Let a and b be two-dimensional vectors. Notice that a and b can also be regarded as points in a 2D coordinate plane.
The parametric equation of a line segment that goes from a to b using a parameter t that ranges from 0 to 1 is:
[tex]r(t)=a+t(b-a)[/tex]And each component would be given by:
[tex]\begin{gathered} x(t)=x_a+t(x_b-x_a)\quad,\quad0\leq t\leq1 \\ y\left(t\right)=y_a+t\lparen y_b-y_a)\quad,\quad0\leq t\leq1 \end{gathered}[/tex]We can use an alternative parameter that moves from 0 to k replacing t for t/k:
[tex]\begin{gathered} x\left(t\right)=x_a+\frac{t}{k}\left(x_b-x_a\right)\quad,\quad0\leq t\leq k \\ \\ y\left(t\right)=y_a+\frac{t}{k}\left(y_b-y_a\right)\quad,\quad0\leq t\leq k \end{gathered}[/tex]Use these expressions to find parametric equations for a line segment that connects the given points.
3) Init: (-5,8), term: (3,7)
For 0≤t≤4:
[tex]\begin{gathered} x(t)=-5+\frac{t}{4}(3--5)=-5+\frac{t}{4}(3+5)=-5+\frac{t}{4}(8)=2t-5 \\ y(t)=8+\frac{t}{4}(7-8)=8+\frac{t}{4}(-1)=-\frac{t}{4}+8 \\ \\ \therefore \\ x\left(t\right)=2t-5 \\ y(t)=-\frac{t}{4}+8 \end{gathered}[/tex]For 0≤t≤12:
[tex]\begin{gathered} x(t)=-5+\frac{t}{12}(3--5)=-5+\frac{t}{12}(3+5)=-5+\frac{t}{12}(8)=\frac{2}{3}t-5 \\ y(t)=8+\frac{t}{12}(7-8)=8+\frac{t}{12}(-1)=-\frac{t}{12}+8 \\ \\ \therefore \\ x\left(t\right)=\frac{2}{3}t-5 \\ y(t)=-\frac{t}{12}+8 \end{gathered}[/tex]4) Init: (3,7), term: (-5,8)
For 0≤t≤4:
[tex]\begin{gathered} x(t)=3+\frac{t}{4}(-5-3)=3+\frac{t}{4}(-8)=-2t+3 \\ y(t)=7+\frac{t}{4}(8-7)=7+\frac{t}{4}(1)=\frac{t}{4}+7 \\ \\ \therefore \\ x\left(t\right)=-2t+3 \\ y(t)=\frac{t}{4}+7 \end{gathered}[/tex]For 0≤t≤12:
[tex]\begin{gathered} x(t)=3+\frac{t}{12}(-5-3)=3+\frac{t}{12}(-8)=-\frac{2}{3}t+3 \\ y(t)=7+\frac{t}{12}(8-7)=7+\frac{t}{12}(1)=\frac{t}{12}+7 \\ \\ \therefore \\ x\left(t\right)=-\frac{2}{3}t+3 \\ y(t)=\frac{t}{12}+7 \end{gathered}[/tex]
