Let the points A (2, 3), B{tex}\\left( {4,k} \\right){/tex} and C{tex}\\left( {6, – 3} \\right){/tex} be collinear.If the points are collinear then area of triangle ABC formed by these three points is 0.{tex}\\therefore {/tex}{tex}{\\text{ar}}\\left( {\\Delta {\\text{ABC}}} \\right) = \\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]{/tex}= 0{tex} \\Rightarrow {/tex}{tex}\\frac{1}{2}\\left[ {2\\left( {k + 3} \\right) + 4\\left( { – 3 – 3} \\right) + 6\\left( {3 – k} \\right)} \\right] = 0{/tex}{tex} \\Rightarrow {/tex}{tex}\\left[ {2k + 6 – 24 + 18 – 6k} \\right] = 0{/tex}{tex} \\Rightarrow {/tex}{tex}\\left[ { – 4k} \\right] = 0{/tex}{tex} \\Rightarrow {/tex}{tex}k = 0{/tex}