Polar Coordinates and Straight Lines

1. Introduction to Polar Coordinates

In the Cartesian coordinate system, a point is located by its perpendicular distances from two fixed perpendicular lines (axes). In the polar coordinate system, a point is located by its distance from a fixed point and the angle made with a fixed line. This system is particularly useful for describing curves that have a rotational symmetry or are centered around a point.

The fixed point is called the pole, which corresponds to the origin (0,0) in the Cartesian system. The fixed line is called the polar axis, which usually corresponds to the positive x-axis in the Cartesian system.

A point P in polar coordinates is represented by an ordered pair (r, θ), where:

• r is the radial distance from the pole to the point P. It is the length of the line segment OP, where O is the pole and P is the point. 'r' can be positive or negative. A negative 'r' means the point is located in the opposite direction of the ray defined by θ.
• θ is the angle measured counterclockwise from the polar axis to the line segment OP. 'θ' is called the angular coordinate or polar angle. It is usually measured in radians or degrees.

2. Relationship between Polar and Cartesian Coordinates

We can establish a relationship between the Cartesian coordinates (x, y) and polar coordinates (r, θ) of a point P. Assume the pole coincides with the origin and the polar axis coincides with the positive x-axis.

If P has Cartesian coordinates (x, y) and polar coordinates (r, θ), then:

• $x = r \cos(\theta)$
• $y = r \sin(\theta)$

Conversely, to find r and θ from x and y:

• $r^2 = x^2 + y^2 \implies r = \pm \sqrt{x^2 + y^2}$
• $\tan(\theta) = \frac{y}{x}$ (provided $x \neq 0$)

The angle θ needs to be determined based on the quadrant of the point (x, y).

3. Converting Equations between Coordinate Systems

We can convert equations from Cartesian to polar form and vice versa using the conversion formulas.

3.1 Cartesian to Polar Conversion

Substitute $x = r \cos(\theta)$ and $y = r \sin(\theta)$ into the Cartesian equation.

Example: Convert the Cartesian equation $x^2 + y^2 = 9$ to polar form.

Substituting $x^2 + y^2 = r^2$, we get $r^2 = 9$, which simplifies to $r = 3$ (since r is usually taken as positive for simplicity, representing a circle of radius 3 centered at the pole).

Example: Convert the Cartesian equation $y = 2x$ to polar form.

Substituting $y = r \sin(\theta)$ and $x = r \cos(\theta)$, we get $r \sin(\theta) = 2(r \cos(\theta))$. If $r \neq 0$, we can divide by r: $\sin(\theta) = 2 \cos(\theta)$. Dividing by $\cos(\theta)$ (assuming $\cos(\theta) \neq 0$), we get $\tan(\theta) = 2$. This represents a straight line passing through the origin with a constant angle.

3.2 Polar to Cartesian Conversion

Substitute $r = \sqrt{x^2 + y^2}$ and $\tan(\theta) = \frac{y}{x}$ (and potentially $r \cos(\theta) = x$, $r \sin(\theta) = y$).

Example: Convert the polar equation $r = 4 \sin(\theta)$ to Cartesian form.

Multiply both sides by r: $r^2 = 4r \sin(\theta)$. Substitute $r^2 = x^2 + y^2$ and $r \sin(\theta) = y$: $x^2 + y^2 = 4y$. Rearranging, we get $x^2 + y^2 - 4y = 0$. Completing the square for y: $x^2 + (y^2 - 4y + 4) = 4$. This gives $x^2 + (y-2)^2 = 2^2$, which is the equation of a circle centered at (0, 2) with radius 2.

Example: Convert the polar equation $\theta = \frac{\pi}{4}$ to Cartesian form.

Taking the tangent of both sides: $\tan(\theta) = \tan(\frac{\pi}{4})$. Substitute $\tan(\theta) = \frac{y}{x}$: $\frac{y}{x} = 1$. This gives $y = x$, which is a straight line passing through the origin with a slope of 1.

4. Straight Lines in Polar Coordinates

A straight line can be represented in polar coordinates. The general form depends on its orientation relative to the pole.

4.1 Line Through the Pole

A line passing through the pole (origin) is defined by a constant angle θ. The equation is simply $\theta = \alpha$, where $\alpha$ is the constant angle the line makes with the polar axis.

For example, $\theta = \frac{\pi}{3}$ represents a line passing through the origin making an angle of 60 degrees with the polar axis.

4.2 Line Not Through the Pole

Consider a line that does not pass through the pole. Let p be the perpendicular distance from the pole to the line, and let $\alpha$ be the angle that this perpendicular makes with the polar axis.

Let P(r, θ) be any point on this line. Consider the triangle formed by the pole O, the point P, and the foot of the perpendicular from the pole to the line (let's call this point Q).

In this triangle OQP, the angle at Q is 90 degrees. The length OQ is p. The angle $\angle QOP$ is the difference between the angle of P (θ) and the angle of the perpendicular (α), i.e., $|θ - \alpha|$.

Using trigonometry in the right-angled triangle OQP: $\cos(|θ - \alpha|) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{p}{r}$

So, $r \cos(|θ - \alpha|) = p$.

Since $\cos(-x) = \cos(x)$, we can write this as $r \cos(θ - \alpha) = p$.

Expanding this using the cosine subtraction formula: $r (\cos(\theta)\cos(\alpha) + \sin(\theta)\sin(\alpha)) = p$

Rearranging: $(r \cos(\theta))\cos(\alpha) + (r \sin(\theta))\sin(\alpha) = p$

Substituting $x = r \cos(\theta)$ and $y = r \sin(\theta)$: $x \cos(\alpha) + y \sin(\alpha) = p$

This is the normal form of a straight line in Cartesian coordinates, where p is the length of the normal from the origin and $\alpha$ is the angle it makes with the positive x-axis. This confirms our polar equation is correct.

4.3 Special Cases for Line Not Through Pole

Let's analyze some specific forms of the equation $r \cos(θ - \alpha) = p$.

4.3.1 Vertical Line

Consider a vertical line $x = k$. In polar coordinates, $r \cos(\theta) = k$. This corresponds to the general form $r \cos(θ - \alpha) = p$ where $\alpha = 0$ (the perpendicular is along the polar axis) and $p = k$. So, $r \cos(\theta) = p$.

If the line is $x=k$ (where $k>0$), then $p=k$ and $\alpha=0$. If the line is $x=-k$ (where $k>0$), then the perpendicular distance is $p=k$. The perpendicular lies on the negative x-axis, so $\alpha = \pi$. The equation becomes $r \cos(θ - \pi) = k$. Since $\cos(θ - \pi) = -\cos(θ)$, this is $-r \cos(θ) = k$, or $r \cos(θ) = -k$, which is consistent.

4.3.2 Horizontal Line

Consider a horizontal line $y = k$. In polar coordinates, $r \sin(\theta) = k$. This corresponds to the general form $r \cos(θ - \alpha) = p$ where $\alpha = \frac{\pi}{2}$ (the perpendicular is along the positive y-axis) and $p = k$. So, $r \cos(θ - \frac{\pi}{2}) = p$. Since $\cos(θ - \frac{\pi}{2}) = \sin(θ)$, this becomes $r \sin(\theta) = p$.

If the line is $y=k$ (where $k>0$), then $p=k$ and $\alpha=\frac{\pi}{2}$. If the line is $y=-k$ (where $k>0$), then $p=k$ and $\alpha=\frac{3\pi}{2}$ (or $-\frac{\pi}{2}$). The equation becomes $r \cos(θ - \frac{3\pi}{2}) = k$. Since $\cos(θ - \frac{3\pi}{2}) = \sin(θ)$, this is $r \sin(\theta) = k$, which is consistent.

5. Intersection of Lines in Polar Coordinates

To find the intersection of two lines given in polar coordinates, we need to solve the system of equations simultaneously.

Example: Find the intersection of the lines $\theta = \frac{\pi}{4}$ and $r \sin(\theta) = 3$.

We are given: 1. $\theta = \frac{\pi}{4}$ 2. $r \sin(\theta) = 3$

Substitute the value of θ from equation (1) into equation (2): $r \sin(\frac{\pi}{4}) = 3$ $r \left(\frac{\sqrt{2}}{2}\right) = 3$ $r = \frac{6}{\sqrt{2}} = 3\sqrt{2}$

So, the point of intersection in polar coordinates is $(3\sqrt{2}, \frac{\pi}{4})$.

To find the Cartesian coordinates of this intersection point: $x = r \cos(\theta) = 3\sqrt{2} \cos(\frac{\pi}{4}) = 3\sqrt{2} \left(\frac{\sqrt{2}}{2}\right) = 3$ $y = r \sin(\theta) = 3\sqrt{2} \sin(\frac{\pi}{4}) = 3\sqrt{2} \left(\frac{\sqrt{2}}{2}\right) = 3$ The Cartesian coordinates are (3, 3). This matches the line $y=x$ (from $\theta=\frac{\pi}{4}$) and $y=3$ (from $r \sin(\theta) = 3$).

6. Graphing Lines in Polar Coordinates

Graphing lines in polar coordinates can be done by understanding their equations.

• $\theta = \alpha$: This is a ray starting from the pole at an angle $\alpha$ with the polar axis. If the equation is just $\theta=\alpha$ without constraints on r, it represents the entire line passing through the pole.
• $r = c$: This represents a circle centered at the pole with radius $|c|$.
• $r = k/\cos(\theta)$: This simplifies to $r \cos(\theta) = k$, which is a vertical line $x=k$.
• $r = k/\sin(\theta)$: This simplifies to $r \sin(\theta) = k$, which is a horizontal line $y=k$.
• $r \cos(θ - \alpha) = p$: This is a line whose shortest distance from the pole is p, and the line segment of length p makes an angle $\alpha$ with the polar axis.

To graph $r \cos(θ - \alpha) = p$:

1. Identify p (perpendicular distance from pole) and $\alpha$ (angle of the perpendicular).
2. Draw a ray from the pole at angle $\alpha$.
3. Mark a point Q on this ray at a distance p from the pole.
4. Draw a line through Q that is perpendicular to the ray OQ. This is the required line.

Example: Graph the line $r \cos(θ - \frac{\pi}{6}) = 2$.

Here, $p=2$ and $\alpha = \frac{\pi}{6}$ (or 30 degrees).

1. Draw the polar axis (initial ray).
2. Draw a ray from the pole at 30 degrees to the polar axis.
3. On this 30-degree ray, mark a point Q at a distance of 2 units from the pole.
4. Draw a line through Q that is perpendicular to the 30-degree ray. This line is the graph of $r \cos(θ - \frac{\pi}{6}) = 2$.

Alternatively, convert to Cartesian: $r (\cos\theta \cos\frac{\pi}{6} + \sin\theta \sin\frac{\pi}{6}) = 2$ $r (\cos\theta \frac{\sqrt{3}}{2} + \sin\theta \frac{1}{2}) = 2$ $\frac{\sqrt{3}}{2} (r \cos\theta) + \frac{1}{2} (r \sin\theta) = 2$ $\frac{\sqrt{3}}{2} x + \frac{1}{2} y = 2$ $\sqrt{3} x + y = 4$ $y = -\sqrt{3} x + 4$. This is a line with slope $-\sqrt{3}$ and y-intercept 4.

7. Area of a Triangle in Polar Coordinates

Consider a triangle with vertices at the pole O(0,0), and two other points $P_1(r_1, \theta_1)$ and $P_2(r_2, \theta_2)$.

The lengths of the two sides from the pole are $r_1$ and $r_2$. The angle between these two sides is the difference between their polar angles, $|\theta_2 - \theta_1|$.

The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$. If we consider $OP_1$ as the base, then $r_1$ is the base length. The height would be $r_2 \sin(|\theta_2 - \theta_1|)$.

So, the area A is: $A = \frac{1}{2} r_1 (r_2 \sin(|\theta_2 - \theta_1|))$ $A = \frac{1}{2} r_1 r_2 \sin(|\theta_2 - \theta_1|)$

Since $\sin(-x) = -\sin(x)$ but area must be positive, we often use $\sin(\theta_2 - \theta_1)$ and take the absolute value, or ensure $\theta_2 > \theta_1$ and $|\theta_2 - \theta_1| < \pi$.

Example: Find the area of the triangle formed by the pole and the points $(4, \frac{\pi}{3})$ and $(6, \frac{\pi}{2})$.

Here, $r_1 = 4$, $\theta_1 = \frac{\pi}{3}$, $r_2 = 6$, $\theta_2 = \frac{\pi}{2}$. Area $A = \frac{1}{2} |4 \times 6 \sin(\frac{\pi}{2} - \frac{\pi}{3})|$ $A = \frac{1}{2} |24 \sin(\frac{3\pi - 2\pi}{6})|$ $A = \frac{1}{2} |24 \sin(\frac{\pi}{6})|$ $A = \frac{1}{2} |24 \times \frac{1}{2}|$ $A = \frac{1}{2} |12| = 6$ square units.

8. Summary and Key Takeaways

Polar coordinates offer an alternative way to describe points and curves, especially those with circular or radial symmetry. The conversion between polar $(r, \theta)$ and Cartesian $(x, y)$ coordinates is fundamental: $x = r \cos(\theta)$, $y = r \sin(\theta)$, and $r^2 = x^2 + y^2$.

Straight lines in polar coordinates have specific forms:

• Lines through the pole are $\theta = \alpha$.
• Lines not through the pole are generally $r \cos(θ - \alpha) = p$, where p is the perpendicular distance from the pole and $\alpha$ is the angle of that perpendicular.
• Vertical lines are $r \cos(\theta) = k$.
• Horizontal lines are $r \sin(\theta) = k$.

Solving systems of polar equations yields intersection points. The area of a triangle formed by the pole and two points $(r_1, \theta_1)$ and $(r_2, \theta_2)$ is given by $\frac{1}{2} |r_1 r_2 \sin(\theta_2 - \theta_1)|$. Understanding these concepts allows for effective representation and analysis of geometric figures in the polar plane.