question_answer

If the points \[A(k+1,2k),B(3k,2k+3)\] and \[C(5k-1,5k)\] are collinear, then find the value of k.

Answer:

Since \[A(k+1,2k),B(3k,2k+3)\] and \[C(5k-1,5k)\] are collinear points, so area of triangle = 0.

\[\Delta =\frac{1}{2}[(k+1)(2k+3)-6{{k}^{2}}+15{{k}^{2}}-(5k-1)(2k+3)+2k(5k-1)-(k+1)(5k)]\]

\[0=\frac{1}{2}[2{{k}^{2}}+5k+3-6{{k}^{2}}+15{{k}^{2}}-10{{k}^{2}}-13k+3+10{{k}^{2}}-2k-5{{k}^{2}}-5k]\]

\[0=\frac{1}{2}[6{{k}^{2}}-15k+6]\]

\[\Rightarrow \]   \[6{{k}^{2}}-15k+6=0\]

\[\Rightarrow \] \[6{{k}^{2}}-12k-3k+6=0\]

\[\Rightarrow \] \[6k(k-2)-3(k-2)=0\]

\[\Rightarrow \]   \[(k-2)(6k-3)=0\]

\[k=2\] or \[k=\frac{1}{2}\]