Given are the points $P=[7; 2]$ and $Q=[-1; 3]$. Write the parametric expression of the half-line opposite to the half-line $QP$.
Linda solved the problem in the following steps:
(1) She wrote the general parametric equation of the half-line opposite to the half-line $QP$: $$X = P + t \cdot\overrightarrow{u},\mbox{ where } \overrightarrow{u} = \overrightarrow{PQ},\quad t\in [ 0;\infty).$$
(2) She calculated the coordinates of the direction vector $\overrightarrow{u}$: $\ \overrightarrow{u}=\overrightarrow{PQ}= Q\ –\ P = (-8; 1)$.
(3) She wrote the parametric expression of the half-line opposite to the half-line $QP$ in coordinates: $$\left. \begin{aligned} x&=7-8t\cr y&=2+t \end{aligned}\right\} \quad t\in[0; \infty)$$
Is Linda's solution correct? If not, identify where the mistake is.
Linda's solution is correct.
The mistake is in step (1): Linda incorrectly determined the admissible values of the parameter $t$.
The mistake is in step (2): Linda incorrectly calculated the coordinates of the vector $\overrightarrow{u}$.
The mistake is in step (3): Linda incorrectly wrote the parametric expressions of the half-line opposite to the half-line $QP$ in coordinates.
The half-line opposite to the half-line $QP$ (marked in red in the figure) is not the half-line $PQ$, but the half-line, which "starts" not at the point $P$ but at the point $Q$, all admissible values of the parameter $t$ are from the interval $[1;\infty)$.
(1) The parametric equation of the half-line opposite to the ray $QP$ is : $$X = P + t \cdot\overrightarrow{u},\mbox{ where } \overrightarrow{u} =\overrightarrow{PQ},\quad t\in [ 1;\infty).$$
(2) Direction vector $\overrightarrow{u}$ : $\ \overrightarrow{u}=\overrightarrow{PQ}= Q\ –\ P = (-8; 1)$.
(3) The parametric expression of the half-line opposite to the half-line $QP$ in coordinates: $$\left. \begin{aligned} x&=7-8t\cr y&=2+t \end{aligned}\right\} \quad t\in[1; \infty)$$
