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

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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.
Let \: L1 \: \epsilon \: L
Since, one line is always parallel to itself.
(L1, L1) \: \epsilon \: R
R is reflexive.
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
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.

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