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### Section 12.2 : Equations of Lines

1. Give the equation of the line through the points $$\left( {2, - 4,1} \right)$$ and $$\left( {0,4, - 10} \right)$$ in vector form, parametric form and symmetric form.

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Okay, regardless of the form of the equation we know that we need a point on the line and a vector that is parallel to the line.

We have two points that are on the line. We can use either point and depending on your choice of points you may have different answers that we get here. We will use the first point listed above for our point for no other reason that it is the first one listed.

The parallel vector is really simple to get as well since we can always form the vector from the first point to the second point and this vector will be on the line and so will also be parallel to the line. The vector is,

$\vec v = \left\langle { - 2,8, - 11} \right\rangle$ Show Step 2

The vector form of the line is,

$\require{bbox} \bbox[2pt,border:1px solid black]{{\vec r\left( t \right) = \left\langle {2, - 4,1} \right\rangle + t\left\langle { - 2,8, - 11} \right\rangle = \left\langle {2 - 2t, - 4 + 8t,1 - 11t} \right\rangle }}$ Show Step 3

The parametric form of the line is,

$\require{bbox} \bbox[2pt,border:1px solid black]{{x = 2 - 2t\hspace{0.5in}y = - 4 + 8t\hspace{0.5in}z = 1 - 11t}}$ Show Step 4

To get the symmetric form all we need to do is solve each of the parametric equations for $$t$$ and then set them all equal to each other. Doing this gives,

$\require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{2 - x}}{2} = \frac{{4 + y}}{8} = \frac{{1 - z}}{{11}}}}$