Expert: Josh - 11/17/2007

Question
QUESTION: I am unsure about how to write equations for lines in polar coordinates and I am unsure about it basic concepts. If you could explain this it would be very helpful. thank you

ANSWER: The vertical (y) axis and horizontal (x) axis splits the x-y plane into four quadrants. Measuring angles from the origin in the anti-clockwise direction, every line passing through (0,0) subtends an angle relative to the positive direction (right hand side) of the x-axis. If we consider the situation depicted in figure 1, the line OA subtends an angle q bounded by BOA. You can mark the angle q in the diagram. In this diagram, B is supposed to be a point directly below point A on the x-axis.

y
|
|
|      A
|
|
O------B-------> x

We learned from trigonometry that sine(q) is the ratio given by length of AB (side opposing the angle), divided by the length of OA (the hypotenuse, or longest side in the OAB triangle). Similarly, cosine(q) is the ratio given by the length of OB (side adjacent to the angle q), divided by the length of OA (the hypotenuse). There are three entities to remember in all, the tangent is derived from sine and cosine. To summarize then, using |OA| to denote the length of OA...,

sin(q)=|AB|/|OA|  (remember opposite-over-hypotheuse)
cos(q)=|OB|/|OA|  (remember adjacent-over-hypotheuse)
tan(q)=sin(q)/cos(q)=|AB|/|OB|  (remember opposite-over-adjacent)

Now, if (x,y) denote the coordinate of point A on the diagram, and the length of the hypotenuse |OA| as described above is given by "R", then, the ratios become
sin(q)=|AB|/|OA|=y/R
cos(q)=|OB|/|OA|=x/R

It follows that the polar coordinates (x,y) for point A with angle q is given by x=R*cos(q) and y=R*sin(q).

Note: (x,y) are called the cartesian coordinates. The polar coordinates are expressed in terms of the hypotenuse (length of OA in this example, which is R) and the angle "q" (subtended by OA in this example).

Example, if length of OA is R=5 and the angle q=30, then (x,y)=(R*cos(30),R*sin(30))=(4.3301,2.5)

Careful: If point A is in the second, third or fourth quadrant, we need to watch out for the sign.

Some useful numbers to remember, as you'll see this over and over again,

sin(30)=0.5
sin(45)=1/square_root_of(2)=0.7071...
sin(60)=sqrt(3)/2=0.8660...

cos(30)=sin(90-30)=sqrt(3)/2=0.8660...
cos(45)=1/square_root_of(2)=0.7071...
sin(60)=sin(90-60)=0.5

It make sense too, that tan(45)=sin(45)/cos(45)=1.


---------- FOLLOW-UP ----------

QUESTION: what happens if know the equation of one line and are asked to find the equation
of another line say 30 degrees away and the two lines do share one point from which
they both extend which would create an angle of 30 degrees. Thank you

Answer
Hi Sean,

One way of looking at this is that we are rotating the line by an angle theta (q). The problem can be set up nicely using linear algebra. Unfortunately, I don't know if you know anything about matrix/vector or not. Not knowing what grade you are in, I will develop the basic ideas from scratch.

WHAT IS VECTOR REPRESENTATION?
In two dimensions, a vector is associated with a point P=(x,y) in the x-y plane. The vector is constructed by drawing an arrow from the origin (0,0) to point P. A vector has length (aka magnitude) d=sqrt(x^2+y^2) and points in the direction given by inverse tangent of (y/x). For example, the point (1,1) has length d=sqrt(2) and orientation of 45 degrees.

VECTOR GEOMETRY:
[Result 1] A line passing through M=(x1,y1) and N=(x2,y2), from M to N points in the direction N-M=(x2-x1,y2-y1).
[Result 2] In general, such a line is described by the vector equation (x,y) = M+k*[N-M], where k can take on any real number (positive or negative, including zero). In fact, the parameter "k" spans along the line. That is to say, we trot along the line as the value of k changes.

NOTATIONS: It is important that I clarify this. When I write (x,y), as in the left hand side of the above equation, I am referring to some arbitrary point in the x-y plane. Everything typed in capital letters OR enclosed in square brackets (such as M, [N-M]) represents a vector. Think of a vector as an arrow from a fixed point if you like. On the right hand side of the equation, M is a vector with coordinates (x1,y1); similarly [N-M] is also a vector, with coordinates (x2-x1,y2-y1). Here, the x position is x2-x1 (some number), and the y position is y2-y1.

Example 1: Consider a line passing through M=(1,1) and N=(4,3). Its equation is given by (x,y) = (1,1)+k*(4-1,3-1). We can simplify this as (x,y) = (1,1)+k*(3,2). Everything here is expressed in terms of x-, y-coordinates, as in (x,y). When k=0, we get a point (1,1) which lies on the line. "k" can take arbitrary values. However, when k=1, we obtain another point (4,3) which coincides with point N. If the line extends to infinity, i.e., if it goes on forever, and ever in both directions, k can take any value from -infinity to +infinity. If we consider only a line segment which extends from point M to point N, it has a vector equation given by (x,y)=M+k*[N-M], where k is some constant bounded between zero and one, viz., 0 <= k <= 1. You can write this as (x,y) = (1-k)*M+k*N if you like. Of course, if point M coincides with the origin, the M disappears from the equation altogether.

[Result 3] If a line originates from M but extends to P, obtained by rotating the vector N-M by an angle "q", then the vector P-M is given by Equation 1 below. Note: N-M is a segment from point M to N. Similarly, P-M is a segment from point M to P. From earlier derivation, the vector N-M=(x2-x1,y2-y1) since M=(x1,y1) and N=(x2,y2). To simplify notations, I will let N-M be (x0,y0), when in actual fact, x0=x2-x1, y0=y2-y1.

ROTATION FORMULA:
The arrow pointing from M to P is given by vector P-M = (x0*cos(q)-y0*sin(q), x0*sin(q)+y0*cos(q)) ...[Equation 1]

If it helps, draw a diagram, labeling the points M, N, and P, with angle q separating the rays P-M and N-M to help you visualize this. [Remember that P-M means a line from point M to P, N-M represents a line from M to N. They both have point M in common.]

Once the vector P-M is known, the new line formed by rotating N-M "q" degrees about point M is simply:
(x,y) = M+k*[P-M], where M=(x1,y1), the coordinates for vector P-M are given in Equation 1.

Example 2: A line passing through M=(1,1) and N=(4,3) is given by (x,y) = M+k*(N-M) = (1,1)+k*(3,2). Find an expression for a line obtained when the said line is rotated by q=30 degrees about the point M.

We identify the part (3,2) to be the same as (x2-x1,y2-y1). This is same as (x0,y0) by definition. Using formula 1,
P-M=(x0*cos(q)-y0*sin(q), x0*sin(q)+y0*cos(q))
  =(3*cos(30)-2*sin(30), 3*sin(30)+2*cos(30))
  =(1.5980,3.2320) ...these are approximate values only

The equation for the rotated line extending from M to P is
(x,y) = M+k*[P-M] = (1,1)+k*(1.5980,3.2320)
where 0<= k <= 1

Note: The point P is given by (1.5980,3.2320)+M = (2.5980,4.2320)

FINALLY
The equation of a line may be represented in various ways. If this is somewhat different to what you have learned, send me a follow-up with a numerical example so I can respond to your question with a clear understanding of your class experience.