The points (K,2-2K), (-K+1,2K) and (-4-K,6-2K) are collinear if $$ \text{(A) \( K=\frac{1}{2} \) } $$ $$ \text{ (B) \( K=\frac{-1}{2}\) } $$ $$ \text{ (C) \( K=\frac{3}{2}\) } $$ $$ \text{ (D) \( K=-1\) } $$ $$ \text{ (E) \( K=1 \) } $$
Three points are collinear when the area of the triangle they form is 0:
\(x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) = 0\)
With \((K, 2-2K)\), \((1-K, 2K)\), \((-4-K, 6-2K)\):
\(K(4K-6) + 4(1-K) + (-4-K)(2-4K) = 0\)
\(4K^2 - 6K + 4 - 4K - 8 + 14K + 4K^2 = 0 \Rightarrow 8K^2 + 4K - 4 = 0\)
\(2K^2 + K - 1 = 0 \Rightarrow (2K-1)(K+1) = 0\), so \(K = \frac{1}{2}\) or \(K = -1\).
These are statements (A) and (D).
Answer: (A) (A) and (D) only
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