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 The set of real values of k for which the lines x+3y+1=0,kx+2y2=0 and 2xy+3=0 form a triangle is 

a
R−−4,23
b
R−−4,−65,23
c
R−−23,4
d
R

detailed solution

Correct option is B

Lines form triangle. Therefore, x+3y+1=0 is not parallel  to kx+2y−2=0⇒ -k2x-y+1=0∴ −k2≠−13⇒k≠23 Also line 2x−y+3=0 is not parallel to kx+2y−2=0 2x3-y3+1=0   -kx2-y+2=0 ⇒-3k4≠3⇒k≠-4 Further lines must not be concurrent. ∴ 131k2−22−13≠0⇒k≠−65

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