Tag: transformation and symmetry in geometrical shapes

Quizzes (2) Questions (65)

Questions Related to transformation and symmetry in geometrical shapes

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
$(a-b) \displaystyle \left [ \left (X + \frac{ab}{a-b} \right )^2 + y^2 \right ] - 2ab \left [ X+ \frac{ab}{a-b} \right ] = 0$
$\Rightarrow (a-b) \displaystyle \left [ X^2 + \frac{a^2 b^2}{(a-b)^2} + Y^2 + \frac{2abX}{a-b} \right ] - 2abX - \frac{2a^2 b^2}{a-b} = 0$
$\Rightarrow (a-b) (X^2 + Y^2) = \displaystyle \frac{a^2 b^2}{a-b}$
$\Rightarrow (a-b)^2 (X^2+Y^2) = a^2 b^2$

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

The new equation of the curve $4(x-2y+1)^{2}+9(2x+y+2)^{2}=25$ if the lines $2x+y+2=0$ and $x-2y+1=0$ are taken as the new $x$ and $y$ axes respectively is

  1. $4X^{2}+9Y^{2}=5$
  2. $4X^{2}+9Y^{2}=25$
  3. $4X^{2}+9Y^{2}=7$
  4. $4X^{2}-9Y^{2}=7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

By substituting the new axes X = 2x + y + 2 and Y = x - 2y + 1 into the equation, the expression simplifies directly to 4Y^2 + 9X^2 = 25. The question asks for the equation in terms of X and Y.

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

The coordinates axes are rotated about the origin $O$ in the counter clockwise direction through an angle of $\dfrac{\pi}{6}$. If $a$ and $b$ are intercepts made on the new axes by a straight line whose equation referred to old the axes is $x+y=1$, then the value of $\displaystyle \frac{1}{a^{2}}+\displaystyle \frac{1}{b^{2}}$ is equal to

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

Given equation is $x+y=1$
We know that 
$\displaystyle x=X\cos\theta-Y\sin\theta$
$y=X\sin\theta+Y\cos\theta$
$\Rightarrow (\cos\theta+\sin\theta)X+(\cos\theta-\sin\theta)Y=1$  
$\displaystyle \Rightarrow \frac{(\sqrt{3}+1)}{2}X+ \frac{(1-\sqrt{3})}{2}Y=1$       .....(i)
According to problem, we have

$\displaystyle\frac{X}{a}+\frac{Y}{b}=1$ .....(ii)
$\displaystyle\Rightarrow \frac { 1 }{ a } =\frac { (\sqrt { 3 } +1) }{ 2 } $
$\Rightarrow \displaystyle \frac { 1 }{ b } =\frac { (1-\sqrt { 3 } ) }{ 2 } $
So, $\displaystyle \frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } =2$

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

Reflection of the line $\dfrac{x-1}{-1}=\dfrac{y-2}{3}=\dfrac{z-4}{1}$ in the plane $x+y+z=7$ is:

  1. $\dfrac{x-1}{3}=\dfrac{y-2}{1}=\dfrac{z-4}{1}$
  2. $\dfrac{x-1}{-3}=\dfrac{y-2}{-1}=\dfrac{z-4}{1}$
  3. $\dfrac{x-1}{-3}=\dfrac{y-2}{1}=\dfrac{z-4}{-1}$
  4. $\dfrac{x-1}{3}=\dfrac{y-2}{1}=\dfrac{z-4}{-2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The line has direction vector (-1, 3, 1) and passes through (1, 2, 4). The reflection of a line in a plane involves reflecting the direction vector and a point on the line; the normal to the plane is (1, 1, 1). Calculating the reflection of the direction vector across the plane normal yields the new direction vector (-3, -1, 1).

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

The image of the line $x-y-1=0$ in the line $2x-3y+1=0$ is

  1. $7x-17y+23=0$
  2. $17x-7y+23=0$
  3. $7x+17y+23=0$
  4. $ 17x+7y+23=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To find the image of a line in another line, find the intersection point and reflect a point from the first line. The intersection of x-y-1=0 and 2x-3y+1=0 is (2, 1). Reflecting a point like (1, 0) from the first line across the second line gives the new line equation.

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 and the image of B by the line mirror $y=0$ is the point $\left(\alpha, \beta \right)$, then :

  1. $\alpha =1,\beta =-2$
  2. $\alpha =0,\beta =0$
  3. $\alpha =2,\beta =-1$

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

The image of (1, 2) in y=x is (2, 1). The image of (2, 1) in y=0 is (2, -1). Thus, alpha=2 and beta=-1.

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

A ray of light travelling along the line $x+\sqrt{3}y=5$ is incident on the $x-axis$ and after refraction it enters the other side of the $x-axis$ by turning $\dfrac{\pi}{6}$ away from the $x-axis$. The equation of the line along which the refracted ray travels is

  1. $x+\sqrt{3}y-5\sqrt{3}=0$
  2. $x-\sqrt{3}y-5\sqrt{3}=0$
  3. $\sqrt{3}x+y-5\sqrt{3}=0$
  4. $\sqrt{3}-y-5\sqrt{3}=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The incident ray x+sqrt(3)y=5 hits the x-axis at (5, 0) with slope -1/sqrt(3), which is -30 degrees. Refraction turns it 30 degrees away from the x-axis, making the new slope tan(30) = 1/sqrt(3). The line passes through (5, 0) with this slope.