Question

Find the value(s) of k for which the points (3k − 1, k − 2), (k, k − 7) and (k − 1, −k − 2) are collinear.     

Answer 1

Let A(3k − 1, k − 2), B(k, k − 7) and C(k − 1, −k − 2) be the given points.
The given points are collinear. Then,

\[\text{ ar } \left( ∆ ABC \right) = 0\]
\[ \Rightarrow \frac{1}{2}\left| x_1 \left( y_2 - y_3 \right) + x_2 \left( y_3 - y_1 \right) + x_3 \left( y_1 - y_2 \right) \right| = 0\]
\[ \Rightarrow x_1 \left( y_2 - y_3 \right) + x_2 \left( y_3 - y_1 \right) + x_3 \left( y_1 - y_2 \right) = 0\]
\[\Rightarrow \left( 3k - 1 \right)\left[ \left( k - 7 \right) - \left( - k - 2 \right) \right] + k\left[ \left( - k - 2 \right) - \left( k - 2 \right) \right] + \left( k - 1 \right)\left[ \left( k - 2 \right) - \left( k - 7 \right) \right] = 0\]
\[ \Rightarrow \left( 3k - 1 \right)\left( 2k - 5 \right) + k\left( - 2k \right) + 5\left( k - 1 \right) = 0\]
\[ \Rightarrow 6 k^2 - 17k + 5 - 2 k^2 + 5k - 5 = 0\]
\[ \Rightarrow 4 k^2 - 12k = 0\]

\[\Rightarrow 4k\left( k - 3 \right) = 0\]
\[ \Rightarrow k = 0 \text{ or } k - 3 = 0\]
\[ \Rightarrow k = 0 \text{ or  } k = 3\]

Hence, the value of k is 0 or 3.