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 point $\mathrm{A}(2,1)$ 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. $(2+2\sqrt{2},2+2\sqrt{2})$
  2. $(-2+\sqrt{2},-1-2\sqrt{2})$
  3. $(2-2\sqrt{2},1-2\sqrt{2})$
  4. $(-2-\sqrt{2},-1-2\sqrt{2})$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$ y = x-3$ $\Rightarrow m=1$ ; $tan \theta =1$
By parametrization we have $x=2 \pm r cos\theta $
$y=1 \pm r sin\theta $

$x =2 \pm 4\times \dfrac{1}{\sqrt{2}} ;\ y=1 \pm 4\times \dfrac{1}{\sqrt{2}}$.....(consider -ive sign for third quadrant)
$x=2-2\sqrt{2};\ y=1-2\sqrt{2}$ since they are in third quadrant.

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

If the points $(5, 5), (7, 7)$ and $(a, 8)$ are collinear then the value of a is

  1. $6$
  2. $3$
  3. $8$
  4. $7$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When three points are collinear, Slope of line joining any two points is same as the slope of line joining any other two points


Slope of line joining two points $ ({x} _{1}, {y} _{1}) $ and $ ({x} _{2}, {y} _{2}) $ is $\dfrac { {y} _{2} - {y} _{1}}{ {x} _{2} - {x} _{1}} $

So, Slope of line joining $ (5,5) ;  (7,7) $ is $ \dfrac {7-5}{7-5} = \dfrac {2}{2} = 1 $

And Slope of line joining $ (a,8) ;  (7,7) $ is $ \dfrac {7-a}{7-8} = a - 7 $

As they are collinear $ a - 7 = 1 => a = 8 $



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

${A}$ line has intercepts $ a$ and ${b}$ on the co ordinate axes. When the axes are rotated through an angle $\alpha$, keeping the origin fixed, the line makes equal intercepts on the coordinate axes, then $\tan\alpha=$ 

  1. $\displaystyle \frac{{a}+b}{{a}-b}$
  2. $\displaystyle \frac{{a}-b}{{a}+b}$
  3. $\dfrac{b}{a}$
  4. $\displaystyle \frac{{a}}{b}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let the equation of line be $ \displaystyle \frac{x}{a}+\frac{y}{b}=1.$

When axes are rotated through an angle $ \alpha$, the new coordinates $XY$ are related to old coordinates $xy$ as follows:

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

$y=X \sin\alpha +Y\ \cos \alpha $

Substituting these values in the equation of line, we get
$ \displaystyle \frac { X\cos { \alpha  } -Y\sin { \alpha  }  }{ a } +\frac { X\sin { \alpha  } +Y\cos { \alpha  }  }{ b } =1\\ \displaystyle \Rightarrow X\left( \frac { \cos { \alpha  }  }{ a } +\frac { \sin { \alpha  }  }{ b }  \right) +Y\left( \frac { \cos { \alpha  }  }{ b } -\frac { \sin { \alpha  }  }{ a }  \right) =1$

As it makes equal intercepts in the new coordinate system, we get

$ \displaystyle \Rightarrow \frac{\cos \alpha }{a}+\frac{\sin \alpha }{b}=\frac{\cos \alpha }{b}-\frac{\sin \alpha }{a}$

$\Rightarrow \cos \alpha \left ( b-a \right )=-\sin \alpha \left ( a+b \right )$

$ \displaystyle \Rightarrow \tan  \alpha =\dfrac{a-b}{a+b}$
Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

The angle of rotation of the axes so that the equation $\sqrt{3}\mathrm{x}-\mathrm{y}+5=0$ may be reduced to the form $\mathrm{Y}=\mathrm{k}$, where $\mathrm{k}$ is a constant is 

  1. $\dfrac{\pi}{6}$
  2. $\dfrac{\pi}{4}$
  3. $\dfrac{\pi}{3}$
  4. $\dfrac{\pi}{12}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let axis be rotated through an angle $\theta $ then 
$ x= x^{1} \cos \theta - y^{1} \sin\theta $
$ y = x^{1} \sin \theta + y^{1} \cos\theta $
$ \sqrt{3}\times x -y +5 = 0$
$ \sqrt{3} (x^{1} \cos \theta - y^{1} \sin\theta) - (x^{1} \sin \theta + y^{1} \cos\theta) +5 = 0$
$x^{1} (\sqrt{3} \cos \theta - \sin\theta ) = 0$
$ \Rightarrow \tan\theta  = \sqrt{3}$
$ \theta = 60^{\circ} = \dfrac{\pi }{3}$
Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

Find the equation of a line whose inclination is $\displaystyle 30^{\circ}$ and making an intercept of -3/5 on the y-axis

  1. $ y = \dfrac {5}{\sqrt{6}}x +\dfrac {2}{5} $
  2. $ y = \dfrac {1}{\sqrt{3}}x -\dfrac {3}{5} $
  3. $ y = \dfrac {3}{\sqrt{7}}x -\dfrac {1}{3} $
  4. $\displaystyle \dfrac{-5}{3}x$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation of any straight line can be written as $ y = mx + c $, where $m$ is its slope and $c$ is its y - intercept.

As inclination is $ 30^o $, slope of the line $ =  tan (30 ^o) = \dfrac {1}{\sqrt{3}} $

So equation of line is $ y = \dfrac {1}{\sqrt{3}}x  -\dfrac {3}{5} $

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

lf the equation $4\mathrm{x}^{2}+2\sqrt{3}\mathrm{x}\mathrm{y}+2\mathrm{y}^{2}-1=0$ becomes $5\mathrm{X}^{2}+\mathrm{Y}^{2}=1$, when the axes are rotated through an angle $\theta$, then $\theta$ is 

  1. $15^{\mathrm{o}}$
  2. $30^{\mathrm{o}}$
  3. $45^{0}$
  4. $60^{\mathrm{o}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
By  rotation  of  axes  through  $\theta $,  co-ordinates  become
$x = x^{1}  \cos\theta  - y^{1}  \sin\theta $
$y = x^{1}  \sin\theta  + y^{1}  \cos\theta $
$4x^{2} + 2\sqrt{3}xy + 2y^{2} - 1=0$
$\Rightarrow 4(x^{1}\cos\theta - y^{1} \sin\theta )^{2} + 2\sqrt{3} (x^{1} \cos\theta  -y^{1} \sin\theta )  (x^{1} \sin\theta + y^{1} \cos\theta )  +2 (x^{1}\sin\theta +y^{1} \cos\theta )^{2} -1 =0$
coeff  of $xy=0$
$\Rightarrow 4(-\sin2\theta )+2\sqrt{3}   \cos20  +  2 \sin2\theta  = 0$
$2\sqrt{3}   \cos2\theta  = 2\sin2\theta $
$\tan2\theta = \sqrt{3}$
$2\theta = \dfrac{\pi }{3}$
$\theta  = \dfrac{\pi }{6} = 30^{\circ}$
$\therefore $ angle  to  be  rotated $= 30^{\circ}$
Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

lf the distance between two given points is $2$ units and the points are transferred by shifting the origin to $(2, 2)$, then the distance between the points in their new position is.

  1. $2$
  2. $5$
  3. $6$
  4. $7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Shifting the origin to $(2 , 2)$ transforms the coordinates to
$(x , y)$ to $(x - 2, y - 2)$.

$\therefore$ distance between two points is

$\sqrt{(x _{1}-x _{2})^{2}+(y _{1}-y _{2})^{2}}$

$=\sqrt{((x _{1}-2)-(x _{2}-2))^{2}+((y _{1}-2)-(y _{2}-2))}^{2}$

$=\sqrt{(x _{1}-x _{2})^{2}+(y _{1}-y _{2})}^{2}$

$\therefore$ distance is unaltered $ = 2$ units.

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

If a point $\mathrm{P}(4,3)$ is shifted by a distances $\sqrt{2}$ units parallel to the line $\mathrm{y}=\mathrm{x}$, then the coordinates of $\mathrm{P}$ in its new position are

  1. $(5,4)$
  2. $(5+\sqrt{2},4+\sqrt{2})$
  3. $(5-\sqrt{2},4-\sqrt{2})$
  4. $(4,5)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

shifted by $\sqrt{2}$ units $\Rightarrow r=\sqrt{2}$

for the line is tan $\theta =1$

$\therefore$ by parametric equations, we have

$x=x^{1}+r\ cos \theta$ ;  $y=y^{1}+r\ sin \theta$

$\Rightarrow x=4+\sqrt{2}\frac{1}{\sqrt{2}}$ ; $y=3+\sqrt{2}\frac{1}{\sqrt{2}}$

$x=5$  ;  $y=4$

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

A line has intercepts a, b on the coordinate axes. If the axes are rotated about the origin through an angle $\displaystyle \alpha$ then the line has intercepts p,q on the new position of the axes respectively. Then 

  1. $\displaystyle \frac{1}{p^{2}}+\frac{1}{q^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$
  2. $\displaystyle \frac{1}{p^{2}}-\frac{1}{q^{2}}=\frac{1}{a^{2}}-\frac{1}{b^{2}}$
  3. $\displaystyle \frac{1}{p^{2}}+\frac{1}{a^{2}}=\frac{1}{q^{2}}+\frac{1}{b^{2}}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since the line has intecepts a and b on the coordinate axes, therefore its equation is 
$\cfrac { x }{ a } +\cfrac { y }{ b } =1$
When the axes are rotated, its equation with respect to the new axes and the same origin will become
$\cfrac { x }{ p } +\cfrac { y }{ q } =1$
In both the cases, the length of the perpendicular from the origin to the line will be same
Therefore
$\cfrac { 1 }{ \sqrt { \cfrac { 1 }{ { a }^{ 2 } } +\cfrac { 1 }{ { b }^{ 2 } }  }  } =\cfrac { 1 }{ \sqrt { \cfrac { 1 }{ { p }^{ 2 } } +\cfrac { 1 }{ { q }^{ 2 } }  }  } \ \Rightarrow \cfrac { 1 }{ { a }^{ 2 } } +\cfrac { 1 }{ { b }^{ 2 } } =\cfrac { 1 }{ { p }^{ 2 } } +\cfrac { 1 }{ { q }^{ 2 } } $

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

If the origin is shifted to the point $(\displaystyle\frac{ab}{a-b}, 0)$ without rotation, then the equation $(a-b)(x^2 + y^2) - 2abx = 0$ becomes

  1. $(a-b) (X^2+Y^2) - (a+b)XY + abX = a^2$
  2. $(a+b) (X^2 + Y^2) = 2ab$
  3. $(X^2 + Y^2) = (a^2 + b^2)$
  4. $(a-b)^2 (X^2 + Y^2) = a^2 b^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The given equation is 
$(a-b) (x^2 + y^2) - 2abx = 0$             ........... (i)
The origin is shifted to (ab/(a-b), 0). Any point (x, y) on the curve (i) must be replaced with a new point (X, Y) with reference to new axes, such that
$\displaystyle x = X + \frac{ab}{a-b}  ,\    y = Y +0$
substituting these in (i), we get
$(a-b) \displaystyle \left [ \left (X + \frac{ab}{a-b} \right )^2 + y^2 \right ] - 2ab \left [ X+ \frac{ab}{a-b} \right ] = 0$
$\Rightarrow (a-b) \displaystyle \left [ X^2 + \frac{a^2 b^2}{(a-b)^2} + Y^2 + \frac{2abX}{a-b} \right ] - 2abX - \frac{2a^2 b^2}{a-b} = 0$
$\Rightarrow (a-b) (X^2 + Y^2) = \displaystyle \frac{a^2 b^2}{a-b}$
$\Rightarrow (a-b)^2 (X^2+Y^2) = a^2 b^2$