## Algebra and Trigonometry 10th Edition

The general form of a matrix of order $3 \times 3$ is: $\begin{bmatrix} a & b & c \\ d & e & f \\ g & h& i \end{bmatrix}=a(ei-fh) -b(di-fg)+c(dh-eg)$ Use the formula for the area of a triangle with the determinant. $D=det \begin{bmatrix} a & x & 1 \\ b & y & 1 \\ c & z & 1 \end{bmatrix}$ $Area=|\dfrac{1}{2} D|$ Now, $D=det \begin{bmatrix} -1 & 7 & 1 \\ 3 & -9 & 1 \\ -3 & 15 & 1 \end{bmatrix} =24-42+18=0$ Because the determinant is zero, the three points are Collinear.