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  • Find the values of k if the points A(k + 1, 2k), B(3k, 2k + 3) and C (5k –1, 5k)are collinear.

Find the values of k if the points A(k + 1, 2k), B(3k, 2k + 3) and C (5k –1, 5k)are collinear.

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Solution.     If points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear then area of triangle is equal to zero
\frac{1}{2}\left [ x_{1}\left ( y_{2}-y_{3} \right )+x_{2}\left ( y_{3}-y_{1} \right ) +x_{3}\left ( y_{1}-y_{2} \right )\right ]= 0
\left [ x_{1}\left ( y_{2}-y_{3} \right )+x_{2}\left ( y_{3}-y_{1} \right ) +x_{3}\left ( y_{1}-y_{2} \right )\right ]= 0
[(k + 1) (2k + 3 – 5k) + 3k(5k – 2k) + (5k – 1)(2k – 2k – 3)] = 0
[(k + 1) (3 – 3k) + 3k(3k) + 5(k – 1) (–3)] = 0
[3k – 3k2 + 3 – 3k + 9k2 – 15k + 3] = 0
6k2 – 15k + 6 = 0
6k2 – 12k – 3k + 6 = 0
6k(k – 2) – 3(k – 2) = 0
(k – 2)(6k – 3 )
k= 2,k= \frac{3}{6}
          = \frac{1}{2}
Hence, values of k are 2, 1/2

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