Answer
(a) See explanations.
(b) $y=\frac{5}{6}x+\frac{100}{3}$
Work Step by Step
(a) Based on the results from Exercise 69(a), we know that if three points are on a line, the determinant formula for the area calculation would be zero.
Let $P(x, y)$ be any point on a line passing $(x_1, y_1)$ and $(x_2, y_2)$, since these three points are collinear, the "triangle" formed by these three points would be zero, which means the following determinant is zero:
$\begin{vmatrix} x&y&1\\ x_1&y_1&1\\ x_2&y_2&1 \end{vmatrix}
\begin{array}( \\=0 \\ \\ \end{array}$
Do the operations $R_1-R_2\to R_2$ and $R_3-R_2\to R_3$, we have:
$\begin{vmatrix} x&y&1\\ x-x_1&y-y_1&0\\ x_2-x_1&y_2-y_1&0 \end{vmatrix}
\begin{array}( \\=0 \\ \\ \end{array}$
Expand with column3, we have:
$( x-x_1)(y_2-y_1)-(x_2-x_1)(y-y_1)=0$ or
$\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$
which is the equation of a line passing points $(x_1, y_1)$ and $(x_2, y_2)$
(b) Given two points $(20, 50)$ and $(-10,25)$, use the above results, we have:
$\begin{vmatrix} x&y&1\\20&50&1\\-10&25&1 \end{vmatrix}
\begin{array}( \\=0 \\ \\ \end{array}$
or $\frac{y-50}{25-50}=\frac{x-20}{-10-20}$ which gives:
$y=\frac{5}{6}x+\frac{100}{3}$
