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Name: Please Solve RD Sharma Class 12 Chapter Relation Exercise 1.2 Question 9 Maths Textbook Solution.
Uploaded: Sun, 2021-07-04 19:50
Description: Please Solve RD Sharma Class 12 Chapter Relation Exercise 1.2 Question 9 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Relation Exercise 1.2 Question 9 Maths Textbook Solution.

Answers (1)

Answer: R is an equivalence relation.
$T\! he\: set\: o\! f \: line\: parallel\: to\: the \: line \: y=2x+4$
$y=2x+C\: f\! or \: all\: C \: \epsilon \: R \: where\: R\: is \: the \: set\: o\! f \: real \: numbers.$
Hint:To prove equivalence relation it is necessary that the given relation should be reflexive, symmetric and transitive.
Given:   We have, L is the set of lines
$R=\! \left \{\left (L1, L2 \right ) :L1\: is\: parallel\: to \: L2 \right \} be\: a \: relation\: on\: L.$
Explanation:
Now,
Reflexivity:
$Let \: L1 \: \epsilon \: L$
Since, one line is always parallel to itself.
$(L1, L1) \: \epsilon \: R$
R is reflexive.
Symmetric:
$Let,L1, L2 \: \epsilon \: L \: and (L1,L2) \: \epsilon \: R$
          $L1\: is \: parallel\: to\: L2$
         $L2\: is\: parallel\: to\: L1$
        $(L2, L1) \: \epsilon \: R$
R is symmetric
Transitive:
$Let,L1, L2 \: and\: L3 \: \epsilon \: L\: such\: that (L1, L2) \: \epsilon \: R\: and (L2, L3) \: \epsilon \: R$
          $L1\: is\: parallel\: to\: L3\: and\: L2\: is\: parallel\: to\: L3$
         $L1\: is \: parallel\: to\: L3$
        $(L1, L3) \: \epsilon \: R$
R is transitive
Therefore, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation.
The set of lines parallel to the line $y=2x+4$ is
$y=2x+C f\! or\: all\: C \: \epsilon \: R$
where R is the set of real numbers.