If a, b, c are three terms of an A.P., then the line ax + by + c = 0

Let a, b, c be pth, qth and rth terms of an A.P. whose first term is A and common difference is d. The given line is ax + by + c = 0⇒[A + (p – 1) d] x + [A + (q – 1) d] y + [A + (r – 1) d] = 0⇒A(x + y + 1) + d((p – 1) x + (q – 1) y + r – 1) = 0⇒The given line passes through the point of intersection of lines x + y + 1 = 0 and ( p – 1) x + (q – 1)y + r – 1 = 0, which is a fixed point.