Polar Lines and Constant-Angle Rays

Introduction

A fixed angle in polar coordinates determines a direction from the pole, so varying the radius traces outward along that direction. This gives a simple description of a ray. A complete line that does not pass through the pole needs a different description: it is determined by a direction perpendicular to the line and its signed distance from the pole.

These two ideas are related but should not be confused. The angle of a ray tells you where the ray points. The angle used in the equation of a general line tells you the direction of a normal vector, which is perpendicular to the line.

Constant-Angle Rays and Lines Through the Pole

Definition: In polar coordinates, a point with coordinates $(r,\theta)$ has Cartesian coordinates $x=r\cos\theta$ and $y=r\sin\theta$. If $r\ge 0$ and $\theta=\alpha$, varying $r$ traces the ray from the pole in direction $\alpha$. Its endpoint at $r=0$ is the pole.

For example, $\theta=\pi/4$, with $r\ge 0$, traces the ray at angle $45^\circ$ from the positive $x$-axis. The radius can be any nonnegative number, so the ray starts at the pole and extends without bound.

A ray is not, by itself, a complete line. To obtain the full line through the pole in that direction, include the opposite ray as well. Its angle is $\alpha+\pi$. Equivalently, if signed radii are allowed, letting $r$ take every real value in $(r,\alpha)$ traces the same complete line: negative values of $r$ place the point in the opposite direction.

Takeaway: With the usual restriction $r\ge 0$, $\theta=\alpha$ describes a ray. The full line through the pole is the union of the rays at angles $\alpha$ and $\alpha+\pi$. Allowing $r$ to be negative also traces that full line.

Worked Example: From a Constant Angle to a Full Line

Consider the ray $\theta=2\pi/3$, $r\ge 0$. Its direction vector is $(\cos(2\pi/3),\sin(2\pi/3))$. Since $\cos(2\pi/3)=-1/2$ and $\sin(2\pi/3)=\sqrt{3}/2$, the ray has the parametrization

(x,y)=r\left(-\frac12,\frac{\sqrt3}{2}\right),\qquad r\ge 0.

To include the opposite half, use angle $2\pi/3+\pi=5\pi/3$, again with a nonnegative radius. The two direction vectors are opposites, and together they trace the full line. Alternatively, allow $r\in\mathbb R$ in the first parametrization.

To check the line, compute the ratio of the two components for $r\ne 0$: $y/x=(r\sqrt3/2)/(-r/2)=-\sqrt3$. Thus the line has equation $y=-\sqrt3x$, which passes through the pole and has the expected direction. Points from the original ray have $x\le 0$ and $y\ge 0$; the opposite ray supplies the other half.

Representing a Line by Its Normal Direction

A line not necessarily passing through the pole can be specified by a perpendicular direction and a signed distance. Let $\alpha$ be the angle of a unit vector perpendicular to the line. That unit normal vector is $(\cos\alpha,\sin\alpha)$. A point on the line has a fixed projection onto this normal vector. Denote that fixed value by $c$.

Definition: A polar equation for a line with unit normal direction $\alpha$ and signed distance $c$ from the pole is $$ r\cos(\theta-\alpha)=c. $$ Here $c$ is signed: its sign indicates which side of the pole the line lies on relative to the chosen normal direction. The perpendicular distance from the pole to the line is $|c|$.

The angle $\alpha$ is the angle of the normal, not the angle along the line. A line direction is perpendicular to the normal, so one direction angle for the line is $\alpha+\pi/2$; the opposite direction is $\alpha-\pi/2$. Since a line has no preferred orientation, those angles describe its same direction as a line.

Why the Polar Line Equation Works

The polar-to-Cartesian formulas give a direct justification. The projection of a point $(x,y)$ onto the unit normal $(\cos\alpha,\sin\alpha)$ is its dot product with that normal. Substituting $x=r\cos\theta$ and $y=r\sin\theta$ gives

x\cos\alpha+y\sin\alpha =r\cos\theta\cos\alpha+r\sin\theta\sin\alpha =r\cos(\theta-\alpha).

Thus the equation $r\cos(\theta-\alpha)=c$ says exactly that the projection onto the normal is fixed at $c$. The points with a fixed projection form a line perpendicular to the normal.

Derivation: Using the cosine difference identity, $$ r\cos(\theta-\alpha) =r\cos\theta\cos\alpha+r\sin\theta\sin\alpha =x\cos\alpha+y\sin\alpha. $$ Therefore the polar equation is equivalent to the Cartesian equation $$ x\cos\alpha+y\sin\alpha=c. $$ Its normal vector is $(\cos\alpha,\sin\alpha)$, which has length $1$. The distance of this line from the origin is consequently $|c|$: moving from the origin in the normal direction by signed distance $c$ reaches the point $c(\cos\alpha,\sin\alpha)$ on the line. Every point on the line differs from that point by a vector perpendicular to the normal.

This form also handles a line through the pole. If $c=0$, the equation becomes $x\cos\alpha+y\sin\alpha=0$. Its points lie along the direction perpendicular to the normal; the line passes through the pole. In contrast, the equation $\theta=\alpha$ specifies a ray along angle $\alpha$, not a line with normal angle $\alpha$.

Worked Examples with the Normal Form

Worked Example: Interpreting a Polar Line Equation

Interpret $r\cos(\theta-\pi/3)=2$, and find two points on the line.

The normal direction has angle $\pi/3$, so its unit vector is $(\cos(\pi/3),\sin(\pi/3))=(1/2,\sqrt3/2)$. Converting to Cartesian form gives

x\left(\frac12\right)+y\left(\frac{\sqrt3}{2}\right)=2, \qquad\text{or}\qquad x+\sqrt3\,y=4.

Setting $y=0$ gives $x=4$, so $(4,0)$ is on the line. Setting $x=0$ gives $\sqrt3\,y=4$, hence $y=4/\sqrt3$, so $(0,4/\sqrt3)$ is another point.

Check them in the Cartesian equation: $4+\sqrt3(0)=4$, and $0+\sqrt3(4/\sqrt3)=4$. The direction along the line is perpendicular to its normal; one such unit direction is $(-\sqrt3/2,1/2)$, whose dot product with the normal is $-(1/2)(\sqrt3/2)+(\sqrt3/2)(1/2)=0$. The pole’s perpendicular distance from the line is $|2|=2$.

Worked Example: Finding the Polar Equation of a Given Line

Write the line $x+y=2$ in the form $r\cos(\theta-\alpha)=c$, choosing a unit normal.

The coefficients of $x$ and $y$ give a normal vector $(1,1)$, whose length is $\sqrt{1^2+1^2}=\sqrt2$. Divide the equation by $\sqrt2$ so its normal has unit length:

\frac{x}{\sqrt2}+\frac{y}{\sqrt2}=\frac{2}{\sqrt2}=\sqrt2.

The unit normal $(1/\sqrt2,1/\sqrt2)$ has angle $\alpha=\pi/4$, since $\cos(\pi/4)=\sin(\pi/4)=1/\sqrt2$. Therefore $c=\sqrt2$, and the polar equation is

r\cos(\theta-\pi/4)=\sqrt2.

Check by expanding: $r\cos(\theta-\pi/4)=(x+y)/\sqrt2$. Setting this equal to $\sqrt2$ gives $x+y=2$, as required. The point $(2,0)$ lies on the line and satisfies the equation because its polar coordinates are $(2,0)$ and $2\cos(-\pi/4)=\sqrt2$.

Worked Example: A Line with a Negative Signed Distance

Convert $r\cos(\theta-\pi/2)=-3$ to Cartesian form and identify its distance from the pole.

The normal is $(\cos(\pi/2),\sin(\pi/2))=(0,1)$. Thus

x\cos(\pi/2)+y\sin(\pi/2)=-3, \qquad 0x+y=-3, \qquad y=-3.

This is a horizontal line three units below the $x$-axis, so its distance from the pole is $|-3|=3$. For a direct check, the point $(0,-3)$ has polar coordinates $(3,3\pi/2)$. Substitution gives $3\cos(3\pi/2-\pi/2)=3\cos\pi=-3$, so the point satisfies the equation. Every point $(x,-3)$ on the line has the same projection onto the upward-pointing unit normal: $x(0)+(-3)(1)=-3$.

Equivalent Choices of Normal

A geometric line does not have only one polar equation of this form. Reversing the normal changes both the normal angle and the sign of the signed distance. Since the normal at angle $\alpha+\pi$ is the negative of the normal at angle $\alpha$, the equations with parameters $(\alpha,c)$ and $(\alpha+\pi,-c)$ describe the same line.

r\cos\bigl(\theta-(\alpha+\pi)\bigr)=-c \quad\Longleftrightarrow\quad -r\cos(\theta-\alpha)=-c \quad\Longleftrightarrow\quad r\cos(\theta-\alpha)=c.

This equivalence is useful when a negative $c$ seems inconvenient: reversing the normal makes the signed distance positive without changing the line. It does not change the actual perpendicular distance, which is always $|c|$.

A Common Mistake: Treating the Normal as the Line Direction

In $r\cos(\theta-\alpha)=c$, the angle $\alpha$ belongs to the normal vector, not to a direction along the line. For instance, for $r\cos(\theta-\pi/3)=2$, it would be wrong to say that the line points at angle $\pi/3$. The Cartesian equation is $x+\sqrt3y=4$, whose line direction is perpendicular to $(1/2,\sqrt3/2)$, such as $(-\sqrt3/2,1/2)$. Its direction angle is $5\pi/6$, not $\pi/3$.

Another error is to treat a fixed-angle equation as a full line when the convention is $r\ge 0$. Under that convention, $\theta=\alpha$ gives only one ray. To include the opposite half, also include $\theta=\alpha+\pi$, or explicitly allow signed $r$.

Practical check: For $r\cos(\theta-\alpha)=c$, convert to $x\cos\alpha+y\sin\alpha=c$. The coefficients reveal the unit normal, and a proposed point can be checked by substitution. The line’s direction must be perpendicular to that normal.

Why These Forms Matter

Polar coordinates are especially useful when direction from the pole is central to the description. A constant angle gives a concise way to represent a ray, while the normal form gives a concise way to represent a line at a specified signed distance. Recognizing which angle is being used prevents confusion when reading or sketching polar graphs.

The normal form is also a reliable bridge between polar and Cartesian descriptions. It lets you check a proposed sketch with familiar line geometry: the coefficient vector is perpendicular to the line, $c=0$ means the line passes through the pole, and $|c|$ is the shortest distance from the pole to the line.

Check Your Understanding

Use the definitions and checks from this tutorial to answer the following.

With $r\ge 0$, what geometric set is described by $\theta=7\pi/6$? How would you describe the complete line through the pole in that direction?
Convert $r\cos(\theta-\pi/4)=3$ to Cartesian form. What is the line’s normal direction, and what is its distance from the pole?
Write $x-y=4$ in the form $r\cos(\theta-\alpha)=c$ using a unit normal with positive $x$-component.
Explain why $(\alpha,c)$ and $(\alpha+\pi,-c)$ give the same geometric line.
For a polar line equation, what geometric direction does $\alpha$ describe: the line’s direction or its normal direction?