Originally Posted by
HallsofIvy You were given the line as
$\displaystyle \frac{x+1}{7} = \frac{y-2}{4} = \frac{z}{-1}$
That's NOT one equation, it is two : $\displaystyle \frac{x+1}{7} = \frac{y-2}{4}$ and $\displaystyle \frac{y-2}{4} = \frac{z}{-1}$
(strictly speaking, there is a third equaton, $\displaystyle \frac{x+1}{7} = \frac{z}{-1}$ but that's obviously not independent of the first two.)
Take x= 6 (so that $\displaystyle \frac{x+1}{7}= \frac{7}{7}= 1$ is easy) and then you have $\displaystyle \frac{y- 2}{4}= 1$ so that y= 6 and $\displaystyle \frac{z}{-1}= 1$ so that z= -1. One point on the line is (6, 6, -1).
Now take x= -8 so that $\displaystyle \frac{x+ 1}{7}= \frac{-7}{7}= -1$. Then $\displaystyle \frac{y- 2}{4}= -1$ so that y= -2 and $\displaystyle \frac{z}{-1}= -1$ so that z= 1. Another point on the line is (-8, -2, 1).