‌2x2+xy−6y2=0 ⇒‌‌2x2+4xy−3xy−6y2=0 ⇒‌‌2x(x+2y)−3y(x+2y)=0 ⇒‌‌(2x−3y)(x+2y)=0 ⇒‌‌2x−3y=0‌‌‌⋅⋅⋅⋅⋅⋅⋅(i) ‌ or ‌‌x+2y=0‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)
Given, x+y−1=0... (iii) Now on solving Eqs. (i) and (ii) we gel. ( 0,0 ). On solving Eqs. (ii) and (iii), we get ( 2,−1 ) and solving Eqs. (i) and (iii) we get, (‌
3
5
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2
5
) As we observe that, AB≠BC≠AC Because, AB=√‌