Class 12 maths ncert solutions chapter 1 ex 1.1

It is given that Relation R in the set \( \mathrm{A}=\{1,2,3, \ldots, 13,14\} \) defined as
\(R=\{(x, y): 3 x-y=0\}\)
Then, \( \mathrm{R}=\{(1,3),(2,6),(3,9),(4,12)\} \)
REFLEXIVE
A relation is said to be reflexive if \( (x, x) \in R \), where \( x \) is from domain.
\( R \) is not reflexive as \( (1,1),(2,2) \ldots \ldots(14,14) \notin R \)
SYMMETRIC
A relation is said to be symmetric if \( (y, x) \in R \) whenever \( (x, y) \in R \).
Also, \( R \) is not symmetric as \( (1,3) \in R \), but \( (3,1) \notin R \)
TRANSITIVE
A relation is said to be transitive if \( (x, z) \in R \) whenever \( ( x, y) \in R \) and \(( y , z) \in \mathrm{R} \)
And, also \( R \) is not transitive as \( (1,3),(3,9) \in R \), but \( (1,9) \notin R \)
Therefore, R is neither reflexive, nor symmetric, nor transitive.

It is given that Relation R in the set N of natural numbers defined as
\( R=\{(x, y): y=x+5 \) and \( x < 4\} \)
Clearly,
\( \mathrm{R}=\{(1,6),(2,7),(3,8)\} \)
REFLEXIVE
A relation is said to be reflexive if \( (x, x) \in R \), where \( x \) is from domain.
we can see that \( (1,1) \notin R \)
\( \Rightarrow R \) is not reflexive.
SYMMETRIC
A relation is said to be symmetric if \( (y, x) \in R \) whenever \( (x, y) \in R \).
Now, \( (1,6) \in r \) but \( (6,1) \notin R \).
\( \Rightarrow \mathrm{R} \) is not symmetric.
TRANSITIVE
Now, since there is no pair in \( R \) such that \( (x, y) \) and \( (y, z) \in R \), then \( (x, z) \) cannot belong to R.
\( \therefore \mathrm{R} \) is not transitive.
Therefore, R is neither reflexive, nor symmetric, nor transitive.

It is given that relation R in the set \( \mathrm{A}=\{1,2,3,4,5,6\} \) as \( \mathrm{R}=\{(x, \mathrm{y}): \mathrm{y} \) is divisible by \( x\} \)
We know that any number \(( x )\) is divisible by itself.
\( \Rightarrow(x, x) \in \mathrm{R} \)
\( \Rightarrow \mathrm{R} \) is reflexive.
Now, \( (2,4) \in R \) but \( (4,2) \notin R \).
\( \Rightarrow R \) is not symmetric.
Let \( (x, \mathrm{y}),(\mathrm{y}, \mathrm{z}) \in \mathrm{R} \). Then, \(y\) is divisible \(x\) and \(z\) is divisible by \(y\).
\( \Rightarrow \mathrm{z} \) is divisible by \(x .\)
\( \Rightarrow(x, \mathrm{z}) \in \mathrm{R} \)
\( \Rightarrow \mathrm{R} \) is transitive.
Therefore, \( R \) is reflexive and transitive but not symmetric.

It is given that Relation \( R \) in the set \( Z \) of all integers defined as
\( \mathrm{R}=\{(x, \mathrm{y}): x-\mathrm{y} \) is an integer\( \} \)
Now, for every \( x \in Z,(x, x) \in R \), as \( x-x=0 \) is an integer.
\( \Rightarrow \mathrm{R} \) is reflexive.
Now, for every \( x, y \in Z \) if \( (x, y) \in R \), then \( x-y \) is an integer.
\( \Rightarrow-(x-\mathrm{y}) \) is also an integer.
\( \Rightarrow(y-x) \) is an integer.
\( \Rightarrow(y, x) \in R \)
\( \Rightarrow \mathrm{R} \) is symmetric.
Now, for every \( (x, y) \) and \( (y, z) \in R \) where \( x, y, z \in R \),
\( \Rightarrow(x-\mathrm{y}) \) and \( (\mathrm{y}-\mathrm{z}) \) is an integer.
\( \Rightarrow(x-\mathrm{z})=(x-\mathrm{y})+(\mathrm{y}-x) \) is an integer.
\( \Rightarrow(x, \mathrm{z}) \in \mathrm{R} \)
\( \Rightarrow \mathrm{R} \) is transitive.
Therefore, R is reflexive, symmetric and transitive.

It is given that \( \mathrm{R}=\{(x, \mathrm{y}) \) : \(x\) and \(y\) work at the same place\( \} \)
\( \Rightarrow(x, x) \in \mathrm{R} \)
\( \Rightarrow \mathrm{R} \) is reflexive.
Now, if \( (x, \mathrm{y}) \in \mathrm{R} \), then \(x\) and \(y\) work on the same place.
\( \Rightarrow \mathrm{y} \) and \(x\) work at the same place.
\( \Rightarrow(y, x) \in R \)
\( \Rightarrow \mathrm{R} \) is symmetric.
Now, let \( (x, y), (y, z) \in \mathrm{R} \)
\( \Rightarrow x \) and \(y\) work at the same place and \(y\) and \(z\) work at the same place.
\( \Rightarrow x \) and \(z\) work at the same place
\( \Rightarrow(x, \mathrm{z}) \in \mathrm{R} \)
\( \Rightarrow \mathrm{R} \) is transitive.
Therefore, \( R \) is reflexive, symmetric and transitive.

It is given that \( \mathrm{R}=\{(x, y) : x\) and \(y\) live in the same locality\( \} \)
\( \Rightarrow(x, x) \in \mathrm{R} \) as x and x live in the same human being.
\( \Rightarrow \mathrm{R} \) is reflexive.
Now, if \( ( x, \mathrm{y}) \in \mathrm{R} \), then \(x\) and \(y\) live in the same locality.
\( \Rightarrow \mathrm{y} \) and \(x\) live in the same locality.
\( \Rightarrow(y, x) \in R \)
\( \Rightarrow \mathrm{R} \) is symmetric.
Now, let ( \( x, y),(y, z) \in R \)
\( \Rightarrow x \) and \(y\) live in the same locality and \(y\) and \(z \) live in the same locality.
\( \Rightarrow x \) and \(z\) live in the same locality
\( \Rightarrow(x, \mathrm{z}) \in \mathrm{R} \)
\( \Rightarrow \mathrm{R} \) is transitive.
Therefore, R is reflexive, symmetric and transitive.

It is given that \( \mathrm{R}=\{(x, \mathrm{y}) : x\) is exactly 7 cm taller than \(y \} \)
\( \Rightarrow(x, x) \notin \mathrm{R} \) as human being \(x\) cannot be taller than himself.
\( \Rightarrow R \) is not reflexive.
Now, if \( (x, y) \in R \), then \( x \) is exactly 7 cm taller than \( y \).
\( \Rightarrow \) But \(y\) is not taller than \(x \).
\( \Rightarrow(\mathrm{y}, x) \notin \mathrm{R} \)
\( \Rightarrow R \) is not symmetric.
Now, let \((x, y), (y, z) \in \mathrm{R} \)
\( \Rightarrow x \) is exactly 7 cm taller than \(y\) and \(y\) is exactly 7 cm taller than \(z\).
\( \Rightarrow x \) is exactly 14 cm taller than \(z\).
\( \Rightarrow(x, \mathrm{z}) \notin \mathrm{R} \)
\( \Rightarrow \mathrm{R} \) is not transitive.
Therefore, R is neither reflexive, nor symmetric, nor transitive.

It is given that \( R=\{(x, y) \) : \( x \) is wife of \( y\} \)
\( \Rightarrow(x, x) \notin \mathrm{R} \) as \(x\) cannot be the wife of herself.
\( \Rightarrow R \) is not reflexive.
Now, if \(( x, y )\) \( \in \mathrm{R} \), then \( x \) is the wife of \( y \).
\( \Rightarrow \) But \(y\) is not wife of \(x \).
\( \Rightarrow(\mathrm{y}, x) \notin \mathrm{R} \)
\( \Rightarrow R \) is not symmetric.
Now, let \( (x, y), (y, z) \in \mathrm{R} \)
\( \Rightarrow x \) is the wife of y and y is the wife of z.
\( \Rightarrow \) This cannot be possible.
\( \Rightarrow(x, \mathrm{z}) \notin \mathrm{R} \)
\( \Rightarrow \mathrm{R} \) is not transitive.
Therefore, R is neither reflexive, nor symmetric, nor transitive.

It is given that \( R=\{(x, y) \) : \( x \) is father of \( y\} \)
\( \Rightarrow(x, x) \notin \mathrm{R} \) as \(x\) cannot be the father of himself.
\( \Rightarrow R \) is not reflexive.
Now, if \( ( x, y )\in \mathrm{R} \), then \( x \) is the father of \( y \).
\( \Rightarrow \) But \( y \) is not father of \( x \).
\( \Rightarrow(\mathrm{y}, x) \notin \mathrm{R} \)
\( \Rightarrow R \) is not symmetric.
Now, let \( (x, y), (y, z) \in \mathrm{R} \)
\( \Rightarrow x \) is the father of y and y is the father of z.
\( \Rightarrow x \) is not the father of z.
\( \Rightarrow \) Indeed \(x\) is the grandfather of z.
\( \Rightarrow(x, \mathrm{z}) \notin \mathrm{R} \)
\( \Rightarrow \mathrm{R} \) is not transitive.
Therefore, R is neither reflexive, nor symmetric, nor transitive.

It is given that \( R=\left\{(a, b): a \leq b^{2}\right\} \)
Check for reflexive:
We can see that \( \left(\frac{1}{2}, \frac{1}{2}\right) \notin R \),
Since, \( \frac{1}{2} > \left(\frac{1}{2}\right)^{2} \)
Therefore, R is not reflexive.
Check for symmetric:
Now, \( (1,4) \in \mathrm{R} \) as \( 1 < 4^{2} \)
But 4 is not less than \( 1^{2} \).
Then, \( (4,1) \notin \mathrm{R} \)
Therefore, R is not symmetric.
Check for transitive:
Now, (3, 2), (2, 1.5) \( \in \mathrm{R} \)
But, \( 3 > (1.5)^{2}=2.25 \).
Then, \( (3,1.5) \notin \mathrm{R} \)
Therefore, R is not transitive.
Therefore, R is neither reflexive, nor symmetric, nor transitive.

Let us take \( A=\{1,2,3,4,5,6\} \)
\( A \) relation \( R \) is defined on set \( A \) as:
\( \mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{b}=\mathrm{a}+1\} \)
Then, \( \mathrm{R}=\{(1,2),(2,3),(3,4),(4,5),(5,6)\} \)
Now, we will find \( (\mathrm{a}, \mathrm{a}) \notin \mathrm{R} \), where \( \mathrm{a} \in \mathrm{A} \)
For instance,
\( (1,1),(2,2),(3,3),(4,4),(5,5),(6,6) \notin R \)
Therefore, R is not reflexive.
We can see that \( (1,2) \in R \), but \( (2,1) \notin R \).
Therefore, R is not symmetric.
And now, \( (1,2),(2,3) \in \mathrm{R} \)
But, \( (1,3) \notin \mathrm{R} \)
Therefore, R is not transitive.
Therefore, R is neither reflexive, nor symmetric, nor transitive.

It is given that \( R=\{(a, b): a \leq b\} \),
It is clear that \( (a, a) \in r \) as \( a=a \).
Therefore, R is reflexive.
Now let us take \( (2,4) \in R(2 < 4) \)
But, \( (4,2) \notin R \) as 4 is greater than 2.
Therefore, R is not symmetric.
Now, let \( (a, b), (b, c) \in \mathrm{R} \)
Then, \( \mathrm{a} \leq \mathrm{b} \) and \( \mathrm{b} \leq \mathrm{c} \)
\(\Rightarrow \mathrm{a} \leq \mathrm{c} \Rightarrow(\mathrm{a}, \mathrm{c}) \in \mathrm{R}
\)
Therefore, \( R \) is a transitive.
Therefore, R is reflexive and transitive but not symmetric.

It is given that \( R=\left\{(a, b): a \leq b^{3}\right\} \)
Now, It can observed that as: \( \left(\frac{1}{3}, \frac{1}{3}\right) \notin \mathrm{R} \) as \( \frac{1}{3} > \left(\frac{1}{3}\right)^{3}=\frac{1}{3} \)
Therefore, R is not reflexive.
Now, \( \left(3, \frac{3}{2}\right),\left(\frac{3}{2}, \frac{6}{5}\right) \in \mathrm{R} \) as \( 3 < \left(\frac{3}{2}\right)^{3} \) and \( \frac{3}{2} < \left(\frac{6}{5}\right)^{3} \)
But, \( \left(3, \frac{6}{5}\right) \notin \mathrm{R} \) as \( 3 > \left(\frac{6}{5}\right)^{3} \)
Therefore, \( R \) is not transitive.
Therefore, \(R\) is neither reflexive, nor symmetric, nor transitive.

Let us take \( \mathrm{A}=\{1,2,3\} \)
A relation R is defined on set A as \( \mathrm{R}=\{(1,2),(2,1)\} \)
It is seen that \( (1,1),(2,2),(3,3) \notin R \).
Therefore, \( R \) is not reflexive.
Now, we can see that \( (1,2) \in R \) and \( (2,1) \in R \)
Therefore, \( R \) is symmetric.
And now, \( (1,2),(2,1) \in \mathrm{R} \)
But, \( (1,1) \notin \mathrm{R} \)
Therefore, R is not transitive.
Therefore, R is symmetric but neither reflexive, nor transitive.

It is given that the set A of all the books in a library of a college. Then, \( R=\{(x, y): x \) and \( y \) have same number of pages\( \} \)
Now, \( R \) is reflexive
Since, \( (x, x) \in R \) as \( x \) and \( x \) have same number of pages.
Let \( (x, x) \in R \)
\( \Rightarrow x \) and \(y\) have the same number of pages
\( \Rightarrow y \) and \(x\) have the same number of pages.
\(\Rightarrow(y, x) \in R
\)
Therefore, \( R \) is symmetric.
Now, let \( ( x, y) \in R \) and \( (y, z) \in R \)
\( \Rightarrow x \) and \(y\) have the same number of pages and y and z have the same number of pages.
\( \Rightarrow x \) and z have the same number of pages.
\(\Rightarrow(x, \mathrm{z}) \in \mathrm{R}\)
Therefore, \( R \) is transitive.
Therefore, R is an equivalence relation.

It is given that the relation R in the set \( \mathrm{A}=\{1,2,3,4,5\} \) given by \( R=\{(a, b):|a-b| \) is even\( \} \),
It is clear that for any element \( a \in A \), we have \( |a-a|=0 \) (which is even)
Therefore, \( R \) is reflexive.
Now, Let \( (a, b) \in R \)
\( \Rightarrow|\mathrm{a}-\mathrm{b}| \) is even
\( \Rightarrow|-(\mathrm{a}-\mathrm{b})|=|\mathrm{b}-\mathrm{a}| \) is also even.
\( \Rightarrow(\mathrm{b}, \mathrm{a}) \in \mathrm{R} \)
Therefore, \( R \) is symmetric.
Now, Let \( (a, b) \in R \) and \( (b, c) \in R \).
\( \Rightarrow|\mathrm{a}-\mathrm{b}| \) is even and \( |\mathrm{b}-\mathrm{c}| \) is even
\( \Rightarrow(\mathrm{a}-\mathrm{b}) \) is even and \( (\mathrm{b}-\mathrm{c}) \) is even
\( \Rightarrow(\mathrm{a}-\mathrm{c})=(\mathrm{a}-\mathrm{b})+(\mathrm{b}-\mathrm{c}) \) is even
\( \Rightarrow|\mathrm{a}-\mathrm{c}| \) is even
\( \Rightarrow(\mathrm{a}, \mathrm{c}) \in \mathrm{R} \)
Therefore, R is transitive.
Therefore, R is an equivalence relation.
Now, all elements of the set \( \{1,3,5\} \) are related to each other as all the elements of this subset are odd.
Therefore, the modules of the difference between any two elements will be even.
Similarly, all elements of the sets \( \{2,4\} \) are related each other as all the elements of this subset are even.
Also, no element of the subset \( \{1,3,5\} \) can be related to any element of \( \{2,4\} \) as all elements of \( \{1,3,5\} \) are odd and all elements of \( \{2,4\} \) are even.
Therefore, the modulus of the difference between the two elements will not be even.

It is given that the relation \( R \) in the set \( A=\{x \in Z: 0 \leq x \leq 12\} \), given by \( \mathrm{R}=\{(\mathrm{a}, \mathrm{b}):|\mathrm{a}-\mathrm{b}| \) is a multiple of \(4 \} \)
For any element \( a \in A \), we have \( (a, a) \in R \) as \( |a-a|=0 \) is a multiple of \(4 \).
Therefore, \( R \) is reflexive.
Now, Let \( (a, a) \in R \)
\( \Rightarrow|\mathrm{a}-\mathrm{b}| \) is a multiple of 4
\( \Rightarrow|\mathrm{b}-\mathrm{a}|=|\mathrm{a}-\mathrm{b}| \) is a multiple of 4
\( \Rightarrow(\mathrm{b}, \mathrm{a}) \in \mathrm{R} \)
Therefore, \( R \) is symmetric.
Now, Let \( (\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{c}) \in \mathrm{R} \)
\( \Rightarrow|\mathrm{a}-\mathrm{b}| \) is a multiple of 4 and \( |\mathrm{b}-\mathrm{c}| \) is a multiple of 4
\( \Rightarrow|\mathrm{a}-\mathrm{c}|=|(\mathrm{a}-\mathrm{b})+(\mathrm{b}-\mathrm{c})| \) is a multiple of 4
\( \Rightarrow(\mathrm{a}, \mathrm{c}) \in \mathrm{R} \)
Therefore, R is transitive.
Therefore, \( R \) is an equivalence relation.
The set of elements related to 1 is \( \{1,5,9\} \)
\( |1-1|=0 \) is multiple of 4
\( |5-1|=4 \) is multiple of 4
\( |9-1|=8 \) is multiple of 4.

It is given that the relation \( R \) in the set \( A=\{x \in Z: 0 \leq x \leq 12\} \), given by \( \mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}=\mathrm{b}\} \)
For any element \( a \in A \), we have \( (a, a) \in R \) as \( a=a \).
Therefore, \( R \) is reflexive.
Now, Let \( (a, a) \in R \)
\(\Rightarrow \mathrm{a}=\mathrm{b}\)
\(\Rightarrow \mathrm{b}=\mathrm{a}\)
\(\Rightarrow(\mathrm{b}, \mathrm{a}) \in \mathrm{R}
\)
Therefore, \( R \) is symmetric.
Now, Let \( (a, b),(b, c) \in R \)
\(\Rightarrow \mathrm{a}=\mathrm{b} \text { and } \mathrm{b}=\mathrm{c}\)
\(\Rightarrow \mathrm{a}=\mathrm{c}\)
\(\Rightarrow(\mathrm{a}, \mathrm{c}) \in \mathrm{R}
\)
Therefore, R is transitive.
Therefore, \( R \) is an equivalence relation.
The set of elements related to 1 will be those elements from set A which are equal to 1.
Therefore, the set of elements related to 1 is \( \{1\} \).

Let \( A=\{3,4,5\} \)
Define a relation R on A as \( \mathrm{R}=\{(3,4),(4,3)\} \)
Relation \( R \) is not reflexive as \( (3,3),(4,4) \) and \( (5,5) \notin R \).
Now, as \( (3,4) \in R \) and also \( (4,3) \in R \),
R is symmetric.
\( \Rightarrow(3,4),(4,3) \in R \), but \( (3,3) \notin R \)
\( \Rightarrow \mathrm{R} \) is not transitive.
Therefore, relation R is symmetric but not reflexive or transitive.

Let a relation R in R defined as:
\( \mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a} < \mathrm{b}\} \)
For any a \( \in R \), we have \( (a, a) \notin R \) as a cannot be strictly less than a itself.
In fact, \( \mathrm{a}=\mathrm{a} \),
Therefore, R is not reflexive.
Now, \( (1,2) \in R \) but \( 2 > 1 \)
\( \Rightarrow(2,1) \notin \mathrm{R} \).
\( \Rightarrow R \) is not symmetric.
Now, let \((a, b), (b, c) \in \mathrm{R} \)
\( \Rightarrow \mathrm{a} < \mathrm{b} \) and \( \mathrm{b} < \mathrm{c} \)
\( \Rightarrow \mathrm{a} < \mathrm{c} \)
\( \Rightarrow(\mathrm{a}, \mathrm{c}) \in \mathrm{R} \)
\( \Rightarrow \mathrm{R} \) is transitive.
Therefore, relation R is transitive but not reflexive and a symmetric.

Let us take \( A=\{2,4,6\} \)
Define a relation R on A as:
\(\mathrm{A}=\{(2,2),(4,4),(6,6),(2,4),(4,2),(4,6),(6,4)\}\)
Relation of \( R \) is reflexive as for every \( a \in A \),
\( (a, a) \in \mathrm{R} \)
\(\Rightarrow(2,2),(4,4),(6,6) \in \mathrm{R}
\)
Relation \( R \) is symmetric as \( (a, b) \in R \)
\(\Rightarrow(b, a) \in R \text { for all } a, b \in R
\)
And Relation \( R \) is not transitive as \( (2,4),(4,6) \in R \), but \( (2,6) \notin \mathrm{R} \)
Therefore, relation \( R \) is reflexive and symmetric but not transitive.

Let us define a relation R in R as
\(R=\left\{(a, b): a^{3} \geq b^{3}\right\}
\)
It is clear that \( (a, a) \in R \) as \( a^{3}=a^{3} \)
\( \Rightarrow \mathrm{R} \) is reflexive.
Now, \( (2,1) \in \mathrm{R} \)
But \( (1,2) \notin \mathrm{R} \)
\( \Rightarrow \mathrm{R} \) is not symmetric.
Now, let \( (a, b)(b, c) \in R \)
\( \Rightarrow a^{3} \geq b^{3} \) and \( b^{3} \geq c^{3} \)
\( \Rightarrow \mathrm{a}^{3} \geq \mathrm{c}^{3} \)
\( \Rightarrow(\mathrm{a}, \mathrm{c}) \in \mathrm{R} \)
\( \Rightarrow R \) is transitive.
Therefore, relation R is reflexive and transitive but not symmetric.

Let \( \mathrm{A}=\{-7,-8\} \)
Define a relation R on A as:
\( \mathrm{R}=\{(-7,-8),(-8,-7),(-7,-7)\} \)
Relation \( R \) is not reflexive as \( (-8,-8) \notin R \)
Relation R is symmetric as \( (-7,-8) \in \mathrm{R} \) and \( (-8,-7) \in \mathrm{R} \)
But it is seen that \( (-7,-8),(-8,-7) \in \mathrm{R} \).
Also, \( (-7,-7) \in \mathrm{R} \).
\( \Rightarrow \mathrm{R} \) is transitive.
Therefore, relation \( R \) is symmetric and transitive but not reflexive.

It is given that
\( R=\{(P, Q) \) : distance of the point \( P \) from the origin is same as the distance of the point Q from the origin\(\}\),
Now, it is clear that
\( (\mathrm{P}, \mathrm{P}) \in \mathrm{R} \) since the distance of point P from origin is always the same as the distance of the same point \( P \) from the origin.
Therefore, R is reflexive.
Now, let us take (P, Q) \( \in \mathrm{R} \),
\( \Rightarrow \) The distance of point P from origin is always the same as the distance of the same point Q from the origin.
\( \Rightarrow \) The distance of point Q from origin is always the same as the distance of the same point \( P \) from the origin.
\(\Rightarrow(\mathrm{Q}, \mathrm{P}) \in \mathrm{R}
\)
Therefore, \( R \) is symmetric.
Now, Let (P, Q), (Q, S) \( \in \mathrm{R} \)
\( \Rightarrow \) The distance of point P and Q from origin is always the same as the distance of the same point Q and S from the origin.
\( \Rightarrow \) The distance of points P and S from the origin is the same.
\( \Rightarrow(\mathrm{P}, \mathrm{S}) \in \mathrm{R} \)
Therefore, \( R \) is transitive.
Therefore, \( R \) is equivalence relation.
The set of all points related to \( \mathrm{P} \neq(0,0) \) will be those points whose distance from the origin is the same as the distance of point \( P \) from the origin.
So, if \( O(0,0) \) is the origin and \( O P=k \), then the set of all points related to \( P \) is at a distance of \( k \) from the origin.
Therefore, this set of points forms a circle with the center as the origin and this circle passes through point \( P \).

It is given that the relation R defined in the set A of all triangles as
\( R=\left\{\left(T_{1}, T_{2}\right): T_{1}\right. \) is similar to \( \left.T_{2}\right\} \),
Now, R is reflexive as every triangle is similar to it self.
Now, if \( \left(T_{1}, T_{2}\right) \in R \), then \( T_{1} \) is similar to \( T_{2} \).
\( \Rightarrow \mathrm{T}_{2} \) is similar to \( \mathrm{T}_{1} \).
\( \Rightarrow\left(\mathrm{T}_{1}, \mathrm{~T}_{2}\right) \in \mathrm{R} \)
Therefore, \( R \) is symmetric.
Now, if \( \left(T_{1}, T_{2}\right),\left(T_{2}, T_{3}\right) \in R \),
\( \Rightarrow \mathrm{T}_{1} \) is similar to \( \mathrm{T}_{2} \) and \( \mathrm{T}_{2} \) is similar to \( \mathrm{T}_{3} \).
\( \Rightarrow \mathrm{T}_{1} \) is similar to \( \mathrm{T}_{3} \).
\(\Rightarrow\left(\mathrm{T}_{1}, \mathrm{~T}_{3}\right) \in \mathrm{R}
\)
Therefore, R is transitive.
Therefore, \( R \) is equivalence relation.
Now, we can see that,
\(\frac{3}{6}=\frac{4}{8}=\frac{5}{10}=\frac{1}{2}
\)
Therefore, the corresponding sides of triangles \( T_{1} \) and \( T_{3} \) are in the same ratio.
Thus, triangle \( T_{1} \) is similar to triangle \( T_{3} \).
Therefore, \( \mathrm{T}_{1} \) is related to \( \mathrm{T}_{3} \).

It is given that the relation R defined in the set A of all polygons as \( R=\left\{\left(P_{1}, P_{2}\right): P_{1}\right. \) and \( P_{2} \) have same number of sides\( \} \),
Then, \( R \) is reflexive since \( \left(P_{1}, P_{2}\right) \in R \) as the same polygon has the same number of sides with itself.
Let \( \left(P_{1}, P_{2}\right) \in R \)
\( \Rightarrow P_{1} \) and \( P_{2} \) have the same number of sides.
\( \Rightarrow P_{2} \) and \( P_{1} \) have the same number of sides.
\(\Rightarrow\left(\mathrm{P}_{2}, \mathrm{P}_{1}\right) \in \mathrm{R}
\)
Therefore, \( R \) is symmetric.
Now, let \( \left(\mathrm{P}_{1}, \mathrm{P}_{2}\right),\left(\mathrm{P}_{2}, \mathrm{P}_{3}\right) \in \mathrm{R} \)
\( \Rightarrow P_{1} \) and \( P_{2} \) have the same number of sides. Also, \( P_{2} \) and \( P_{3} \) have the same number of sides.
\( \Rightarrow P_{1} \) and \( P_{3} \) have the same number of sides.
\(\Rightarrow\left(\mathrm{P}_{1}, \mathrm{P}_{3}\right) \in \mathrm{R}
\)
Therefore, R is transitive.
Thus, \( R \) is an equivalence relation.
The elements in A related to the right-angled triangle (T) with sides 3, 4 and 5 are those polygons which have 3 sides.
Therefore, the set of all elements in A related to triangle T is the set of all triangles.

It is given that the relation in \( L \) defined as
\( R=\left\{\left(L_{1}, L_{2}\right): L_{1}\right. \) is parallel to \( \left.L_{2}\right\} \)
\( R \) is reflexive as any line \( L_{1} \) is parallel to itself
\(\Rightarrow\left(\mathrm{L}_{1}, \mathrm{~L}_{2}\right) \in \mathrm{R}
\)
Now, Let \( \left(\mathrm{L}_{1}, \mathrm{~L}_{2}\right) \in \mathrm{R} \)
\( \Rightarrow L_{1} \) is parallel to \( L_{2} \).
\( \Rightarrow L_{2} \) is parallel to \( L_{1} \).
\(\Rightarrow\left(\mathrm{L}_{2}, \mathrm{~L}_{1}\right) \in \mathrm{R}\)
Therefore, R is symmetric.
Now, Let \( \left(L_{1}, L_{2}\right),\left(L_{2}, L_{3}\right) \in R \)
\( \Rightarrow L_{1} \) is parallel to \( L_{2} \). Also, \( L_{2} \) is parallel to \( L_{3} \).
\( \Rightarrow L_{1} \) is parallel to \( L_{3} \).
\(\Rightarrow\left(\mathrm{L}_{1}, \mathrm{~L}_{3}\right) \in \mathrm{R}
\)
Therefore, \( R \) is transitive.
Therefore, \( R \) is an equivalence relation.
The set of all lines related to the line \( y=2 x+4 \) is the set of all lines that are parallel to the line
\(y=2 x+4\)
Slope of line \( y=2 x+4 \) is \( m=2 \)
We know that parallel lines have the same slopes.
The line parallel to the given line is of the form \( y=2 x+c \) where, \( c \in R \).
Therefore, the set of all lines related to the given line by \( y=2 x+c \), where \( \mathrm{c} \in \mathrm{R} \).

It is given that the relation in the set \( \{1,2,3,4\} \) given by \( \mathrm{R}=\{(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2)\} \)
Check symmetric:
As \( (1,1),(2,2),(3,3),(4,4) \in \mathrm{R} \)
It is seen that \( (a, a) \in R \), for every \(a \in\{1,2,3,4\} \)
Therefore, \( R \) is reflexive.
Check symmetric:
We can see that \( (1,2) \in R \), but \( (2,1) \notin R \).
Therefore, R is not symmetric.
Check transitive:
\((a, b),(b, c) \in R\)
\(\Rightarrow(a, c) \in R
\)
here \( (1,3) \in R,(3,2) \in R \) and \( (1,2) \in R \)
Therefore, \( R \) is transitive.
Therefore, \( R \) is reflexive and transitive but not symmetric.

It is given that the relation in the set N given by
\(R=\{(a, b): a=b-2, b > 6\}\)
Now, \( b > 6,(2,4) \notin R \)
Also, as \( 3 \neq 8-2,(3,8) \notin R \)
And \( 8 \neq 7-2 \)
Therefore, \( (8,7) \notin \mathrm{R} \)
Now, consider \( (6,8) \)
We have \( 8 > 6 \) and also \( 6=8-2 \)
Therefore, \( (6,8) \in \mathrm{R} \).