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

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Solution

Given, the points A(k+1,2k),B(3k,2k+3) and C(5k-1,5k) are collinear.

$∵$ The point to be collinear.

$∴ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0$

(k+1)(2k+3-5k)+3k(5k-2k)+(5k-1)(2k-2k-3)=0

(k+1)(3-3k)+3k(3k)+(5k-1)(-3)=0

$3 k + 3 - 3 k^{2} - 3 k + 9 k^{2} - 15 k + 3 = 0$

$6 k^{2} - 15 k + 6 = 0$

$2 k^{2} - 5 k + 2 = 0$

$2 k^{2} - 4 k - k + 2 = 0$

2k(k-2)-1(k-2)=0

(2k-1)(k-2)=0

$k = \frac{1}{2}$ or k=2

Hence, $k = \frac{1}{2}$ or k=2.

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