Determine which points are on a line containing (2,-1) and (-4,2)

Let $L$ be a line containing the points $P = (2,-1)$ and $Q = (-4,2)$ . Determine which of the following points are also on $L$ .

$(0,0)$ ;
$(0,1)$ ;
$(1,2)$ ;
$(2,1)$ ;
$(-2,1)$ ;

First, we have $P -Q = (6,-3)$ , so the line containing the points $P$ and $Q$ is the set

$L (P;P-Q) = \{ P + t(6,-3) \mid t \in \mathbb{R} \} = \{ (2+6t, -1-3t) \mid t \in \mathbb{R} \}.$

The point $(0,0)$ is on $L$ since
$\begin{align*} (0,0) &= (6t + 2, -1-3t) & \implies && 6t+2 &=0 & \implies && t = -\frac{1}{3} \\ && \implies && -1-3t &= 0. \end{align*}$

Thus, $(0,0) = (2+6t,-1-3t)$ for $t = -\frac{1}{3}$ , so $(0,0)$ is on $L$ .
The point $(0,1)$ is not on $L$ since
$\begin{align*} (0,1) &= (6t + 2, -1-3t) & \implies && 6t+2 &=0 & \implies && t = -\frac{1}{3} \\ && \implies && -1-3t &= 0 \neq 1. \end{align*}$

Thus, $(0,1) \neq (2+6t,-1-3t)$ for any $t \in \mathbb{R}$ , so $(0,1)$ is not on $L$ .
The point $(1,2)$ is not on $L$ since
$\begin{align*} (1,2) &= (6t + 2, -1-3t) & \implies && 6t+2 &=1 & \implies && t = -\frac{1}{6} \\ && \implies && -1-3t &= -\frac{1}{2} \neq 2. \end{align*}$

Thus, $(1,2) \neq (2+6t,-1-3t)$ for any $t \in \mathbb{R}$ , so $(1,2)$ is not on $L$ .
The point $(2,1)$ is not on $L$ since
$\begin{align*} (2,1) &= (6t + 2, -1-3t) & \implies && 6t+2 &=2 & \implies && t = 0 \\ && \implies && -1-3t &= -1 \neq 1. \end{align*}$

Thus, $(2,1) \neq (2+6t,-1-3t)$ for any $t \in \mathbb{R}$ , so $(2,1)$ is not on $L$ .
The point $(-2,1)$ is on $L$ since
$\begin{align*} (-2,1) &= (6t + 2, -1-3t) & \implies && 6t+2 &=-2 & \implies && t = -\frac{2}{3} \\ && \implies && -1-3t &= 1. \end{align*}$

Thus, $(-2,1) = (2+6t,-1-3t)$ for $t = -\frac{2}{3}$ , so $(-2,1)$ is on $L$ .