## Thomas' Calculus 13th Edition

$-x-y=1$
The result of exercise 32 tells us that ${\bf v}=a{\bf i} + b{\bf j}$ is parallel to the lines $bx-ay=c$ Here, ${\bf v}=1\cdot{\bf i} + (-1)\cdot{\bf j}$, $a=1,\quad b=-1,$ and the lines are $-x-y=c.$ $P(-2,1)$ is on the line $\Rightarrow \left\{\begin{array}{l} -(-2)-(1)=c\\ c=1 \end{array}\right.$ so the line equation is $\quad -x-y=1$ (a second point on the line is $(-1,0)$ )