Parameterizing a line that passes through points

Question

I've been having a lot of trouble. I know you have to set up an equation where f(15)=P and f(17)=Q, but I'm not sure how to do it.

Thank you.

Answer 1

A line in general has the equation $$r(t)=a+t b,$$ where $a=(a_1,a_2)$ is the initial point and $b=(b_1,b_2)$ is the direction. You want that $r(15)=(4,2)$ and $r(17)=(11,9)$, i.e., that $$\begin{cases}(a_1+15b_1,a_2+15b_2)=(4,2)\\(a_1+17b_1,a_2+17b_2)=(11,9).\end{cases}$$

Comparing the first entries gives two simultaneous equations for $a_1$ and $b_1$, and comparing the second entries gives the same for $a_2$ and $b_2$. Solving these simultaneous equations gives that $a=-\tfrac12(97,101)$ and $b=\tfrac72(1,1)$, so the line is $$\boxed{r(t)=\tfrac12(7t-97,7t-101)}$$

Answer 2

A line is parametrized in the following way:

$$ \begin{bmatrix} x \\ y \end{bmatrix}= \begin{bmatrix} x_0 \\ y_0 \end{bmatrix}+t \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} $$

Now we know that $$ \begin{bmatrix} 4 \\ 2 \end{bmatrix}= \begin{bmatrix} x_0 \\ y_0 \end{bmatrix}+15 \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} $$ and $$ \begin{bmatrix} 11 \\ 9 \end{bmatrix}= \begin{bmatrix} x_0 \\ y_0 \end{bmatrix}+17 \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} $$ hence $$ \begin{bmatrix} 11 \\ 9 \end{bmatrix}- \begin{bmatrix} 4 \\ 2 \end{bmatrix}= 2 \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} $$ From which we get $$\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}= \begin{bmatrix} 3.5 \\ 3.5 \end{bmatrix}$$ So now we can write our line in the form $$ \begin{bmatrix} x \\ y \end{bmatrix}= \begin{bmatrix} x_0 \\ y_0 \end{bmatrix}+t \begin{bmatrix} 3.5 \\ 3.5 \end{bmatrix} $$ Substituting either $P$ or $Q$ we get $$ \begin{bmatrix} 4 \\ 2 \end{bmatrix}= \begin{bmatrix} x_0 \\ y_0 \end{bmatrix}+15 \begin{bmatrix} 3.5 \\ 3.5 \end{bmatrix} $$ and we conclude that $$ \begin{bmatrix} x_0 \\ y_0 \end{bmatrix}= \begin{bmatrix} -48.5 \\ -50.5 \end{bmatrix} $$ and the parametrization of our line is $$ \begin{bmatrix} x \\ y \end{bmatrix}= \begin{bmatrix} -48.5 \\ -50.5 \end{bmatrix}+t \begin{bmatrix} 3.5 \\ 3.5 \end{bmatrix} $$