer
Correct Option: B