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Question : 1 of 120
Marks:
+1,
-0
Solution:
(−−2x−3y)(−3x−2y) =0 ⇒(−1)(1)+(−2x)(−3x) +(−3y)(−2y)=0 ⇒6x2+6y2=1 ⇒x2+y2=(1∕√6)2 Locus of
(x,y) is a circle
(−−2x−3y)(−3x−2y) =0 ⇒(−1)(1)+(−2x)(−3x) +(−3y)(−2y)=0 ⇒6x2+6y2=1 ⇒x2+y2=(1∕√6)2 Locus of
(x,y) is a circle
Hence, option (3) is correct.
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