Tag: transformations

Questions Related to transformations

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$
$\Rightarrow \dfrac { 3 }{ 2 } \left( { x }^{ 2 }+{ y }^{ 2 }+2xy \right) +\dfrac { 3 }{ 2 } \left( { x }^{ 2 }+{ y }^{ 2 }-2xy \right) +\dfrac { 2 }{ 2 } \left( { x }^{ 2 }-{ y }^{ 2 } \right) =2$
$\Rightarrow  4{ x }^{ 2 }=2{ y }^{ 2 }=2$
$\Rightarrow  2{ x }^{ 2 }+{ y }^{ 2 }=1$

The ordinate of the point which divides the lines joining the origin and the point $(1,2)  $ externally in the ratio of $3:2$ is

  1. $-2$

  2. $ \displaystyle \frac{3}{5} $

  3. $ \displaystyle \frac{2}{5} $

  4. $6$


Correct Option: D
Explanation:

Co-ordinates of the required point will be
$\displaystyle y=\frac{m _{1}y _{2}-m _{2}y _{1}}{m _{1}-m _{2}}=\frac{3\times 2-2\times 0}{3-2}=6$

The points (22,23) divides the join of P (7,5) and Q externally in the ratio 3:5, then Q=

  1. $(3,7)$

  2. $(-3,7)$

  3. $(3,-7)$

  4. $(-3,-7)$


Correct Option: D
Explanation:
$\left(22,23\right)$dives $P\left(7,5\right)$ and $Q$ in the ratio $3:5$ externally.
Let $Q\left(x,y\right)$
$22=\dfrac{3x-5\times 7}{-2}$
$-44=3x-35$
$x=-3$
$23=\dfrac{3y-5\times 5}{-2}$
$y=-7$
$Q\left(-3,-7\right)$
$D$ is correct.

The point $(22, 33)$ divides the join of $P(7, 5)$ and $Q$ externally in the ratio $3 : 5$, then coordinates of $Q$ are

  1. $(3, 7)$

  2. $( - 3, -7)$

  3. $(3, - 7)$

  4. $( - 3, \frac{ - 41}{3})$


Correct Option: B
Explanation:
Let the point $Q$ be $\left(x,y \right)$
As the point $\left(22,23 \right)$ diverts the line joining point $P\left(7,5 \right)$ and $Q \left(x,y \right)$ in $3:5$ externally.
So.  
$22=\dfrac{ 3\left(x\right)-5\left(7\right)} {3-5}\quad 23=\dfrac{ 3\left(y\right)-5\left(5\right)} {3-5}$
$22=\dfrac{3x-35}{-2}\quad 23=\dfrac{3y-25}{-2}$
$-44=3x-35\quad -46=3y-25$
$-=3x\quad -21=3y$
$x=-3\quad y=\dfrac{-21}{3}=-7$
Point $Q=\left(-3,-7\right)$