Tag: position and movement

Questions Related to position and movement

Find the image of the point $\displaystyle \left ( -2, -7 \right )$ under the transformation
$\left ( x, y \right )\rightarrow \left ( x-2y,-3x+y \right ).$

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

  1. $\displaystyle \left ( 12, -1 \right )$

  2. $\displaystyle \left ( -12, 1 \right )$

  3. $\displaystyle \left ( 2, 7 \right )$

  4. $\displaystyle \left ( -2, 7 \right )$

Show answer
Correct Option: A
Explanation:

Given transformation is 
$\left ( x, y \right )\rightarrow \left ( x-2y,-3x+y \right )$
So, the point $(-2,-7)$ $\rightarrow (-2-2(-7),-3(-2)-7)$
So, the image of the point $(-2,-7)$ under the given transformation is $ (12,-1)$

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

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

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

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

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

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

Show answer
Correct Option: C
Explanation:

Let points are $A(-2,6)$ & $B(4,2)$

$\therefore$    $L=0$ is $\bot $ to $AB$ & passes through the mid point slope $AB=\dfrac { -4 }{ 6 } =\dfrac { -2 }{ 3 } $
$\therefore$    slope of $L=0=3/2$
passes through $(1,4)$
$\therefore$    required equation is $\left( y-4 \right) =\frac { 3 }{ 2 } \left( x-1 \right) $
                                     $\Rightarrow 2y-8=3x-3$
                                     $\Rightarrow 3x-2y+5=0$        [C]

The image of the pair of lines represented by$\displaystyle :ax^{2}+2hxy+by^{2}= 0 $ by the line $ y= 0 $ mirror is:

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

  1. $\displaystyle :ax^{2}-2hxy-by^{2}= 0$

  2. $\displaystyle :bx^{2}-2hxy+ay^{2}= 0$

  3. $\displaystyle :bx^{2}+2hxy+ay^{2}= 0$

  4. $\displaystyle :ax^{2}-2hxy+by^{2}= 0$

Show answer
Correct Option: D
Explanation:

In ${ ax }^{ 2 }+2hxy+{ by }^{ 2 }=0 $ replacing $y$ by $-y$
we get, ${ ax }^{ 2 }+2hx\left( -y \right) +{ by }^{ 2 }=0 $
$\Rightarrow { ax }^{ 2 }-2hxy+{ by }^{ 2 }=0 $

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. $(-1,-1)$

Show answer
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. $\alpha =1, \beta =-1$

Show answer
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. $-x+2y=0$

Show answer
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. $\displaystyle (-3,2)$

Show answer
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