Parametric equation of line

Question

Write an equation for the line through A=$\left( \begin{matrix} 3\\ 1\end{matrix} \right)$ and B=$\left( \begin{matrix} 1\\ 2\end{matrix} \right)$.

Hint: This line will go through B and be parallel to the vector B $-$ A.

My answer: X $=$ B $+$ t A = $\left( \begin{matrix} 1\\ 2\end{matrix} \right) +t\left( \begin{matrix} 3\\ 1\end{matrix} \right) =\left( \begin{matrix} 1+3t\\ 2+t\end{matrix} \right)$.

But, the hint says ''This line will go through B and be parallel to the vector B $-$ A''. Can you help me?

Answer 1

we get $$OB-OA=[-2,1]$$ and our equation is $$x=[1,2]+t[-2,1]$$ and $$t\in \mathbb{R}$$

Answer 2

If you write ${\bf X} = {\bf B}+t{\bf A}$, your line will be parallel to the vector ${\bf A}$, and not to ${\bf B}-{\bf A}$. The correct line is $${\bf X} = {\bf B} + t({\bf B}-{\bf A}).$$The vector starting at ${\bf A}$ and with endpoint at ${\bf B}$ is ${\bf B}-{\bf A}$. Now picture all the multiples of that last vector starting at ${\bf B}$: it gives exactly the points ${\bf X} = {\bf B}+t({\bf B}-{\bf A})$.

Answer 3

Hint. Consider ${\bf X} = {\bf B} + t({\bf A}-{\bf B})$.

What happens when $t=0$ and $t=1$? Compare with your answer.

Answer 4

$\mathbf x = \mathbf B + (\mathbf A - \mathbf B) t$

this way when $t = 0, \mathbf x = \mathbf B$ and when $t = 1, \mathbf x = \mathbf A$

$\mathbf x = \begin{pmatrix}1\\2\end{pmatrix} + \begin{pmatrix}2\\-1\end{pmatrix} t$

Answer 5

It should be intuitive that the middle of $AB$ lies on the line, and that the middle corresponds to the arithmetic averages of the coordinates. In vector terms,

$$M=\frac12A+\frac12B.$$

Now for the same reason, the middle of $AM$ is

$$N=\frac12A+\frac12M=\frac14A+\frac34B.$$

Continuing this reasoning, you conclude that any point on the line is of the form

$$P=tA+uB$$ with $t+u=1$, or

$$P=tA+(1-t)B=B+t(A-B).$$