If `a ,\ b ,\ c` are in A.P. prove that the straight lines `a x+2y+1=0,\ b x+3y+1=0\ a n d\ c x+4y+1=0` are concurrent.

Explanation

Given `ax + 2y + 1 = 0, bx + 3y + 1 = 0 `and` cx + 4y + 1 = 0` are concurrent.


So determinant = 0

=`|[a,2,1],[b,3,1],[c,4,1]|`



`=> a(3 – 4) – 2(b – c) + 1(4b – 3c) = 0`


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