question_answer

Find the values of k for which the points \[A(k+1,2k),B(3k,2k+3)\] and  \[C(5k-1,5k)\] are collinear.

Answer:

Given, the points \[A(k+1,2k),B(3k,2k+3)\]and \[C(5k-1,5k)\]

\[\because \] The point to be collinear

\[\therefore {{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}})=0\]

\[(k+1)(2k+3-5k)+3k(5k-2k)+(5k-1)(2k-2k-3)=0\]

\[(k+1)(3-3k)+3k(3k)+(5k-1)(-3)=0\]

\[3k+3-3{{k}^{2}}-3k+9{{k}^{2}}-15k+3=0\]

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

\[2{{k}^{2}}-5k+2=0\]

\[2{{k}^{2}}-4k-k+2=0\]

\[2k(k-2)-1(k-2)=0\]

\[(2k-1)(k-2)=0\]

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

Hence, \[k=2\] or \[k=\frac{1}{2}\].