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Question: If the points (2k + 1, k), (2, 3) and (k, 2k − 1) are collinear then sum of possible real values of ...

If the points (2k + 1, k), (2, 3) and (k, 2k − 1) are collinear then sum of possible real values of k, is

A

-1/3

B

2

C

5/3

D

11/3

Answer

5/3

Explanation

Solution

For collinearity, the area of the triangle formed by the points must be zero. This can be expressed using a determinant:

∣2k+1k1231k2k−11∣=0\begin{vmatrix} 2k+1 & k & 1 \\ 2 & 3 & 1 \\ k & 2k-1 & 1 \end{vmatrix} = 0

Expanding the determinant:

(2k+1)[3−(2k−1)]−k[2−k]+1[2(2k−1)−3k]=0(2k+1)[3 - (2k-1)] - k[2 - k] + 1[2(2k-1) - 3k] = 0

Simplifying step by step:

  1. 3−(2k−1)=4−2k3 - (2k-1) = 4 - 2k, so the first term is (2k+1)(4−2k)(2k+1)(4-2k).
  2. Expanding (2k+1)(4−2k)=8k+4−4k2−2k=−4k2+6k+4(2k+1)(4-2k) = 8k + 4 - 4k^2 -2k = -4k^2 + 6k + 4.
  3. The second term is −k(2−k)=−2k+k2-k(2-k) = -2k + k^2.
  4. The third term is 2(2k−1)−3k=4k−2−3k=k−22(2k-1)-3k = 4k -2 -3k = k-2.

Combining the terms:

−4k2+6k+4−2k+k2+k−2=0-4k^2 + 6k + 4 - 2k + k^2 + k - 2 = 0

Combining like terms:

−3k2+5k+2=0-3k^2 + 5k + 2 = 0

Multiplying by -1:

3k2−5k−2=03k^2 - 5k - 2 = 0

Solving the quadratic equation using the quadratic formula:

k=−b±b2−4ac2a=5±(−5)2−4⋅3⋅(−2)2⋅3=5±25+246=5±76k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 3 \cdot (-2)}}{2 \cdot 3} = \frac{5 \pm \sqrt{25 + 24}}{6} = \frac{5 \pm 7}{6}

Thus, the roots are:

k=126=2andk=−26=−13k = \frac{12}{6} = 2 \quad \text{and} \quad k = \frac{-2}{6} = -\frac{1}{3}

The sum of the possible real values of kk is:

2+(−13)=63−13=532 + \left(-\frac{1}{3}\right) = \frac{6}{3} - \frac{1}{3} = \frac{5}{3}