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Write a vector equation of the line that passes through point P and is parallel to vector a

Point $P=(-2,1)$
Vector $a=(3,-4)$

Can someone just explain this to me? I know how to write vector equations but not with only one point and when I have to make it parallel to something else!

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  • $\begingroup$ What kind of vector equations do you know how to write? $\endgroup$ – pjs36 Mar 4 '15 at 0:04
  • $\begingroup$ I know how to write 4 different kinds but my preferred form is <x,y> = <r,s> + t<a,b> $\endgroup$ – user220798 Mar 4 '15 at 0:27
  • $\begingroup$ Would you know how to write the equation for the line that's parallel to (pointing in the same direction as) $(3, -4)$? If so, you could just translate it by the vector $(-2, 1)$. $\endgroup$ – pjs36 Mar 4 '15 at 0:32
  • $\begingroup$ It sounds like you know how find the vector equation of the line when given some other information. What information? $\endgroup$ – user147263 Mar 4 '15 at 0:43
  • $\begingroup$ To do that equation I need two points. $\endgroup$ – user220798 Mar 4 '15 at 1:10
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so the general solution here would be r(t) = (a,b) + t, or x = a + ct, y = b + dt. In your case you have the equation r(t) = (-2,1) + t<3,-4>. Because (-2,1) is the point you want to go through, that'll be your point, and t is multiplied by the vector <3,-4> because that's the vector you want yours to be parallel to. The equation I gave you for r(t) above should be acceptable, but it can also be x = -2 + 3t, y = 1 - 4t. Hope this helps!

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Choose between the following formulations:

  • $\boldsymbol{r} = \boldsymbol{P} + t \boldsymbol{a}$

  • $ \frac{x - \boldsymbol{P}_x}{\boldsymbol{a}_x}=\frac{y - \boldsymbol{P}_y}{\boldsymbol{a}_y} = const. $

  • $ (-\boldsymbol{a}_y) x + ( \boldsymbol{a}_x) y + (\boldsymbol{a}_y \boldsymbol{P}_x - \boldsymbol{a}_x \boldsymbol{P}_y) = 0$

All of them are equations of a line on a plane.

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Pick an arbitrary point $Q=(x,y)$ in the plane (different from $P$). This point is in the line you want if the vector $Q-P = (x+2, y-1)$ is parallel to the vector $a$. This gives us $(x+2, y-1) = t (3,-4)$ for some $t\in \mathbb{R}$.

Drawing this picture may clarify the relation between the geometric and analytic definitions.

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