if points A(x1,y1), B(x2,y2),and C(x3,y3) are such that x1 ,x2 ,x3 and y1, y2 ,y3 are in g.p. with same c
if points in g.p then-
(B) A, B and C are collinear points.
Consider the determinant, with first, second and third row respectively,
x1 y1 1
x2 y2 1
x3 y3 1
which gives the area of triangle formed by the points.
We find the determinant,
x1(y2-y3) - y1(x2-x3) + (x2y3 - y2x3)
= (x1/y2)(1 - y3/y2) - (y1/x2)(1 - x3/x2) + (1/x2y3)(1 - y2x3/x2y3)
= (x1/y2)(1 - b) - (y1/x2)(1 - b) + (1/x2y3)(1 - b/b) ... where x3/x2 = y3/y2 = b = common ratio.
= (x1/y2)(1 - b) - (y1/x2)(1 - b)
= (1 - b) ((x1/y2) - (y1/x2))
= (1 - b)(x1/y2) (1 - (y1/x2)/(x1/y2))
= (1 - b)(x1/y2)( 1 - b/b) = 0