# Parameterize a straight line using polar coordinates… without angle.

I had to parameterize a straight line with starting point in $A=(-3,7)\\$ and endpoint in $B=(4,1)$.

My idea was to use the equation for the line that goes through two points. That is:

$$\frac { y\quad -\quad { y }^{ 1 } }{ x\quad -\quad { x }^{ 1 } } =\frac { { y }^{ 2 }\quad -\quad { y }^{ 1 } }{ { x }^{ 2 }\quad -\quad { x }^{ 1 } }$$

$$y\quad =\quad -\left( \frac { 8x\quad -\quad 25 }{ 7 } \right)$$

Which after I plotted, ended up with a straight line passing by the given points of the exercice. I was happy, until my teacher advised me of a "very easy way" to do this, which I should "already know", and it's to find out the position of the line using polar coordinates.

I have been reading about the subject, but as far as I understand, I need two things: the length from $A$ to $B$ and then the angle of said line against the X axis, am I right? I find this a bit confusing. I would like to know if my equation is really a bad way to do this, and in case that polar coordinates are the best way to parameterize this exercise, I don't understand which operation should I do to find out the angle, assuming I already know the starting and endpoint of AB.

Thank you for any help! I have been one whole day thinking about this and I am sure I am missing something very obvious.

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Your line passes through $A = (-3,7)$ and $B = (4,1).$ The gradient of the line is:

$$\frac{B_y-A_y}{B_x-A_x} = -\frac{6}{7} \, .$$

The cartesian equation of the line is thus:

$$\frac{y-B_y}{x-B_x} = -\frac{6}{7}$$

which simplifies to give $6x+7y=31.$ Next, we substitute $x=r\cos\theta$ and $y=r\sin\theta$ to give:

$$6r\cos\theta + 7r\sin\theta = 31 \iff r= \frac{31}{7\sin\theta+6\cos\theta} \, .$$

This expression is valid provided $7\sin\theta+6\cos\theta \neq 0.$ In this case, the ray joining $(0,0)$ to $(r,\theta)$ is parallel to the line. The parametrisation is thus $\theta \mapsto (r(\theta),\theta)$, where $r$ is as above and

$$-\arctan\left(\frac{6}{7}\right) < \theta < \pi - \arctan\left(\frac{6}{7}\right) .$$

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Thanks, your example made it look much easier than in my textbook. What a fascinating subject! – telex-wap Oct 16 '12 at 18:34