Answer
(a) 0, see explanations.
(b) (i) Yes, (ii) No. see graph and explanations.
Work Step by Step
(a) If three points lie on a line, the area of the "triangle" that they determine will be zero. Based on the determinant formula for a triangle area, the determinant
$\begin{vmatrix} a_1&b_1&1\\a_2&b_2&1\\a_3&b_3&1 \end{vmatrix}\begin{array}( \\=2A=0 \\ \\ \end{array}$
Where $A$ is the area of the triangle, thus if the three points are on a line, the determinant should be zero.
(b) Use the determinant area formula to check if each set of points is collinear:
(i) Do the operations $R_2-R_1\to R_2$ and $R_3-R_1\to R_3$
$\begin{vmatrix} -6&4&1\\2&10&1\\6&13&1 \end{vmatrix}\begin{array}( \\= \\ \\ \end{array}
\begin{vmatrix} -6&4&1\\8&6&0\\12&9&0 \end{vmatrix}$
Use column3 to expand, we have $2A=(1)(8\times9-6\times12)=0$, thus the points are on a line.
(ii) Repeat the above process for the second point set:
$\begin{vmatrix} -5&10&1\\2&6&1\\15&-2&1 \end{vmatrix}\begin{array}( \\= \\ \\ \end{array}
\begin{vmatrix} -6&4&1\\7&-4&0\\20&-12&0 \end{vmatrix}$
Use column3 to expand, we have $|2A|=|(1)(7(12)-20(-4))|=4\ne0$, thus the points are not on a line.
See graph for the locations of the vertices.
