Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 10 - Review Exercises - Page 763: 113

Answer

Collinear.

Work Step by Step

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.
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