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The set of real values of k for which th...

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

A

A. `R-{-4,(2)/(3)}`

B

B. `R-{-4,(-6)/(5),(2)/(3)}`

C

C. `R-{(-2)/(3),4}`

D

D. R

Text Solution

Verified by Experts

The correct Answer is:
B

Lines form triangle. Therefore, `x +3y +1 = 0` is not parallel to `kx +2y -2 = 0`
`:. (-k)/(2) ne (-1)/(3) rArr k ne (2)/(3)`
Also line `2x - y+3 =0` is not parallel to `kx +2y -2 = 0`
`:. (-k)/(2) ne 2 rArr k ne -4`
Further lines must not be concurrent
`:. |{:(1,3,1),(k,2,-2),(2,-1,3):}|ne0rArrkne(-6)/(5)`
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