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# if points in g.p then-

Asked by ankitmathur 24th March 2010, 7:59 PM

(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

Regards,

Team,

TopperLearning.

Answered by Expert 28th March 2010, 7:39 PM
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