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Combining transformations - class-X

Description: combining transformations
Number of Questions: 17
Created by:
Tags: transformations vectors and transformations maths
Attempted 0/17 Correct 0 Score 0

When the axes are rotated through an angle $\dfrac{\pi}{6}$ , find the new coordinate for $(1,0)$

  1. $(\dfrac{\sqrt3}{2},\dfrac{-1}{2})$

  2. $(\dfrac{\sqrt4}{2},\dfrac{-1}{2})$

  3. $(\dfrac{\sqrt5}{2},\dfrac{-1}{2})$

  4. $(\dfrac{\sqrt3}{2},\dfrac{-1}{3})$


Correct Option: A
Explanation:

$\\ sin(\frac{\pi}{6})=(\frac{b}{1})\\\therefore b= (\frac{1}{2})\\ so new y-coordinate wil be  = (\frac{-1}{2})\\cos(\frac{\pi}{6})=(\frac{a}{1})\\\therefore a=(\frac{\sqrt3}{2})\\\therefore new x-coordinate =(\frac{\sqrt3}{2})$

The point to which is shifted in order to remove the first degree terms in $ 2x^{ 2 }+5xy+3y^{ 2 }+6x+7y+1=0 $ is

  1. (2,1)

  2. (1,-2)

  3. (2,-1)

  4. (1,2)


Correct Option: A

If the transformed equation of a curve is $9x^{2}+16y^{2}=144$ when the axes rotated through an angle of $45^{o}$ then the original equation of a curve is:

  1. $25x^{2}+14yxy+25y^{2}=228$

  2. $25x^{2}-14yxy+25y^{2}=228$

  3. $25x^{2}+14yxy-25y^{2}=228$

  4. $25x^{2}-14yxy-25y^{2}=228$


Correct Option: A

By translating the axes the equation $xy-x+2y=6$ has changed to $XY=C$, then $C=$

  1. 4

  2. 5

  3. 6

  4. 7


Correct Option: A
Explanation:

We have

$xy-x+2y=6$


Now,

$ xy-x+2y=6 $

$ \Rightarrow xy-x+2y-6=0 $

$ \Rightarrow xy-x+2y-2=4 $

$ \Rightarrow x\left( y-1 \right)+2\left( y-1 \right)=4 $

$ \Rightarrow \left( x+2 \right)\left( y-1 \right)=4 $


This equation is the form of $XY=C$

Then,

$C=4$


Hence, this is the answer.

lf the axes are translated to the point $(-2, -3)$ , then the equation $\mathrm{x}^{2}+3\mathrm{y}^{2}+4\mathrm{x}+18\mathrm{y}+30=0$ transforms to

  1. $\mathrm{X}^{2}+\mathrm{Y}^{2}=4$

  2. $\mathrm{X}^{2}+3\mathrm{Y}^{2}=1$

  3. $\mathrm{X}^{2}-\mathrm{Y}^{2}=4$

  4. $\mathrm{X}^{2} - 3 \mathrm{Y}^{2}=1$


Correct Option: B
Explanation:

When the point $(x,y)$ changes to $(X,Y)$ on shifting the origin to $(h,k)$
Then, $x=X+h,y=Y+k$
$x=X-2 , y=Y-3$
So, the equation transform to 
$(X-2)^2+3(Y-3)^2+4(X-2)+18(Y-3)+30=0$
$\Rightarrow X^2+3Y^2-1=0$
So, the transformed eqn is $X^2+3Y^2-1=0$

lf the origin is shifted to the point $(-1, 2)$ without changing the direction of axes, the equation ${x}^{2} -{y}^{2}+2{x}+4{y}=0$ becomes 

  1. ${X}^{2}+{Y}^{2}+3=0$

  2. ${X}^{2}+{Y}^{2}-3=0$

  3. ${X}^{2}-{Y}^{2}+3=0$

  4. ${X}^{2}-{Y}^{2}-3=0$


Correct Option: C
Explanation:

Let the point $(x,y)$ on the line changes to $(X,Y)$ on shifting the origin to $(h,k)$
Then, $x=X+h,y=Y+k$
$x=X-1 , y=Y+2$
So, the equation transform to 
$(X-1)^2-(Y+2)^2+2(X-1)+4(Y+2)=0$
$\Rightarrow X^2-Y^2+3=0$
So, the transformed eqn is $X^2-Y^2+3=0$

lf the axes are rotated through an angle $60^{\mathrm{o}}$, then the transformed equation of  $\mathrm{x}^{2}+\mathrm{y}^{2}=25$ is 

  1. $\mathrm{X}^{2}+\mathrm{Y}^{2}=1$

  2. $\mathrm{X}^{2}+\mathrm{Y}^{2}=9$

  3. $\mathrm{X}^{2}+\mathrm{Y}^{2}=16$

  4. $\mathrm{X}^{2}+\mathrm{Y}^{2}=25$


Correct Option: D
Explanation:

The radius of the circle is $5$. Despite rotating the reference frame, the origin still remains the same. Hence, the equation of the circle remains unchanged. 
Hence the new equation is:
$ X^2 + Y^2 =25 $

The transformed equation of $\mathrm{x}\mathrm{c}\mathrm{o}\mathrm{s}\alpha+\mathrm{y}\mathrm{s}\mathrm{i}\mathrm{n}\alpha = \mathrm{P}$ when the axes are rotated through an angle $\alpha$ is

  1. $\mathrm{X}=\mathrm{P}$

  2. $\mathrm{X}+\mathrm{P}=0$

  3. $\mathrm{Y}=\mathrm{P}$

  4. $\mathrm{Y}+\mathrm{P}=0$


Correct Option: A
Explanation:

By  rotation  of  axes,
$ x = x _{1} \cos\alpha -y _{1}\sin\alpha $
$ y = x _{1} \sin \alpha + y _{1} \cos \alpha $
$ x \cos  \alpha  + y \sin  \alpha = P$..........(given)
$ \Rightarrow  x _{1} (\cos^2 \alpha + \sin^{2}\alpha ) + y (\sin  \alpha  \cos  \alpha - \cos  \alpha  \sin  \alpha ) = P$..............(substitute the values of x and y) 
$x _{1} = P$ => X=P.

When axes are rotated by an angle of $135^{0}$, initial coordinates of the new coordinate $(4, -3)$ are             

  1. $\left(\displaystyle \frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$

  2. $\left(\displaystyle \frac{1}{\sqrt{2}}, \frac{-7}{\sqrt{2}}\right)$

  3. $\left(\displaystyle \frac{-1}{\sqrt{2}}, \frac{-7}{\sqrt{2}}\right)$

  4. $\left(\displaystyle \frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$


Correct Option: D
Explanation:

When axes are rotated through an angle $\theta $, then 

$x=X \cos\theta -Y \sin\theta$

$y=Y \cos\theta -X \sin\theta$

where(x,y) are initial coordinates and (X,Y) are new coordinates

$\therefore x=4\left(\dfrac{-1}{\sqrt{2}}\right)-\left(-3\right)\left(\dfrac{1}{\sqrt{2}}\right)=\dfrac{-1}{\sqrt{2}}$

$y=4\left(\dfrac{1}{\sqrt{2}}\right)-3\left(\dfrac{-1}{\sqrt{2}}\right)=\dfrac{7}{\sqrt{2}}$


$\therefore (x,y)=\left(\displaystyle \frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$

The point $(4,3)$ is translated to the point $(3,1)$ and then axes are rotated through $30^{\mathrm{o}}$ about the origin, then the new position of the point is 

  1. $\left(\displaystyle \frac{2\sqrt{3}+1}{2},\frac{\sqrt{3}-2}{2}\right)$

  2. $\left(\displaystyle \frac{\sqrt{3}+1}{2},\frac{2\sqrt{3}+1}{2}\right)$

  3. $\left(\displaystyle \frac{\sqrt{3}+2}{2},\frac{2\sqrt{3}-1}{2}\right)$

  4. $\left(\displaystyle \frac{\sqrt{3}-2}{2},\frac{\sqrt{3}+1}{2}\right)$


Correct Option: C
Explanation:

When axes are rotated through an angle $\theta $, then

$x=\alpha +x^{1}\cos\theta -y^{1}\sin\theta  ; y=\beta +x^{1}\sin\theta +y^{1}\cos\theta $

$\Rightarrow 4=3+\dfrac{x^{1}\sqrt{3}}{2}-\dfrac{y^{1}}{2} ;3=1+\dfrac{x^{1}}{2}+\dfrac{y^{1}\sqrt{3}}{2}$

$\Rightarrow x^{1}\sqrt{3}-y^{1}=2$

$ x^{1}+y^{1}\sqrt{3}=4$

$\Rightarrow x=\dfrac{\sqrt{3}+2}{2}$

$y=\dfrac{2\sqrt{3}-1}{2}$

The area of triangle formed by the lines $x+y-3=0$, $x-3y+9=0$ and $3x-2y+1=0$ is:

  1. $\cfrac { 16 }{ 7 } $ sq. units

  2. $\cfrac { 10 }{ 7 } $ sq. units

  3. $4$ sq. units

  4. $9$ sq. units


Correct Option: B
Explanation:

$L _1:x+y-3=0$

$L _2:x-3y+9=0$
$L _3:3x-2y+1=0$

$A$ is intersection point of $L _1$ and $L _2$ which is $(0,3)$ on solving $L _1$ and $L _2$.

$B$ is intersection point of $L _2$ and $L _3$ which is $(1,2)$ on solving $L _2$ and $L _3$.

$C$ is intersection point of $L _3$ and $L _1$ which is $(\dfrac{15}{7},\dfrac{26}{7})$ on solving $L _3$ and $L _1$.

Now, area of $\Delta ABC=\dfrac{1}{2}|(x _1y _2-x _2y _1)+(x _2y _3-x _3y _2)+(x _3y _1-x _1y _3)|$

$=\dfrac{1}{2}|(0-3)+(\dfrac{26}{7}-\dfrac{30}{7})+(\dfrac{45}{7}-0)|$

$=\dfrac{1}{2}|-3-\dfrac{4}{7}+\dfrac{45}{7}|=\dfrac{1}{2}\times \dfrac{20}{7}$

$=\dfrac{10}{7} sq. units$

if the equation $4{x^2} + 2xy + 2{y^2} - 1 = 0$ becomes $5{x^2} + {y^2} = 1,$ when the axes are rotate through an angle ${45^ \circ }\,$  , then the original equation of the curve is :

  1. $\,{15^ \circ }$

  2. $\,{30^ \circ }\,$

  3. ${45^ \circ }$

  4. ${60^ \circ }$


Correct Option: D

If the axes are shifted to $(-2, -3)$ and rotated $\dfrac{\pi}{4}$ then Transformed equation of $2x^{2}+4xy-5y^{2}+20x-22y-14=0$ is 

  1. $X^{2}-14XY-7Y^{2}=2$

  2. $X^{2}-14XY-7Y^{2}=4$

  3. $X^{2}-14XY+7Y^{2}=2$

  4. $X^{2}+14XY+7Y^{2}=2$


Correct Option: A

The point $A(2, 1)$ is translated parallel to the line $x- y = 3$ by a distance $4$ units. If the new position $A'$ is in third quadrant, then the coordinates of $A'$ are

  1. $(2 + 2 \sqrt{2}, 1 + 2\sqrt{2})$

  2. $(-2 + \sqrt{2}, -1 -2 \sqrt{2})$

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

  4. none of these


Correct Option: C
Explanation:

Since the point $A(2, 1)$ is translated parallel to $x-y =3$, therefore $AA'$ has the same slope as that of $x-y =3$. Therefore, $AA'$ passes through $(2, 1)$ and has the slope of $1$. 

Here, $\tan  \theta = 1 $
$\Rightarrow \cos \theta = \displaystyle\frac{1 }{\sqrt{2}}, \sin \theta = \displaystyle \frac {1 }{ \sqrt{2}}$
Using the parametric form of the line, the coordinates of $A'$ can be written as $\left(2\pm \displaystyle\frac {4}{\sqrt 2}, 1\pm \displaystyle\frac{4}{\sqrt 2}\right) $. 
Now, $A'$ is in the third quadrant. 
Hence, the coordinates of $A'$ are $(2- 2 \sqrt{2}, 1 - 2 \sqrt{2})$.

If the axes are rotated through an angle of ${30}^{o}$ in the anti-clockwise direction, the coordinates of point $(4,-2\sqrt{3})$ with respect to new axes are-

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

  2. $(\sqrt{3}, -5)$

  3. $(2,3)$

  4. $(\sqrt{3},2)$


Correct Option: B
Explanation:

$x=r\cos \alpha=4$

$y=r\sin\alpha=-2\sqrt3$
$x^1=r\cos(\alpha-30^o)$
$y^1=r\sin(\alpha-30^o)$
$x^1=r[\cos\alpha\cos 30^o+\sin\alpha\sin 30^o]$
$x^1=\dfrac{4\sqrt{3}}{2}-2\sqrt{3}\times \dfrac{1}{2}$
$x^1=2\sqrt{3}-\sqrt{3}$
$\therefore x^1=\sqrt{3}$
$y^1=r[\sin\alpha\cos 30^o-\cos \alpha\sin 30^o]$
      $=-2\sqrt{3}\times \dfrac{\sqrt{3}}{2}-4\times \dfrac{1}{2}$
$y^1=-5$
$\therefore \alpha(\sqrt{3}, -5)$

Let $\displaystyle A=(1,0)$ and $\displaystyle B=(2,1).$ The line $AB$ turns about $A$ through an angle $ \dfrac{\pi}6$ in the clockwise sense, and the new position of $B$ is $B'$. Then $B'$ has the coordinates

  1. $\displaystyle \left ( \frac{3+\sqrt{3}}{2},\frac{\sqrt{3}-1}{2} \right )$

  2. $\displaystyle \left ( \frac{3\sqrt{3}}{2},\frac{\sqrt{3}+1}{2} \right )$

  3. $\displaystyle \left ( \frac{1-\sqrt{3}}{2},\frac{1+\sqrt{3}}{2} \right )$

  4. none of these


Correct Option: A
Explanation:

Given, points are $A=(1,0)$ and $B=(2,1)$

Slope of $AB=\cfrac { 1-0 }{ 2-1 } =1$
Then angle of $AB$ with $x$-axis is 
$\angle BAX={ 45 }^{ 0 }$
Hence, $\angle B'AX={ 45 }^{ 0 }-{ 30 }^{ 0 }={ 15 }^{ 0 }$
Therefore for $B'\left( h,k \right) $
$h=1+\sqrt{2}\cos{ 15 }^{ 0 },k= \sqrt{2}\sin{ 15 }^{ 0 }\$
We have, $\sin \left(15^{0} \right) = \dfrac{\sqrt 6 -\sqrt 2 }{4} $
$\Rightarrow \cos \left (15^{0} \right) = \dfrac{\sqrt6 + \sqrt 2 }{4}$
$ \Rightarrow h=\cfrac { 3+\sqrt { 3 }  }{ 2 } ,k=\cfrac { \sqrt { 3 } -1 }{ 2 } $

The transformed equation of $3{ x }^{ 2 }+3{ y }^{ 2 }+2xy=2$. When the coordinate axes are rotated through an angle of $45$, is

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

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

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

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


Correct Option: B
Explanation:

Since, the axes are rotated through an angle $45$, then we replace $\left( x,y \right) $ by
$\left( x\cos { 45 } -y\sin { 45 } ,x\sin { 45 } +y\cos { 45 }  \right) $
Given equation is $3{ x }^{ 2 }+3{ y }^{ 2 }+2xy=2$
Therefore, $ 3{ \left( \dfrac { x }{ \sqrt { 2 }  } -\dfrac { y }{ \sqrt { 2 }  }  \right)  }^{ 2 }+3\dfrac { x+y }{ \sqrt { 2 }  } +2\dfrac { x-y }{ \sqrt { 2 }  } \dfrac { x+y }{ \sqrt { 2 }  } =2$
$\Rightarrow \dfrac { 3 }{ 2 } \left( { x }^{ 2 }+{ y }^{ 2 }+2xy \right) +\dfrac { 3 }{ 2 } \left( { x }^{ 2 }+{ y }^{ 2 }-2xy \right) +\dfrac { 2 }{ 2 } \left( { x }^{ 2 }-{ y }^{ 2 } \right) =2$
$\Rightarrow  4{ x }^{ 2 }=2{ y }^{ 2 }=2$
$\Rightarrow  2{ x }^{ 2 }+{ y }^{ 2 }=1$

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