The points (k+1,1) (2k+1,3) and (2k+2,2k) …

{tex}{/tex}We know that if three points A, B, C are collinear, then area of {tex}\Delta ABC =0{/tex}
Given that points A(k + 1, 1), B(2k + 1, 3) and C(2k + 2, 2k) are collinear.
Here, x₁= k+1 , x₂ =2k+1, x₃= 2k+2, y₁= 1, y₂= 3, y₃= 2k

{tex}\Rightarrow {/tex} Area of {tex}\Delta ABC =0{/tex}

{tex}\Rightarrow {/tex} {tex}\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|=0{/tex}
{tex}\Rightarrow {/tex} {tex}\frac{1}{2}|(k+1)(3-2k)+(2k+1)(2k-1)+(2k+2)(1-3)|=0{/tex}
{tex}\Rightarrow{/tex} {tex}|k(3-2k)+1(3-2k)+(2k)^2-(1)^2+(2k+2)(-2)|=0{/tex}

{tex}\Rightarrow{/tex} |3k - 2k² + 3 - 2k + 4k² - 1 - 4k - 4 |= 0
{tex}\Rightarrow{/tex} |2k² - 3k - 2 |= 0
{tex}\Rightarrow{/tex} 2k² - 4k + k - 2 = 0
{tex}\Rightarrow{/tex} 2k(k - 2) + 1(k - 2) = 0
{tex}\Rightarrow{/tex} (2k + 1) (k - 2) = 0
{tex}\Rightarrow{/tex} k = - {tex}\frac { 1 } { 2 }{/tex}, k = 2