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If 1a,1b,1c are in A.P., then the straight line xa+yb+1c=0 always passes through a fixed point, that point is

a
(-1, -2)
b
(-1, 2)
c
(1, -2)
d
1,-12

detailed solution

Correct option is C

Since  1a,1b,1c are in A.P.⇒1a+1c=2b  -----------(1)The given line is  xa+yb+1c=0⇒xa+yb+2b-1a=0    [Using (1)]⇒1ax-1+1ay+2=0⇒The given line passes through the point of intersection of x – 1 = 0 and y + 2 = 0 i.e., (1, –2) which is a fixed point.

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