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 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.

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

If $B$ is reflection of $A(a,5)$ about line $4x-3y=0$, then area of triangle $ABC$ is equal to

  1. $\dfrac{253}{50}$
  2. $\dfrac{506}{25}$
  3. $\dfrac{253}{25}$
  4. $\dfrac{506}{50}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The reflection B of A(a, 5) across 4x-3y=0 is found using the reflection formula. The area of triangle ABC (where C is the origin or a fixed point) is calculated using the coordinates of A, B, and the intersection point.

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

Locus of the image of the point (2, 3) in the line (2x - 3y + 4) + k(x - 2y + 3) = 0, k $\in $ R, is a 

  1. straight line parallel to x-axis

  2. straight line parallel to y-axis
  3. Circle of radius $\sqrt { 2 } $

  4. circle of radius 3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The family of lines (2x-3y+4) + k(x-2y+3) = 0 passes through a fixed point (intersection of the two lines). The locus of the image of a point reflected across a family of lines passing through a fixed point is a circle centered at that fixed point.

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

The distance of the image of a point (or an object) from the line of symmetry (mirror) is  ----- as that of the point (object )from the line (mirror).

  1. same

  2. double

  3. triple

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
To make the above statement true, the word to be placed in the blank is : “same”
So, the true statement becomes :
The distance of the image of a point (or an object) from the line of symmetry (mirror) is same as that of the point (object) from the line (mirror).
Hence, option A is the correct answer.