Tag: reflection w.r.t a line

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Questions Related to reflection w.r.t a line

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

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

  1. $\left ( \displaystyle -\frac{1}{2},\displaystyle -\frac{1}{2} \right )$
  2. $(-2,-2)$
  3. $(1,1)$
  4. $(-1,-1)$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $(-1,-1)$

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

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   _____________.

  1. x + y - 2 = 0

  2. 8x + y - 9 = 0

  3. 7x - y - 6 = 0

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Reflecting the line y=x across 2x-y=1 involves reflecting two points on the line (e.g., (0,0) and (1,1)) across the mirror line and finding the equation of the line passing through the images.

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

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?

  1. $\alpha =1, \beta =-2$
  2. $\alpha =0, \beta =0$
  3. $\alpha =2, \beta =-1$
  4. $\alpha =1, \beta =-1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Point A(1,2) reflected in y=x gives B(2,1) (coordinates swap). B(2,1) reflected in y=0 gives (2,-1) (y-coordinate negated). Therefore alpha=2, beta=-1. This is identical to question 463562, testing the same two-step reflection concept.

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

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

  1. $\displaystyle (5,-2)$
  2. $\displaystyle (-1,4)$
  3. $\displaystyle (3,-4)$
  4. $\displaystyle (-3,2)$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $D$.

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

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

  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 }

    $
Reveal answer Fill a bubble to check yourself
A Correct answer
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.
Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

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

  1. $(1\sqrt { 2 } ,7/2)$
  2. $(1/2,7\sqrt { 2 } )$
  3. $(1\sqrt { 2 } ,7/\sqrt { 2 } )$
  4. $(1/2,7/2)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Follow the steps: (4, 1) -> (1, 4) -> (3, 4) -> rotate by 45 deg -> reflection about x=0.

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

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 

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

  1. $ \displaystyle 4x+3y-5= 0 $
  2. $ \displaystyle x-4y= 11 $
  3. $ \displaystyle x+4y-5= 0 $
  4. $ \displaystyle -x+y+11= 0 $
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\displaystyle \frac { y+2 }{ x-3 } =\frac { 1 }{ 4 } \Longrightarrow x-4y=11$

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

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

  1. $\left(\dfrac {13}{5}, 0\right)$
  2. $\left(\dfrac {5}{13}, 0\right)$
  3. $(-7, 0)$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Reflect (1, 2) across the x-axis to get (1, -2). The line connecting (1, -2) and (5, 3) intersects the x-axis at the point of reflection A.