determining k

Asked by meghapte | 18th Jan, 2010, 09:21: PM

Expert Answer:

For the three points to be collinear, the following determinant must be zero, whose elements are,

First row:       5 2 1

Second row:  k 3 1

Third row:      G k 1

5(3-k)-2(k-G)+(k2-3G) = 0

k2 -7k +15-G = 0

k = [7±(49-4(15-G))]/2

= [7±(4G - 11)]/2

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Answered by  | 19th Jan, 2010, 10:02: AM

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