Question

If the points (k,2k),(3k,3k) and (3,1) are collinear,then the value of k is ______

Answer 1

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Answer 2

Answer:

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

Step-by-step explanation:

We know that the area of triangle made by three collinear point is zero.

The area of the triangle formed by these points must be zero. Hence, we have

$A=\frac{1}{2}\begin{vmatrix}x_1&y_1&1 \\x_2&y_2&1\\x_3&y_3&1 \end{vmatrix}=0\\\text{On substituting the known values, we get}\frac{1}{2}\begin{vmatrix}k&2k &1\\ 3k&3k&1 \\3&1&1\end{vmatrix}=0\\\\\left ( k(3k-1)-2k(3k-3)+1(3k-9k) \right )=0\\\\-3k^2-k=0\\\\-k(3k+1)=0\\\\k=-\frac{1}{3},0\\\\\text{K can't be zero because for k=0, we have only two points.}\\\text{Thus, the value of k is }k=-\frac{1}{3}$