jmc

Setting $t=0,$ we get $$\left(\begin{array}{c}x\\ y\end{array}\right)=\mathbf{v}.$$But the distance between and is $t=0,$ so $\mathbf{v}=$ Thus, $$\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}3\\ 1\end{array}\right)+t\mathbf{d}.$$Then for $x\ge 3,$ $$\Vert \left(\begin{array}{c}x-3\\ y-1\end{array}\right)\Vert =\Vert \left(\begin{array}{c}x-3\\ \frac{3x-9}{4}\end{array}\right)\Vert =\Vert \left(\begin{array}{c}1\\ \frac{3}{4}\end{array}\right)\Vert (x-3)=\frac{5}{4}(x-3).$$We want this to be $t,$ so $t=\frac{5}{4}(x-3).$ Then $x=\frac{4}{5}t+3,$ and $y=\frac{3x-5}{4}=\frac{3}{5}t+1,$ so $$\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}\frac{4}{5}t+3\\ \frac{3}{5}t+1\end{array}\right)=\left(\begin{array}{c}3\\ 1\end{array}\right)+t\left(\begin{array}{c}4/5\\ 3/5\end{array}\right).$$Thus, $\left(\begin{array}{c}4/5\\ 3/5\end{array}\right)$}." style="color:#cc0000">\mathbf{d} = \boxed{%%DISP_3%%}.