Tag: reflection w.r.t a line

Questions Related to reflection w.r.t a line

The coordinates of the image of the origin $O$ with respect to the line $x+y+1=0$ are

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

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

  2. $(-2,-2)$

  3. $(1,1)$

  4. $(-1,-1)$

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Correct Option: D
Explanation:

Let $Q(h,k)$ be the image of $O(0,0)$ w.r.t the line mirror $x+y+1=0$

$\displaystyle \frac{h-0}{1}=\frac{k-0}{1}=\frac{-2(1)}{2}$

$\displaystyle \Rightarrow h=k=-1$

$\Rightarrow h=-1, k=-1$
So, the image is at $(-1,-1)$

The equation of the line AB is y = x. if And B lie on the same side of the line mirror 2x - y = 1, then the equation of the image of AB is   _____________.

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

  1. x + y - 2 = 0

  2. 8x + y - 9 = 0

  3. 7x - y - 6 = 0

  4. none of these

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Correct Option: C

The image of the point $A(1, 2)$ by the line mirror $y=x$ is the point $B$ and the image of $B$ by the line mirror $y=0$ is the point $(\alpha, \beta)$, then?

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

  1. $\alpha =1, \beta =-2$

  2. $\alpha =0, \beta =0$

  3. $\alpha =2, \beta =-1$

  4. $\alpha =1, \beta =-1$

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Correct Option: C

$P(2,1)$ is image of the point $Q(4,3)$ about the line

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

  1. $x+y=3$

  2. $x-y=1$

  3. $3x+y=5$

  4. $-x+2y=0$

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Correct Option: B

The point $P(2, 1)$ is shifted by $\displaystyle 3\sqrt{2}$ parallel to the line $\displaystyle x+y=1,$ in the direction of increasing ordinate, to reach $Q$.The image of $Q$ by the line $\displaystyle x+y=1$ is

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

  1. $\displaystyle (5,-2)$

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

  3. $\displaystyle (3,-4)$

  4. $\displaystyle (-3,2)$

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Correct Option: D
Explanation:

The point $P(2,1)$ if shifted by $3\sqrt2$ parallel to the line $x+y =1$, in the direction of increasing ordinate, to reach point $Q(\alpha, \beta)$.


The equation of line produced from $P(2,1)$ to $Q(\alpha, \beta)$ parallel to line $x+y = 1$ in parametric form is,

$\Rightarrow \dfrac { \alpha -2}{cos \theta} = \dfrac{ \beta -1}{sin \theta} = -3\sqrt2$   .....$(1)$

Where $\theta$ is slope of line $PQ$, which is parallel to line $x +y = 1$

Hence Slope of $PQ$ $= -1$

$\Rightarrow\text{ if} \   tan \theta = -1 $

$\Rightarrow cos \theta = \dfrac {1}{\sqrt2}$

$\Rightarrow sin \theta = - \dfrac{1}{\sqrt2}$

Hence from equation $(1)$, $\alpha = -1, \beta = 4$ and $Q(-1,4)$

Now image of $Q(-1,4)$ in the line $x+y = 1$ is given by,

$\Rightarrow \dfrac {x - x _1 } {a} = \dfrac{y -y _1} {b} = \dfrac {-2 (ax _1 +by _1 +c)}{a^2 +b^2} $

$\Rightarrow \dfrac{x+1}{1} = \dfrac{y-4}{1} = \dfrac {-2 (-1 +4 -1)}{1^2 +1^2}$

Hence $x = -3$ and $y = 2$

Correct option is $D$.

The image of the line $\displaystyle  \frac { x - 1 } { 3 } = \frac { y - 3 } { 1 } = \frac { z - 4 } { - 5 } $ in the plane $2 x - y + z + 3 = 0 $ is the line

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

  1. $

    \displaystyle\frac { x + 3 } { 3 } = \frac { y - 3 } { 1 } = \frac { z - 2 } { - 5 }

    $

  2. $\displaystyle 

    \frac { x + 3 } { - 3 } = \frac { y - 5 } { - 1 } = \frac { z + 2 } { 5 }

    $

  3. $\displaystyle 

    \frac { x - 3 } { 3 } = \frac { y + 5 } { 1 } = \frac { z - 2 } { - 5 }

    $

  4. $\displaystyle 

    \frac { x - 3 } { - 3 } = \frac { y + 5 } { - 1 } = \frac { z - 2 } { 5 }

    $

Show answer
Correct Option: A
Explanation:
Let $A\left(1,3,4\right)$ be a point.

Let $M$ be the point on the plane.

Given equation of the plane is $2x-y+z+3=0$

Thus, the equation of the plane is

$\dfrac{x-1}{2}=\dfrac{y-3}{-1}=\dfrac{z-4}{3}=k$

Any point on the above line, $AM$ is of the form

$x=2k+1,y=-k+3,z=k+4$

Substituting the above values in the equation of the plane we have

$2\left(2k+1\right)-\left(-k+3\right)+\left(k+4\right)+3=0$

$\Rightarrow\,4k+2+k-3+k+4+3=0$

$\Rightarrow\,6k+6=0$

$\Rightarrow\,k=-1$

Thus the coordinates of $M$ are

$x=2\times -1+1=-2+1=-1$

$y=-\left(-1\right)+3=1+3=4$

$z=-1+4=3$

Let $B\left({x}^{\prime},{y}^{\prime},{z}^{\prime}\right)$ be the image of $A$

Thus, $M$ is the midpoint of $AB$

$\therefore\,-1=\dfrac{1+{x}^{\prime}}{2},\,4=\dfrac{3+{y}^{\prime}}{2},\,3=\dfrac{4+{z}^{\prime}}{2}$ 

$\Rightarrow\,1+{x}^{\prime}=-2,\,\,3+{y}^{\prime}=8,\,\,4+{z}^{\prime}=6$

$\Rightarrow\,{x}^{\prime}=-2-1=-3,\,\,{y}^{\prime}=8-3=5,\,\,{z}^{\prime}=6-4=2$

$\therefore\,\dfrac{x+3}{3}=\dfrac{y-3}{1}=\dfrac{z-2}{-5}$

Hence the image of the given line.

The point (4, 1) undergoes the following transformation successively.
(i)reflection about the line y=x
(ii)translation through a distance 2 units along the positive direction of x-axes.
(iii)rotation through an angle ${ \pi  }/{ 4 }$ about the origin in the anticlockwise direction.
(iv) reflection about x=0
The final position of the given point is

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

  1. $(1\sqrt { 2 } ,7/2)$

  2. $(1/2,7\sqrt { 2 } )$

  3. $(1\sqrt { 2 } ,7/\sqrt { 2 } )$

  4. $(1/2,7/2)$

Show answer
Correct Option: A

Let  $ABC$ be triangle. Let $A$ be the point $(1,2),y=x$be the perpendicular bisector of $AB$ and $x-2y+1=0$ be the angle bisector of $\angle C$. If equation of $BC$ is given by $ax+by-5=0$, then the value of $a+b$ is 

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

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: B
Explanation:

$\begin{array}{l} { m _{ AB } }=-1 \ y-2=-1\left( { x-1 } \right)  \ y-2=-x+1 \ x+y=3 \ \underline { x-y=0 }  \ 2x=3 \ \therefore x=\frac { 3 }{ 2 } \, \, \, \, \, ,y=\frac { 3 }{ 2 }  \ \frac { { 1+h } }{ 2 } =\frac { 3 }{ 2 } \, \, \,  \ h=2 \ \frac { { 2+k } }{ 2 } =\frac { 3 }{ 2 }  \ k=1\, \, \, \, \, \, \, \, \, \, \, B\left( { 2,1 } \right)  \ As\, \, image\, \, through\, \, x-2y+1=0 \ { m _{ AD } }=-2 \ y-2=-2\left( { x-1 } \right)  \ y-2=-2x+2 \ 2x+y=4\times 2\, \, \, \, \, \, \, x-2y+1=0 \ 4x+2y=8 \ \underline { x-2y=-1 }  \ 5x=7 \ x=\frac { 7 }{ 5 } \, \, \, \, \, \, y=\frac { { x+1 } }{ 2 } =\frac { { 12 } }{ { 5\times 2 } } =\frac { 6 }{ 5 }  \ \frac { { m+1 } }{ 2 } =\frac { 7 }{ 5 } \, \, \, \, \, \, \, \, \frac { { n+1 } }{ 2 } =\frac { 6 }{ 5 }  \ m+1=\frac { { 14 } }{ 5 } \, \, \, \, \, \, \, n+2=\frac { { 12 } }{ 5 }  \ m=\frac { 9 }{ 5 } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, n=\frac { { 12 } }{ 5 } -2=\frac { 2 }{ 5 }  \ E\left( { \frac { 9 }{ 5 } ,\frac { 2 }{ 5 }  } \right)  \ BC=BE\, \, \, \, \, \, B\left( { 2,1 } \right) \, \, \, E\left( { \frac { 9 }{ 5 } ,\frac { 2 }{ 5 }  } \right)  \ y-1=\left( { \frac { { 1-\frac { 2 }{ 5 }  } }{ { 2-\frac { 9 }{ 5 }  } }  } \right) \left( { x-2 } \right)  \ 3x-y=5 \ a=3\, \, \, \, b=-1 \ a+b=2 \end{array}$

If the image of the point $ \displaystyle \left ( 4,-6 \right ) $ by a line is the point $(2,2)$, then the equation of the mirror is

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

  1. $ \displaystyle 4x+3y-5= 0 $

  2. $ \displaystyle x-4y= 11 $

  3. $ \displaystyle x+4y-5= 0 $

  4. $ \displaystyle -x+y+11= 0 $

Show answer
Correct Option: B
Explanation:

From given points, required line and line joining points are perpendicular to each other,


$\displaystyle \therefore$ slope of required line is $\displaystyle m=-\frac { 2-4 }{ 2-(-6) } =\frac { 1 }{ 4 } $

The midpoint of the given points will be on our line.

$\Rightarrow$$X=\left ( \dfrac{x _1+x _2}{2}\right ) and \ Y=\left ( \dfrac{y _1+y _2}{2}\right )$

$\Rightarrow$$X=\left ( \dfrac{4+2}{2}\right ) =3 \ and \ Y=\left ( \dfrac{-6+2}{2}\right )=-2$

$\displaystyle \therefore$ point is $\displaystyle (3,-2)\equiv(x _1,y _1)$.

$\therefore$ Equation of line is given by, 

$\Rightarrow$$y-y _1=m(x-x _1)$

Then equation of required line is $\displaystyle \frac { y+2 }{ x-3 } =\frac { 1 }{ 4 } \Longrightarrow x-4y=11$

A ray light comming from the point $(1,2)$ is reflected at a point $A$ on the $x-$axis and then passes through the point $(5,3)$. The co-ordinates of the point $A$ is

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

  1. $\left(\dfrac {13}{5}, 0\right)$

  2. $\left(\dfrac {5}{13}, 0\right)$

  3. $(-7, 0)$

  4. $None\ of\ these$

Show answer
Correct Option: A