The points (k+1,1) (2k+1,3) and (2k+2,2k) are collinear. Then find k

1.	The points (k+1,1) (2k+1,3) and (2k+2,2k) are collinear. Then find k
Answer» {tex}{/tex}We know that if three points A, B, C are collinear, then area of\xa0{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\xa0=2k+1, x3= 2k+2, y1= 1, y2= 3, y3= 2k{tex}\\Rightarrow {/tex}\xa0Area of\xa0{tex}\\Delta ABC =0{/tex}{tex}\\Rightarrow {/tex}\xa0{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}\xa0{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 -\xa04k - 4 \|= 0{tex}\\Rightarrow{/tex}\xa0\|2k2 - 3k - 2 \|= 0{tex}\\Rightarrow{/tex}\xa02k2 - 4k + k - 2 = 0{tex}\\Rightarrow{/tex}\xa02k(k - 2) + 1(k - 2) = 0{tex}\\Rightarrow{/tex}\xa0(2k + 1) (k - 2) = 0{tex}\\Rightarrow{/tex}\xa0k = -\xa0{tex}\\frac { 1 } { 2 }{/tex}, k = 2

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