Given are the line segment $AB$:
$$\left.\begin{aligned} x&= 7-3t\cr y&=-5+9t \end{aligned} \right\rbrace\ \mbox{for } t\in[0;1]$$ and the half-line $KL$: $$\left.\begin{aligned} x&=1-2r\cr y&=3+r\cr \end{aligned}\right\rbrace\ \mbox{for } r\in[0;\infty).$$
Determine their intersection.
Charles solved the problem in the following steps:
(1) He assumed that the intersection is a point $P=[p_1; p_2]$. $$\begin{aligned}&P \in AB:\quad\begin{aligned} p_1&=7-3t\cr p_2&=-5+9t\end{aligned}\cr\cr &P\in\mapsto KL: \begin{aligned} p_1&=1-2r\cr p_2&=3+r \end{aligned}\end{aligned}$$
(2) Using a comparison method, he set up a system of equations: $$\begin{aligned} 7-3t&= 1-2r\cr -5+9t&= 3+r\end{aligned}$$
(3) He solved this system using the substitution method. As a first step, he expressed the variable $r$ from the second equation: $r = 9t-8$
(4) After substituting the variable $r$ into the first equation, he solved the equation with the variable $t$:
$$\begin{alignat}2 7-3t&=1-2(9t-8)&&\cr 7-3t&=1-18t+16\quad&&\big/+ 18t-7\cr 15t &= 10\quad&&\big/ : 15\cr t &=\frac{10}{15}=\frac23 \end{alignat}$$
(5) He checked if $t$ satisfies the condition and found the intersection. $$t\in [0;1] \Rightarrow P=\left[7-3\cdot\frac23; -5+9\cdot\frac23\right]\Rightarrow P=[5; 1] $$
Is Charles's solution correct? If not, determine where Charles made a mistake.
Charles's solution is correct.
The mistake is in step (3). Charles incorrectly expressed the variable $r$.
The mistake is in step (4). Charles incorrectly solved the equation with the variable $t$, and therefore, he incorrectly determined the intersection point $P$.
The mistake is in step (5). Charles did not calculate the variable $r$ and did not check the conditions it should satisfy.
After calculating the variable $r$ ($r = 9\cdot\frac23-8 = -2$), we find that $r\notin [0;\infty)$, therefore the line segment $AB$ and the half-line $KL$ have no intersection.