This is the first year linear algebra question. Which of the following points is on the line containing the points $(1,0,5)$ and $(3,1,-2)$?
a). $( 0,-3,17)$
b). $(9,4,-23)$
c). $(-9,-4,23)$
d). $(0,3,-17)$
e). $(-1,-1,-12)$
The answer is b).
This is the first year linear algebra question. Which of the following points is on the line containing the points $(1,0,5)$ and $(3,1,-2)$?
a). $( 0,-3,17)$
b). $(9,4,-23)$
c). $(-9,-4,23)$
d). $(0,3,-17)$
e). $(-1,-1,-12)$
The answer is b).
The line $L$ through the points $(1,0,5)$ and $(3,1,-2)$ is parallel to the line $L'$ through the origin and
$(3,1,-2)-(1,0,5)=(2,1,-7)\;.$
Because it does pass through the origin, $L'$ is just the set of scalar multiples of $(2,0,-7)$, i.e.,
$L'=\big\{t(2,1,-7):t\in\Bbb R\big\}\;.$
$L$ is parallel to this and passes through $(1,0,5)$, so it’s just $L'$ shifted by $(1,0,5)$:
$L=\big\{(1,0,5)+t(2,1,-7):t\in\Bbb R\big\}\;.$
Or you can simplify $(1,0,5)+t(2,1,-7)$ to get
$L=\big\{(1+2t,t,5-7t):t\in\Bbb R\big\}\;.$
Now just check to see which of your points actually fit this description. Take the point $(0,3,-17)$, for instance: if it’s on $L$, the second coordinate tells you that $t$ must be $3$, but then the first coordinate ought to be $1+2\cdot3=7$. It isn’t, so this point isn’t on $L$.