Find the value of\xa0k\xa0for which the points A (-1,3),B (2,k) and C (5,-1) are collinear.\xa0

Find the value of\xa0k\xa0for which the points A (-1,3),B (2,k) and C

1.	Find the value of\xa0k\xa0for which the points A (-1,3),B (2,k) and C (5,-1) are collinear.\xa0
Answer» According to question,Area of triangle ABC = 0Given: A (-1, 3), B (2, k) and C (5, -1)=> {tex}{1 \\over 2}\\left\| {\\left( { - 1} \\right)\\left( {k + 1} \\right) + 2\\left( { - 1 - 3} \\right) + 5\\left( {3 - k} \\right)} \\right\| = 0{/tex}=> {tex}\\left\| { - k - 1 - 8 + 15 - 5k} \\right\| = 0{/tex}=> {tex}\\left\| { - 6k + 6} \\right\| = 0{/tex}=> {tex}-6k=-6{/tex}=> {tex}k=1{/tex}