If t_1, t_2 and t_3 are distinct,then the poi | vedclass

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(C) For the given points to be collinear,the area of the triangle formed by them must be zero,or the determinant of the coordinates must be zero:$\beg

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$t_1 + t_2 + t_3 = 0$

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(C) For the given points to be collinear,the area of the triangle formed by them must be zero,or the determinant of the coordinates must be zero:$\begin{vmatrix} t_1 & 2at_1 + at_1^3 & 1 \ t_2 & 2at_2 + at_2^3 & 1 \ t_3 & 2at_3 + at_3^3 & 1 \end{vmatrix} = 0$Taking $a$ common from the second column:$a \begin{vmatrix} t_1 & 2t_1 + t_1^3 & 1 \ t_2 & 2t_2 + t_2^3 & 1 \ t_3 & 2t_3 + t_3^3 & 1 \end{vmatrix} = 0$Applying row operations $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$:$\begin{vmatrix} t_1 & 2t_1 + t_1^3 & 1 \ t_2 - t_1 & 2(t_2 - t_1) + (t_2^3 - t_1^3) & 0 \ t_3 - t_1 & 2(t_3 - t_1) + (t_3^3 - t_1^3) & 0 \end{vmatrix} = 0$Expanding along the third column:$(t_2 - t_1)(t_3 - t_1) \begin{vmatrix} 1 & 2 + (t_2^2 + t_1^2 + t_2 t_1) \ 1 & 2 + (t_3^2 + t_1^2 + t_3 t_1) \end{vmatrix} = 0$$(t_2 - t_1)(t_3 - t_1) [(2 + t_3^2 + t_1^2 + t_3 t_1) - (2 + t_2^2 + t_1^2 + t_2 t_1)] = 0$$(t_2 - t_1)(t_3 - t_1) [t_3^2 - t_2^2 + t_1(t_3 - t_2)] = 0$$(t_2 - t_1)(t_3 - t_1)(t_3 - t_2)(t_3 + t_2 + t_1) = 0$Since $t_1, t_2, t_3$ are distinct,$(t_2 - t_1) 
eq 0$,$(t_3 - t_1) 
eq 0$,and $(t_3 - t_2) 
eq 0$.Therefore,$t_1 + t_2 + t_3 = 0$.

If $t_1, t_2$ and $t_3$ are distinct,then the points $(t_1, 2at_1 + at_1^3)$,$(t_2, 2at_2 + at_2^3)$ and $(t_3, 2at_3 + at_3^3)$ are collinear if:

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