Tag: reflection

Questions Related to reflection

If P is (-3, 4) and My Mx (P)  shows the reflection of the  point P in the x"axis and then  the reflection of the image in the  y"axis, then My Mx (P) is

  1. (3, 4)

  2. (- 3, - 4)

  3. (`3, 4)

  4. (3, - 4)


Correct Option: D

When an object is reflected in the mirror, the corresponding lengths and angles of the object and the image

  1. remain same

  2. increase in the image

  3. decrease in the image

  4. cannot be determined


Correct Option: A
Explanation:

When an object is reflected, there is no change in the lengths and angles; i.e. the lengths and angles of the object and the corresponding lengths and angles of the image are the same. However, in one aspect there is a change, i.e. there is a difference between the object and the image. The left and the right sides of an object appear inverted in a mirror
hence option A is correct

If a boy, standing in front of a mirror has his right hand in the pocket, then in his mirror reflection, it will appear as

  1. his left hand in the pocket

  2. missing hand

  3. his right hand in the pocket

  4. can't say


Correct Option: A
Explanation:

A plane mirror has a property of lateral inversion, which states that reflection of an object in the mirror will always be laterally inverted.

$\Rightarrow$ Right hand in the pocket will appear as left hand in the pocket.

Let R be the relation on the set R of all real numbers defined by a R b if $|a-b|\leq \dfrac{1}{2}$. Then R is?

  1. Reflexive and symmetric but not transitive

  2. Symmetric and transitive but not reflexive

  3. Transitive but neither reflexive nor symmetric

  4. Reflexive, symmetric and transitive


Correct Option: A
Explanation:
R is reflexive since $|a-a|=0\leq \dfrac{1}{2}\forall a\in R$.
Also, R is symmetric as $|a-b|\leq \dfrac{1}{2}$
$\Rightarrow |b-a|\leq \dfrac{1}{2}$
But R is not transitive.
e.g.: If we take three numbers $\dfrac{3}{4}, \dfrac{1}{3}, \dfrac{1}{8}$ then $\left|\dfrac{3}{4}-\dfrac{1}{3}\right|=\dfrac{5}{12}\leq \dfrac{1}{2}$ and $\left|\dfrac{1}{3}-\dfrac{1}{8}\right|=\dfrac{5}{24}\leq \dfrac{1}{2}$
But $\left|\dfrac{3}{4}-\dfrac{1}{8}\right|=\dfrac{5}{8} > \dfrac{1}{2}$
Thus, $\dfrac{3}{4}R\dfrac{1}{3}$ and $\dfrac{1}{3}R\dfrac{1}{8}$ but $\dfrac{3}{7}R\dfrac{1}{8}$.

The symbol  '$\geq $'  used in relations is know .............symbol.

  1. transitive

  2. reflective

  3. symmetric

  4. asymmetric


Correct Option: B
Explanation:
Symbol has pre defined meaning.
Tt is known as reflective symbol.

A ray of light passing through the point $(1, 2)$ is reflected on the $x$-axis at a point $P$ and passes through the point $(5, 3)$. The abscissa of the point $P$ is

  1. $3$

  2. $\dfrac {13}{3}$

  3. $\dfrac {13}{5}$

  4. $\dfrac {13}{4}$


Correct Option: C
Explanation:
The ray passes through $A(1,2)$ and reflects at $B$ and passes through $(5,3)$

The image of $(5,3)$ with respect to $x-axis$ is $(5,-3)$

The line $AB$ passes through the image of $(5,3)$

The slope of line is $\dfrac{2+3}{1-5}=\dfrac {-5}4$

So the equation of line is $y-2=\dfrac {-5}4\left( x-1\right)$ 

$4y-8=-5x+5$

$\implies 5x+4y-13=0$

The line touches $x-axis $ at $y=0$

$\implies 5x+4(0)-13=0$

$\implies 5x=13$

$\implies x=\dfrac {13}5$

The coordinates of a point P are $(2, 5).$ Given that the image of P under a reflection in the 'x-axis is P, find the coordinates of P.

  1. $(-2,-5)$

  2. $(-2, 5)$

  3. $(2,-5)$

  4. $(-5, 2)$


Correct Option: C
Explanation:

for reflation in $x-axis$, sing of $y$ coordinets changes and $x-coordinete$ remaing wnstant,

Do,coordinetes of image will be $(2,-5)$

State the following statement is true(T) or false(F).
By lines in geometry, we mean only straight lines.

  1. True

  2. False


Correct Option: A
Explanation:

In geometry, line means straight line not a curved line.

When $ABCD$ is reflected over the $y$-axis to ${ A }^{ \prime  }{ B }^{ \prime  }{ C }^{ \prime  }{ D }^{ \prime  }$ , what can be the coordinates of ${ D }^{ \prime  }$ given that D lies in the $1^{st}$ quadrant?

  1. $\left( -12,1 \right) $

  2. $\left( -12,-1 \right) $

  3. $\left( 12,-1 \right) $

  4. $\left( 1,12 \right) $

  5. $\left( 1,-12 \right) $


Correct Option: A

A circle with centre $(1,1)$ intersects X axis at $(1,0)$ and  Y axis at $(0,1)$. Find the centre of the circle when reflected through Y axis.

  1. $(-1,1)$

  2. $(-1,-1)$

  3. $(1,0)$

  4. $(-1,0)$


Correct Option: A
Explanation:

The circle center is $(1,1)$ and reflection over y axis of point $(x,y)\rightarrow (-x,y)$
therefore $(1,1)$ is $(-1,1)$