# [SOLVED]311.1.5.19 parametric equation of the line through a parallel to b.

#### karush

##### Well-known member
$\tiny{311.1.5.19}$
find the parametric equation of the line through a parallel to b.
$a=\left[\begin{array}{rr} -2\\0 \end{array}\right], \, b=\left[\begin{array}{rr} -5\\3 \end{array}\right]$

ok I know this like a line from 0,0 to -5,3 and $m=dfrac{-5}{3}$
so we could get line eq with point slope formula

but this is be done by parametric eq

this was book section I tried to follow..

anyway.....

$x=a+tb= \left[\begin{array}{rr} -2\\0 \end{array}\right]+t \left[\begin{array}{rr} -5\\3 \end{array}\right]$

maybe

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#### Country Boy

##### Well-known member
MHB Math Helper
Didn't you check this yourself?

When t= 0, $$\displaystyle \begin{bmatrix}-2 \\0 \end{bmatrix}+ 0\begin{bmatrix}-5 \\ 3\end{bmatrix}=\begin{bmatrix}-2 \\0 \end{bmatrix}$$ so it goes through the right point, a= (-2, 0).

When t= 1, $$\displaystyle \begin{bmatrix}-2 \\0 \end{bmatrix}+ 1\begin{bmatrix}-5 \\ 3\end{bmatrix}=\begin{bmatrix}-7 \\3 \end{bmatrix}$$ so it also goes through (-7, 3).

And the vector from (-2, 0) to (-7, 3) is $$\displaystyle \begin{bmatrix}-7- (-2) \\ 3- 0\end{bmatrix}=\begin{bmatrix}-5 \\ 3 \end{bmatrix}$$ as desired.

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