NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.1 - Relations and Functions

Question1(i) . Determine whether each of the following relations are reflexive, symmetric and transitive:

(i) Relation $R$ in the set $A=\{1,2,3...,13,14\}$ defined as $R=\{(x,y):3x-y=0\}$

Answer:

$A=\{1,2,3,\dots ,13,14\}$

$R=\{(x,y):3x-y=0\}=\{(1,3),(2,6),(3,9),(4,12)\}$

Since $(1,1),(2,2),(3,3),(4,4),(5,5),\dots ,(14,14)\notin R$, so $R$ is not reflexive.

Since $(1,3)\in R$ but $(3,1)\notin R$, so $R$ is not symmetric.

Since $(1,3),(3,9)\in R$ but $(1,9)\notin R$, so $R$ is not transitive.

Hence, $R$ is neither reflexive, nor symmetric, nor transitive.

Question 1(ii) . Determine whether each of the following relations are reflexive, symmetric and
transitive:

(ii) Relation R in the set N of natural numbers defined as
$R=\{(x,y):y=x+5\text{and}x<4\}$

Answer:

$R=\{(x,y):y=x+5\text{and}x<4\}$ $=\{(1,6),(2,7),(3,8)\}$

Since, $(1,1)\notin R$

so $R$ is not reflexive.

Since, $(1,6)\in R$ but $(6,1)\notin R$

so $R$ is not symmetric.

Since there is no pair in $R$ such that $(x,y),(y,x)\in R$ so this is not transitive.

Hence, $R$ is neither reflexive nor symmetric and
nor transitive.

Question1(iii) Determine whether each of the following relations are reflexive, symmetric and
transitive:

(iii) Relation R in the set $A=\{1,2,3,4,5,6\}$ as $R=\{(x,y):y\text{is}divisiblebyx\}$

Answer:

$A=\{1,2,3,4,5,6\}$

$R=\{(2,4),(3,6),(2,6),(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}$

Any number is divisible by itself and $(x,x)\in R$ .So it is reflexive.

$(2,4)\in R$ but $(4,2)\notin R$ .Hence,it is not symmetric.

$(2,4),(4,4)\in R$ and 4 is divisible by 2 and 4 is divisible by 4.

Hence, it is transitive.

Hence, it is reflexive and transitive but not symmetric.

Question.1(iv) Determine whether each of the following relations are reflexive, symmetric and
transitive:

(iv). Relation R in the set Z of all integers defined as $R=\{(x,y):x-yisaninteger\}$

Answer:

$R=\{(x,y):x-yisaninteger\}$

For $x\in Z$ , $(x,x)\in R$ as $x-x=0$ which is an integer.

So,it is reflexive.

For $x,y\in Z$ , $(x,y)\in R$ and $(y,x)\in R$ because $x-yandy-x$ are both integers.

So, it is symmetric.

For $x,y,z\in Z$ , $(x,y),(y,z)\in R$ as $x-yandy-z$ are both integers.

Now, $x-z=(x-y)+(y-z)$ is also an integer.

So, $(x,z)\in R$ and hence it is transitive.

Hence, it is reflexive, symmetric and transitive.

Question:1(v) Determine whether each of the following relations are reflexive, symmetric and
transitive:

(v) Relation R in the set A of human beings in a town at a particular time given by

(a) $R=\{(x,y):xandyworkatthesameplace\}$

Answer:

$R=\{(x,y):x\text{ and }y\text{ work at the same place}\}$

$(x,x)\in R$ ,so it is reflexive

$(x,y)\in R$ means $xandyworkatthesameplace$ .

$yandxworkatthesameplace$ i.e. $(y,x)\in R$ so it is symmetric.

$(x,y),(y,z)\in R$ means $xandyworkatthesameplace$ also $yandzworkatthesameplace$ .It states that $xandzworkatthesameplace$ i.e. $(x,z)\in R$ .So, it is transitive.

Hence, it is reflexive, symmetric and transitive.

Question 1 (v) Determine whether each of the following relations are reflexive, symmetric and
transitive:

(v) Relation R in the set A of human beings in a town at a particular time given by

(b) $R=\{(x,y):xandyliveinthesamelocality\}$

Answer:

$R=\{(x,y):xandyliveinthesamelocality\}$

$(x,x)\in R$ as $x$ and $x$ is same human being.So, it is reflexive.

$(x,y)\in R$ means $xandyliveinthesamelocality$ .

It is same as $yandxliveinthesamelocality$ i.e. $(y,x)\in R$ .

So,it is symmetric.

$(x,y),(y,z)\in R$ means $xandyliveinthesamelocality$ and $yandzliveinthesamelocality$ .

It implies that $xandzliveinthesamelocality$ i.e. $(x,z)\in R$ .

Hence, it is reflexive, symmetric and transitive.

Question 1 (v) Determine whether each of the following relations are reflexive, symmetric and
transitive:

(v) Relation R in the set A of human beings in a town at a particular time given by

(c) $R=\{(x,y):xisexactly7cmtallerthany\}$

Answer:

$R=\{(x,y):xisexactly7cmtallerthany\}$

$(x,y)\in R$ means $xisexactly7cmtallerthany$ but $xisnottallerthanx$ i.e. $(x,x)\notin R$ .So, it is not reflexive.

$(x,y)\in R$ means $xisexactly7cmtallerthany$ but $yisnottallerthanx$ i.e $(y,x)\notin R$ .So, it is not symmetric.

$(x,y),(y,z)\in R$ means $xisexactly7cmtallerthany$ and $yisexactly7cmtallerthanz$ .

$xisexactly14cmtallerthanz$ i.e. $(x,z)\notin R$ .

Hence, it is not reflexive, symmetric, or transitive.

Question:1(v) Determine whether each of the following relations are reflexive, symmetric and
transitive:

(v). Relation R in the set A of human beings in a town at a particular time given by

(d) $R=\{(x,y):xiswifeofy\}$

Answer:

$R=\{(x,y):xiswifeofy\}$

$(x,y)\in R$ means $xiswifeofy$ but $xisnotwifeofx$ i.e. $(x,x)\notin R$ .

So, it is not reflexive.

$(x,y)\in R$ means $xiswifeofy$ but $yisnotwifeofx$ i.e. $(y,x)\notin R$ .

So, it is not symmetric.

Let, $(x,y),(y,z)\in R$ means $xiswifeofy$ and $yiswifeofz$ .

This case is not possible, so it is not transitive.

Hence, it is not reflexive, symmetric or transitive.

Question 1 (v) Determine whether each of the following relations are reflexive, symmetric and
transitive:

(v) Relation R in the set A of human beings in a town at a particular time given by

(e) $R=\{(x,y):xisfatherofy\}$

Answer:

If $(x,y)\in R$, then $x$ is the father of $y$. \

But a person cannot be their own father, so $(x,x)\notin R$. \

$\Rightarrow R\text{ is not reflexive.}$

If $(x,y)\in R$, then $x$ is the father of $y$. \

This does not imply that $y$ is the father of $x$, so $(y,x)\notin R$. \

$\Rightarrow R\text{ is not symmetric.}$

Suppose $(x,y),(y,z)\in R$. \

Then $x$ is the father of $y$, and $y$ is the father of $z$. \

This means $x$ is the grandfather of $z$, not the father. \

So, $(x,z)\notin R$. \

$\Rightarrow R\text{ is not transitive.}$

So, it is not transitive.

Hence, it is neither reflexive nor symmetric nor transitive.

Question:2 Show that the relation R in the set R of real numbers defined as
$R=\{(a,b):a\le b^2\}$ is neither reflexive nor symmetric nor transitive.

Answer:

$R=\{(a,b):a\le b^2\}$

Taking

$(\frac{1}{2},\frac{1}{2})\notin R$

and

$\left(\frac{1}{2}\right)>{\left(\frac{1}{2}\right)}^2$

So, R is not reflexive.

Now,

$(1,2)\in R$ because $1<4$ .

But, $4\nless 1$ i.e. 4 is not less than 1

So, $(2,1)\notin R$

Hence, it is not symmetric.

$(3,2)\in Rand(2,1.5)\in R$ as $3<4and2<2.25$

Since $(3,1.5)\notin R$ because $3\nless 2.25$

Hence, it is not transitive.

Thus, we can conclude that it is neither reflexive, nor symmetric, nor transitive.

Question:3 Check whether the relation R defined in the set $\{1,2,3,4,5,6\}$ as
$R=\{(a,b):b=a+1\}$ is reflexive, symmetric or transitive.

Answer:

$R$ defined on the set $\{1,2,3,4,5,6\}$

$R=\{(a,b):b=a+1\}$

$R=\{(1,2),(2,3),(3,4),(4,5),(5,6)\}$

Since $\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}\notin R$, so it is not reflexive.

$\{(1,2),(2,3),(3,4),(4,5),(5,6)\}\in R$ but $\{(2,1),(3,2),(4,3),(5,4),(6,5)\}\notin R$, so it is not symmetric.

$\{(1,2),(2,3),(3,4),(4,5),(5,6)\}\in R$ but $\{(1,3),(2,4),(3,5),(4,6)\}\notin R$, so it is not transitive.

So, it is not transitive.

Hence, it is neither reflexive, nor symmetric, nor transitive

Question:4 Show that the relation R in R defined as $R=\{(a,b):a\le b\}$ , is reflexive and

transitive but not symmetric.

Answer:

$R=\{(a,b):a\le b\}$

As $(a,a)\in R$ so it is reflexive.

Now we take an example

$(2,3)\in R$ as $2<3$

But $(3,2)\notin R$ because $2\nless 3$ .

So,it is not symmetric.

Now if we take, $(2,3)\in Rand(3,4)\in R$

Than, $(2,4)\in R$ because $2<4$

So, it is transitive.

Hence, we can say that it is reflexive and transitive but not symmetric.

Question:5 Check whether the relation R in R defined by $R=\{(a,b):a\le b^3\}$ is reflexive,
symmetric or transitive.

Answer:

$R=\{(a,b):a\le b^3\}$

$(\frac{1}{2},\frac{1}{2})\notin R$ because $\frac{1}{2}⪇(\frac{1}{2}{)}^3$

So, it is not symmetric

Now, $(1,2)\in R$ because $1<2^3$

but $(2,1)\notin R$ because $2⪇1^3$

It is not symmetric

$(3,1.5)\in Rand(1.5,1.2)\in R$ as $3<{1.5}^{3}and1.5<{1.2}^3$ .

But, $(3,1.2)\notin R$ because $3⪇{1.2}^3$

So it is not transitive

Thus, it is neither reflexive, nor symmetric, nor transitive.

Question:6 Show that the relation R in the set $\{1,2,3\}$ given by $R=\{(1,2),(2,1)\}$ is
symmetric but neither reflexive nor transitive.

Answer:

Let A= $\{1,2,3\}$

$R=\{(1,2),(2,1)\}$

We can see $(1,1),(2,2),(3,3)\notin R$ so it is not reflexive.

As $(1,2)\in Rand(2,1)\in R$ so it is symmetric.

$(1,2)\in Rand(2,1)\in R$

But $(1,1)\notin R$ so it is not transitive.

Hence, R is symmetric but neither reflexive nor transitive.

Question:7 Show that the relation R in the set A of all the books in a library of a college,
given by $R=\{(x,y):xandyhavesamenumberofpages\}$ is an equivalence
relation.?

Answer:

A = all the books in a library of a college

$R=\{(x,y):xandyhavesamenumberofpages\}$

$(x,x)\in R$ because x and x have the same number of pages, so it is reflexive.

Let $(x,y)\in R$ means x and y have the same number of pages.

Since y and x have the same number of pages, so $(y,x)\in R$.

Hence, it is symmetric.

Let $(x,y)\in R$ means x and y have the same number of pages.

and $(y,z)\in R$ means y and z have the same number of pages.

This states,x and z also have the same number of pages i.e. $(x,z)\in R$

Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive, i.e. it is an equivalence
relation.

Question:8 Show that the relation R in the set $A=\{1,2,3,4,5\}$ given by $R=\{(a,b):|a-b|iseven\}$ , is an equivalence relation. Show that all the elements of $\{1,3,5\}$ are related to each other and all the elements of $\{2,4\}$ are related to each other. But no element of $\{1,3,5\}$ is related to any element of $\{2,4\}$ .

Answer:

$A=\{1,2,3,4,5\}$

$R=\{(a,b):|a-b|iseven\}$

$R=\{(1,1),(2,2),(3,3),(4,4),(5,5),(1,3),(2,4),(3,5),(3,1),(5,1),(4,2),(5,3)\}$

Let there be $a\in A$ then $(a,a)\in R$ as $|a-a|=0$ which is even number. Hence, it is reflexive

Let $(a,b)\in R$ where $a,b\in A$ then $(b,a)\in R$ as $|a-b|=|b-a|$

Hence, it is symmetric

Now, let $(a,b)\in Rand(b,c)\in R$

$|a-b|and|b-c|$ are even number i.e. $(a-b)and(b-c)$ are even

then, $(a-c)=(a-b)+(b-c)$ is even (sum of even integer is even)

So, $(a,c)\in R$ . Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive, i.e. it is an equivalence relation.

The elements of $\{1,3,5\}$ are related to each other because the difference of odd numbers gives even number and in this set all numbers are odd.

The elements of $\{2,4\}$ are related to each other because the difference of even number is even number and in this set, all numbers are even.

The element of $\{1,3,5\}$ is not related to $\{2,4\}$ because a difference of odd and even number is not even.

Question:9(i) Show that each of the relation R in the set $A=\{x\in Z:0\le x\le 12\}$ , given by

(i) $R=\{(a,b):|a-b|isamultipleof4\}$ is an equivalence relation. Find the set of all elements related to 1 in each case.

Answer:

$A=\{x\in Z:0\le x\le 12\}$

$A=\{0,1,2,3,4,5,6,7,8,9,10,11,12\}$

$R=\{(a,b):|a-b|isamultipleof4\}$

For $a\in A$ , $(a,a)\in R$ as $|a-a|=0$ which is multiple of 4.

Henec, it is reflexive.

Let, $(a,b)\in R$ i.e. $|a-b|$ is multiple of 4.

then $|b-a|$ is also multiple of 4 because $|a-b|$ = $|b-a|$ i.e. $(b,a)\in R$

Hence, it is symmetric.

Let, $(a,b)\in R$ i.e. $|a-b|$ is multiple of 4 and $(b,c)\in R$ i.e. $|b-c|$ is multiple of 4 .

$(a-b)$ is multiple of 4 and $(b-c)$ is multiple of 4

$(a-c)=(a-b)+(b-c)$ is multiple of 4

$|a-c|$ is multiple of 4 i.e. $(a,c)\in R$

Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive i.e. it is an equivalence relation.

The set of all 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.

Question:9(ii) Show that each of the relation R in the set $A=\{x\in Z:0\le x\le 12\}$ , given by

(ii) $R=\{(a,b):a=b\}$ is an equivalence relation. Find the set of all elements related to 1 in each case.

Answer:

$A=\{x\in Z:0\le x\le 12\}$

$A=\{0,1,2,3,4,5,6,7,8,9,10,11,12\}$

$R=\{(a,b):a=b\}$

For $a\in A$ , $(a,a)\in R$ as $a=a$

Henec, it is reflexive.

Let, $(a,b)\in R$ i.e. $a=b$

$a=b$ $\Rightarrow$ $b=a$ i.e. $(b,a)\in R$

Hence, it is symmetric.

Let, $(a,b)\in R$ i.e. $a=b$ and $(b,c)\in R$ i.e. $b=c$

$\therefore$ $a=b=c$

$a=c$ i.e. $(a,c)\in R$

Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive i.e. it is an equivalence relation.

The set of all elements related to 1 is {1}

Question:10(i) Give an example of a relation.

(i) Which is Symmetric but neither reflexive nor transitive.

Answer:

Let

$A=\{1,2,3\}$

$R=\{(1,2),(2,1)\}$

$(1,1),(2,2),(3,3)\notin R$ so it is not reflexive.

$(1,2)\in R$ and $(2,1)\in R$ so it is symmetric.

$(1,2)\in Rand(2,1)\in R$ but $(1,1)\notin R$ so it is not transitive.

Hence, symmetric but neither reflexive nor transitive.

Question:10(ii) Give an example of a relation.

(ii) Which is transitive but neither reflexive nor symmetric.

Answer:

Let

$R=\{(x,y):x>y\}$

Now for $x\in R$ , $(x,x)\notin R$ so it is not reflexive.

Let $(x,y)\in R$ i.e. $x>y$

Then $y>x$ is not possible i.e. $(y,x)\notin R$ . So it is not symmetric.

Let $(x,y)\in R$ i.e. $x>y$ and $(y,z)\in R$ i.e. $y>z$

we can write this as $x>y>z$

Hence, $x>z$ i.e. $(x,z)\in R$ . So it is transitive.

Hence, it is transitive but neither reflexive nor symmetric.

Question:10(iii) Give an example of a relation.

(iii) Which is Reflexive and symmetric but not transitive.

Answer:

Let

$A=\{1,2,3\}$

Define a relation R on A as

$R=\{(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)\}$

If $x\in A$ , $(x,x)\in R$ i.e. $\{(1,1),(2,2),(3,3)\}\in R$ . So it is reflexive.

If $x,y\in A$ , $(x,y)\in R$ and $(y,x)\in R$ i.e. $\{(1,2),(2,1),(2,3),(3,2)\}\in R$ . So it is symmetric.

$(x,y)\in R$ and $(y,z)\in R$ i.e. $(1,2)\in R$ . and $(2,3)\in R$

But $(1,3)\notin R$, So it is not transitive.

Hence, it is Reflexive and symmetric but not transitive.

Question 10 (iv) Give an example of a relation.

(iv) Which is Reflexive and transitive but not symmetric.

Answer:

Let there be a relation R in R

$R=\{(a,b):a\le b\}$

$(a,a)\in R$ because $a=a$

Let $(a,b)\in R$ i.e. $a\le b$

But $(b,a)\notin R$ i.e. $b⪇a$

So it is not symmetric.

Let $(a,b)\in R$ i.e. $a\le b$ and $(b,c)\in R$ i.e. $b\le c$

This can be written as $a\le b\le c$ i.e. $a\le c$ implies $(a,c)\in R$

Hence, it is transitive.

Thus, it is Reflexive and transitive but not symmetric.

Question 10 (v) Give an example of a relation.

(v) Which is Symmetric and transitive but not reflexive.

Answer:

Let there be a relation A in R

$A=\{1,2\}$

$R=\{(1,2),(2,1),(2,2)\}$

$(1,1)\notin R$ So R is not reflexive.

We can see $(1,2)\in R$ and $(2,1)\in R$

So it is symmetric.

Let $(1,2)\in R$ and $(2,1)\in R$

Also $(2,2)\in R$

Hence, it is transitive.

Thus, it Symmetric and transitive but not reflexive.

Question:11 Show that the relation R in the set A of points in a plane given by
$R=\{(P,Q):distanceofthepointPfromtheoriginissameasthedistanceofthepointQfromtheorigin\}$ , is an equivalence relation. Further, show that the set of
all points related to a point $P\ne (0,0)$ is the circle passing through P with origin as
centre.

Answer:

$R=\{(P,Q):distanceofthepointPfromtheoriginissameasthedistanceofthepointQfromtheorigin\}$

The distance of point P from the origin is always the same as the distance of same point P from origin i.e. $(P,P)\in R$

$\therefore$ R is reflexive.

Let $(P,Q)\in R$ i.e. the distance of the point P from the origin is the same as the distance of the point Q from the origin.

this is the same as distance of the point Q from the origin is the same as the distance of the point P from the origin i.e. $(Q,P)\in R$

$\therefore$ R is symmetric.

Let $(P,Q)\in R$ and $(Q,S)\in R$

i.e. the distance of point P from the origin is the same as the distance of point Q from the origin, and also the distance of point Q from the origin is the same as the distance of the point S from the origin.

We can say that the distance of point P, Q, S from the origin is the same. Means distance of point P from the origin is the same as the distance of point S from origin i.e. $(P,S)\in R$

$\therefore$ R is transitive.

Hence, R is an equivalence relation.

The set of all points related to a point $P\ne (0,0)$ are points whose distance from the origin is the same as the distance of point P from the origin.

In other words, we can say there be a point O(0,0) as origin and distance between point O and point P be k=OP then set of all points related to P is at distance k from the origin.

Hence, these sets of points form a circle with the centre as the origin and this circle passes through the point.

Question:12 Show that the relation R defined in the set A of all triangles as $R=\{(T_1,T_2):T_{1}issimilarto{T}_2\}$ , is equivalence relation. Consider three right angle triangles T 1 with sides 3, 4, 5, T 2 with sides 5, 12, 13 and T 3 with sides 6, 8, 10. Which triangles among T 1 , T 2 and T 3 are related?

Answer:

$R=\{(T_1,T_2):T_{1}issimilarto{T}_2\}$

All triangles are similar to itself, so it is reflexive.

Let,

$(T_1,T_2)\in R$ i.e.T 1 is similar to T2

T 1 is similar to T2 is the same asT2 is similar to T 1 i.e. $(T_2,T_1)\in R$

Hence, it is symmetric.

Let,

$(T_1,T_2)\in R$ and $(T_2,T_3)\in R$ i.e. T 1 is similar to T2 and T2 is similar toT 3 .

$\Rightarrow$ T 1 is similar toT 3 i.e. $(T_1,T_3)\in R$

Hence, it is transitive,

Thus, $R=\{(T_1,T_2):T_{1}issimilarto{T}_2\}$ , is equivalence relation.

Now, we see the ratio of sides of triangle T 1 andT 3 are as shown

$\frac{3}{6}=\frac{4}{8}=\frac{5}{10}=\frac{1}{2}$

i.e. ratios of sides of T 1 and T 3 are equal.Hence, T 1 and T 3 are related.

Question:13 Show that the relation R defined in the set A of all polygons as $R=\{(P_1,P_2):P_{1}and{P}_{2}havesamenumberofsides\}$ , is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?

Answer:

$R=\{(P_1,P_2):P_{1}and{P}_{2}havesamenumberofsides\}$

The same polygon has the same number of sides with itself,i.e. $(P_1,P_2)\in R$ , so it is reflexive.

Let,

$(P_1,P_2)\in R$ i.e.P 1 have same number of sides as P 2

P 1 have the same number of sides as P 2 is the same as P 2 have same number of sides as P 1 i.e. $(P_2,P_1)\in R$

Hence,it is symmetric.

Let,

$(P_1,P_2)\in R$ and $(P_2,P_3)\in R$ i.e. P 1 have the same number of sides as P 2 and P 2 have same number of sides as P 3

$\Rightarrow$ P 1 have same number of sides as P 3 i.e. $(P_1,P_3)\in R$

Hence, it is transitive,

Thus, $R=\{(P_1,P_2):P_{1}and{P}_{2}havesamenumberofsides\}$ , is an equivalence relation.

The elements in A related to the right angle triangle T with sides 3, 4 and 5 are those polygons which have 3 sides.

Hence, the set of all elements in A related to the right angle triangle T is set of all triangles.

Question:14 Let L be the set of all lines in XY plane and R be the relation in L defined as $R=\{(L_1,L_2):L_{1}isparallelto{L}_2\}$ . Show that R is an equivalence relation. Find the set of all lines related to the line $y=2x+4.$

Answer:

$R=\{(L_1,L_2):L_{1}isparallelto{L}_2\}$

All lines are parallel to itself, so it is reflexive.

Let,

$(L_1,L_2)\in R$ i.e.L 1 is parallel to L 2 .

L1 is parallel to L 2 is same as L 2 is parallel to L 1 i.e. $(L_2,L_1)\in R$

Hence, it is symmetric.

Let,

$(L_1,L_2)\in R$ and $(L_2,L_3)\in R$ i.e. L1 is parallel to L 2 and L 2 is parallel to L 3 .

$\Rightarrow$ L 1 is parallel to L 3 i.e. $(L_1,L_3)\in R$

Hence, it is transitive,

Thus, $R=\{(L_1,L_2):L_{1}isparallelto{L}_2\}$ , is equivalence relation.

The set of all lines related to the line $y=2x+4.$ are lines parallel to $y=2x+4.$

Here, Slope = m = 2 and constant = c = 4

It is known that the slope of parallel lines are equal.

Lines parallel to this ( $y=2x+4.$ ) line are $y=2x+c$ , $c\in R$

Hence, set of all parallel lines to $y=2x+4.$ are $y=2x+c$ .

Question:15 Let R be the relation in the set A= {1,2,3,4}

given by $R=\{(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2)\}$ . Choose the correct answer.