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

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

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k = 3, $\frac{1}{3}$

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k = 4, $\frac{1}{4}$

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k = 5, $\frac{1}{5}$

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Solution

The correct option is A
k = 2, $\frac{1}{2}$

We know that, if three points are collinear, then the area of the triangle formed by these points is zero.
Since, the points A(k+1,2k), B(3k,2k+3) and C(5k-1,5k) are collinear.
Then, area of $Δ$ ABC = 0.
$\Rightarrow \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = 0$
Multiplying above expression by 2, we get
$[x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = 0$
Here, $x_{1} = k + 1, x_{2} = 3 k, x_{3} = 5 k - 1$
$a n d y_{1} = 2 k, y_{2} = 2 k + 3, y_{3} = 5 k \Rightarrow [(k + 1) (2 k + 3 - 5 k) + 3 k (5 k - 2 k) + (5 k - 1) (2 k - (2 k + 3))] = 0 \Rightarrow [(k + 1) (3 - 3 k) + 3 k (3 k) + (5 k - 1) (2 k - 2 k - 3)] = 0 \Rightarrow [- 3 k^{2} + 3 k - 3 k + 3 + 9 k^{2} - 15 k + 3] = 0 \Rightarrow (6 k^{2} - 15 k + 6) = 0$
$\Rightarrow 6 k^{2} - 15 k + 6 = 0$
Dividing the equation by 3, we get
$\Rightarrow 2 k^{2} - 5 k + 2 = 0 \Rightarrow 2 k^{2} - 4 k - k + 2 = 0$ [by factorization method]
$\Rightarrow 2 k (k - 2) - 1 (k - 2) = 0 \Rightarrow (k - 2) (2 k - 1) = 0$
If k – 2 = 0, then k = 2
If 2k – 1, then k = $\frac{1}{2}$
$∴ k = 2, \frac{1}{2}$
Hence, the required values of k are 2 and $\frac{1}{2}$ .

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