Coordinate Geometry

Where algebra meets geometry

Have you ever given someone directions by saying “go three blocks east, then two blocks north”? Or used a map app that shows your location as a little blue dot? Then you have already experienced the core idea behind coordinate geometry: using numbers to describe positions in space.

Coordinate geometry is where algebra and geometry shake hands. Instead of just drawing shapes and eyeballing measurements, we place everything on a grid with numbered axes. Suddenly, we can calculate exact distances, find precise midpoints, and prove geometric facts using equations. It is like upgrading from a sketch to a blueprint.

The beautiful thing is that this connection goes both ways. Geometric ideas like “how far apart are these two points?” become algebraic formulas. And algebraic equations like $x^2 + y^2 = 25$ turn into perfect geometric shapes (that one is a circle, by the way). This back-and-forth between pictures and equations is one of the most powerful tools in all of mathematics.

Core Concepts

The Coordinate Plane: Your Mathematical Map

The coordinate plane is simply two number lines that cross at right angles. The horizontal line is the x-axis, and the vertical line is the y-axis. Where they cross is called the origin, labeled as the point $(0, 0)$.

Every point on this plane has an address written as an ordered pair $(x, y)$. The first number tells you how far to go horizontally from the origin (positive means right, negative means left). The second number tells you how far to go vertically (positive means up, negative means down).

Think of it like a city laid out on a grid: “Go 3 blocks east and 4 blocks north” translates to the point $(3, 4)$.

Distance Formula: How Far Apart Are Two Points?

This is where the Pythagorean theorem makes a stunning comeback. Suppose you want to find the distance between two points, $A = (x_1, y_1)$ and $B = (x_2, y_2)$.

Draw a horizontal line from $A$ and a vertical line from $B$. These lines meet to form a right triangle, with the segment $\overline{AB}$ as the hypotenuse. The horizontal leg has length $|x_2 - x_1|$, and the vertical leg has length $|y_2 - y_1|$.

By the Pythagorean theorem:

$$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$$

Taking the square root:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

This is the distance formula. Notice that squaring automatically handles any negative signs, so you do not need the absolute value bars once you square the differences.

Midpoint Formula: Finding the Exact Middle

Sometimes you need to find the point that is exactly halfway between two other points. Maybe you are meeting a friend and want to find a restaurant equidistant from both of you. That is a midpoint problem.

The midpoint of a segment connecting $(x_1, y_1)$ and $(x_2, y_2)$ is found by averaging the x-coordinates and averaging the y-coordinates:

$$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$

This makes intuitive sense: to get halfway between two values, you add them up and divide by 2. The midpoint formula just does that for both coordinates at once.

Slope: The Steepness of a Line

The slope of a line measures how steeply it rises or falls as you move from left to right. It is the ratio of vertical change to horizontal change, often called “rise over run.”

For a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$:

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}}$$

What slope tells you:

  • Positive slope: The line goes uphill from left to right
  • Negative slope: The line goes downhill from left to right
  • Zero slope: The line is horizontal (no rise at all)
  • Undefined slope: The line is vertical (no run, so you would divide by zero)

The larger the absolute value of the slope, the steeper the line. A slope of 3 is steeper than a slope of 1. A slope of $-5$ is steeper than a slope of $-2$.

Parallel and Perpendicular Lines

Here is where coordinate geometry reveals something elegant about geometric relationships.

Parallel lines never intersect. In coordinate geometry, this means they have the same slope. If two lines have slopes $m_1$ and $m_2$, they are parallel if and only if:

$$m_1 = m_2$$

Think about it: if two lines rise at the same rate, they will never catch up to each other.

Perpendicular lines intersect at right angles ($90°$). Their slopes have a special relationship: they are negative reciprocals of each other. If two lines have slopes $m_1$ and $m_2$, they are perpendicular if and only if:

$$m_1 \cdot m_2 = -1$$

Or equivalently: $m_2 = -\frac{1}{m_1}$

For example, if one line has slope $\frac{2}{3}$, any line perpendicular to it has slope $-\frac{3}{2}$. Flip the fraction and change the sign.

Special cases: A horizontal line (slope $0$) is perpendicular to a vertical line (undefined slope). Since you cannot multiply by undefined, we handle this case separately.

Equations of Lines

There are several ways to write the equation of a line, each useful in different situations.

Slope-Intercept Form: $y = mx + b$

This is the most common form. Here, $m$ is the slope and $b$ is the y-intercept (where the line crosses the y-axis). If you know the slope and y-intercept, you can write the equation immediately.

Point-Slope Form: $y - y_1 = m(x - x_1)$

This form is handy when you know the slope and any point $(x_1, y_1)$ on the line. Just plug in the values and simplify.

Standard Form: $Ax + By = C$

In standard form, $A$, $B$, and $C$ are integers, and by convention, $A$ is positive. This form is useful for certain calculations and for finding intercepts quickly.

Equations of Circles

A circle is the set of all points that are a fixed distance (the radius) from a center point. Using the distance formula, we can write this definition as an equation.

If the center is at $(h, k)$ and the radius is $r$, then a point $(x, y)$ is on the circle if and only if its distance from the center equals $r$:

$$\sqrt{(x - h)^2 + (y - k)^2} = r$$

Squaring both sides gives us the standard form of a circle:

$$(x - h)^2 + (y - k)^2 = r^2$$

Notice the subtraction signs: if the center is at $(3, -2)$, the equation is $(x - 3)^2 + (y - (-2))^2 = r^2$, which simplifies to $(x - 3)^2 + (y + 2)^2 = r^2$.

Special case: If the center is at the origin $(0, 0)$, the equation simplifies to:

$$x^2 + y^2 = r^2$$

Coordinate Proofs

One of the most powerful applications of coordinate geometry is proving geometric facts using algebra. This is called a coordinate proof.

The strategy is:

  1. Place the figure on a coordinate plane (choose coordinates that make calculations easy)
  2. Use formulas (distance, midpoint, slope) to establish relationships
  3. Draw conclusions based on algebraic results

Good coordinate choices can make proofs much simpler. For example, when proving something about a rectangle, placing one corner at the origin and aligning sides with the axes eliminates unnecessary complexity.

Classifying Figures Using Coordinates

Given the coordinates of vertices, you can determine exactly what type of figure you have:

To show a quadrilateral is a parallelogram:

  • Show that both pairs of opposite sides have equal slopes (parallel), OR
  • Show that both pairs of opposite sides have equal lengths, OR
  • Show that the diagonals bisect each other (same midpoint)

To show a parallelogram is a rectangle:

  • Show that it has a right angle (perpendicular sides, so slopes multiply to $-1$), OR
  • Show that the diagonals have equal length

To show a parallelogram is a rhombus:

  • Show that all four sides have equal length, OR
  • Show that the diagonals are perpendicular

To show a quadrilateral is a square:

  • Show it is both a rectangle and a rhombus

Notation and Terminology

Term Meaning Example
Distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ Distance between $(1, 2)$ and $(4, 6)$
Midpoint formula $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ Midpoint of segment from $(0, 0)$ to $(6, 8)$ is $(3, 4)$
Slope $m = \frac{y_2-y_1}{x_2-x_1}$ Rise over run
Parallel slopes Equal slopes $m_1 = m_2$
Perpendicular slopes Negative reciprocals $m_1 \cdot m_2 = -1$
Slope-intercept form $y = mx + b$ $m$ = slope, $b$ = y-intercept
Point-slope form $y - y_1 = m(x - x_1)$ Uses a point and slope
Standard form $Ax + By = C$ $A$, $B$, $C$ are integers
Circle equation $(x-h)^2 + (y-k)^2 = r^2$ Center $(h,k)$, radius $r$
Origin The point $(0, 0)$ Where the axes cross

Examples

Example 1: Finding the Distance Between Two Points

Find the distance between the points $P = (2, 3)$ and $Q = (7, 15)$.

Solution:

We use the distance formula with $(x_1, y_1) = (2, 3)$ and $(x_2, y_2) = (7, 15)$:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

$$d = \sqrt{(7 - 2)^2 + (15 - 3)^2}$$

$$d = \sqrt{5^2 + 12^2}$$

$$d = \sqrt{25 + 144}$$

$$d = \sqrt{169}$$

$$d = 13$$

The distance between the two points is 13 units.

Note: This is a 5-12-13 right triangle, one of the common Pythagorean triples. Recognizing these can save you time: 3-4-5, 5-12-13, 8-15-17, and 7-24-25 are worth memorizing.

Example 2: Finding the Midpoint of a Segment

Find the midpoint of the segment connecting $A = (-4, 7)$ and $B = (6, -3)$.

Solution:

We use the midpoint formula:

$$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$

$$M = \left(\frac{-4 + 6}{2}, \frac{7 + (-3)}{2}\right)$$

$$M = \left(\frac{2}{2}, \frac{4}{2}\right)$$

$$M = (1, 2)$$

The midpoint is (1, 2).

Check: The midpoint should be equidistant from both endpoints. You can verify this by calculating the distance from $M$ to $A$ and from $M$ to $B$ - they should be equal.

Example 3: Determining if Lines are Parallel or Perpendicular

Line $\ell_1$ passes through $(1, 2)$ and $(4, 8)$. Line $\ell_2$ passes through $(0, 5)$ and $(2, 4)$.

Are these lines parallel, perpendicular, or neither?

Solution:

First, find the slope of each line.

Slope of $\ell_1$: $$m_1 = \frac{8 - 2}{4 - 1} = \frac{6}{3} = 2$$

Slope of $\ell_2$: $$m_2 = \frac{4 - 5}{2 - 0} = \frac{-1}{2} = -\frac{1}{2}$$

Now check the relationships:

  • Are they parallel? Parallel lines have equal slopes. Is $2 = -\frac{1}{2}$? No.
  • Are they perpendicular? Perpendicular lines have slopes that multiply to $-1$. Is $2 \times (-\frac{1}{2}) = -1$? Yes!

$$m_1 \cdot m_2 = 2 \cdot \left(-\frac{1}{2}\right) = -1$$

The lines are perpendicular.

Notice that $-\frac{1}{2}$ is indeed the negative reciprocal of $2$ (flip $\frac{2}{1}$ to get $\frac{1}{2}$, then change the sign to get $-\frac{1}{2}$).

Example 4: Writing the Equation of a Circle

Write the equation of a circle with center $(-3, 5)$ and radius $4$.

Solution:

The standard form of a circle with center $(h, k)$ and radius $r$ is:

$$(x - h)^2 + (y - k)^2 = r^2$$

Here, $(h, k) = (-3, 5)$ and $r = 4$.

Substituting:

$$(x - (-3))^2 + (y - 5)^2 = 4^2$$

$$(x + 3)^2 + (y - 5)^2 = 16$$

The equation of the circle is $(x + 3)^2 + (y - 5)^2 = 16$.

Reading the equation backwards: If you see an equation like $(x - 2)^2 + (y + 7)^2 = 49$, you can identify the center as $(2, -7)$ (watch the signs!) and the radius as $\sqrt{49} = 7$.

Example 5: Proving a Quadrilateral is a Parallelogram

Prove that the quadrilateral with vertices $A = (1, 1)$, $B = (4, 5)$, $C = (9, 6)$, and $D = (6, 2)$ is a parallelogram.

Solution:

A quadrilateral is a parallelogram if both pairs of opposite sides are parallel (have equal slopes). Let us find the slopes of all four sides.

Slope of $\overline{AB}$: $$m_{AB} = \frac{5 - 1}{4 - 1} = \frac{4}{3}$$

Slope of $\overline{DC}$: $$m_{DC} = \frac{6 - 2}{9 - 6} = \frac{4}{3}$$

Since $m_{AB} = m_{DC} = \frac{4}{3}$, sides $\overline{AB}$ and $\overline{DC}$ are parallel.

Slope of $\overline{BC}$: $$m_{BC} = \frac{6 - 5}{9 - 4} = \frac{1}{5}$$

Slope of $\overline{AD}$: $$m_{AD} = \frac{2 - 1}{6 - 1} = \frac{1}{5}$$

Since $m_{BC} = m_{AD} = \frac{1}{5}$, sides $\overline{BC}$ and $\overline{AD}$ are parallel.

Conclusion: Both pairs of opposite sides are parallel, so quadrilateral $ABCD$ is a parallelogram.

Alternative method: We could also prove this by showing that the diagonals bisect each other (have the same midpoint):

Midpoint of $\overline{AC}$: $\left(\frac{1+9}{2}, \frac{1+6}{2}\right) = (5, 3.5)$

Midpoint of $\overline{BD}$: $\left(\frac{4+6}{2}, \frac{5+2}{2}\right) = (5, 3.5)$

Same midpoint, so the diagonals bisect each other, confirming it is a parallelogram.

Example 6: Writing the Equation of a Line

Write the equation of the line that passes through $(2, -1)$ and is perpendicular to the line $3x + 4y = 12$.

Solution:

Step 1: Find the slope of the given line.

First, rewrite $3x + 4y = 12$ in slope-intercept form: $$4y = -3x + 12$$ $$y = -\frac{3}{4}x + 3$$

The slope of the given line is $m = -\frac{3}{4}$.

Step 2: Find the slope of the perpendicular line.

Perpendicular lines have slopes that are negative reciprocals. The negative reciprocal of $-\frac{3}{4}$ is:

$$m_\perp = -\frac{1}{-\frac{3}{4}} = \frac{4}{3}$$

Step 3: Use point-slope form.

With slope $\frac{4}{3}$ and point $(2, -1)$:

$$y - (-1) = \frac{4}{3}(x - 2)$$

$$y + 1 = \frac{4}{3}x - \frac{8}{3}$$

$$y = \frac{4}{3}x - \frac{8}{3} - 1$$

$$y = \frac{4}{3}x - \frac{8}{3} - \frac{3}{3}$$

$$y = \frac{4}{3}x - \frac{11}{3}$$

The equation of the perpendicular line is $y = \frac{4}{3}x - \frac{11}{3}$.

Or in standard form: $4x - 3y = 11$.

Key Properties and Rules

Distance and Midpoint

Distance Formula: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Midpoint Formula: $$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$

Slope

Slope Formula: $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}}$$

Slope Interpretations:

  • $m > 0$: Line rises from left to right
  • $m < 0$: Line falls from left to right
  • $m = 0$: Horizontal line
  • $m$ undefined: Vertical line

Parallel and Perpendicular

Parallel lines: $m_1 = m_2$

Perpendicular lines: $m_1 \cdot m_2 = -1$ (or $m_2 = -\frac{1}{m_1}$)

Equations of Lines

Slope-Intercept Form: $y = mx + b$

Point-Slope Form: $y - y_1 = m(x - x_1)$

Standard Form: $Ax + By = C$

Horizontal line through $(a, b)$: $y = b$

Vertical line through $(a, b)$: $x = a$

Equation of a Circle

Standard Form: $$(x - h)^2 + (y - k)^2 = r^2$$

  • Center: $(h, k)$
  • Radius: $r$

Circle centered at origin: $$x^2 + y^2 = r^2$$

Classifying Quadrilaterals

To prove… Show that…
Parallelogram Both pairs of opposite sides parallel (equal slopes)
Parallelogram Both pairs of opposite sides equal (same length)
Parallelogram Diagonals bisect each other (same midpoint)
Rectangle Parallelogram with one right angle (perpendicular sides)
Rectangle Parallelogram with equal diagonals
Rhombus Parallelogram with all sides equal
Rhombus Parallelogram with perpendicular diagonals
Square Rectangle with all sides equal
Square Rhombus with a right angle

Real-World Applications

GPS and Navigation

Your phone’s GPS works by determining your coordinates on Earth’s surface. Navigation apps constantly calculate distances between your current location and your destination using formulas that are essentially three-dimensional versions of the distance formula. When the app says “3.2 miles to destination,” it has computed that distance from coordinates.

Computer Graphics and Video Games

Every image on your screen is made of pixels, each with $(x, y)$ coordinates. When a video game character moves, the computer is updating coordinates and recalculating distances constantly. Collision detection (knowing when two objects touch) relies heavily on distance calculations. Drawing circles, lines, and other shapes means translating equations into colored pixels.

Urban Planning and Architecture

City planners use coordinate systems to design road layouts, determine property boundaries, and plan utility routes. When designing a new neighborhood, they might need to ensure that no house is more than a certain distance from an emergency services station - a problem solved using the distance formula. Architects use coordinate geometry to create precise blueprints and ensure structures are properly aligned.

Surveying and Mapping

Surveyors establish property boundaries and create maps using coordinate systems. They place reference points (called control points) with known coordinates, then measure distances and angles to determine the coordinates of other points. This is how we know exactly where property lines are and how large parcels of land are.

Robotics and Autonomous Vehicles

Self-driving cars and robots navigate by constantly tracking their position in a coordinate system. They calculate distances to obstacles, determine paths to destinations, and make decisions about when to turn based on coordinate geometry. The midpoint formula might help a robot find the center of a doorway to pass through.

Game Development

When you play a strategy game and click to move a character, the game calculates the distance from the character’s current position to where you clicked. Many games use coordinate-based systems to determine line of sight, attack range, and movement paths. Even simple mobile games use these calculations constantly.

Self-Test Problems

Problem 1: Find the distance between the points $(3, -2)$ and $(-1, 5)$.

Show Answer

Using the distance formula:

$$d = \sqrt{(-1 - 3)^2 + (5 - (-2))^2}$$ $$d = \sqrt{(-4)^2 + (7)^2}$$ $$d = \sqrt{16 + 49}$$ $$d = \sqrt{65}$$

The distance is $\sqrt{65} \approx 8.06$ units.

Problem 2: The endpoints of a diameter of a circle are $(-2, 3)$ and $(6, 9)$. Find the center and radius of the circle.

Show Answer

The center is the midpoint of the diameter:

$$\text{Center} = \left(\frac{-2 + 6}{2}, \frac{3 + 9}{2}\right) = \left(\frac{4}{2}, \frac{12}{2}\right) = (2, 6)$$

The radius is half the length of the diameter. First find the diameter length:

$$d = \sqrt{(6 - (-2))^2 + (9 - 3)^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$$

So the radius is $r = \frac{10}{2} = 5$.

Center: $(2, 6)$, Radius: $5$

The equation of the circle would be $(x - 2)^2 + (y - 6)^2 = 25$.

Problem 3: Line $\ell_1$ has equation $y = 2x + 5$. Line $\ell_2$ passes through $(4, 1)$ and $(6, 2)$. Are these lines parallel, perpendicular, or neither?

Show Answer

From the equation $y = 2x + 5$, line $\ell_1$ has slope $m_1 = 2$.

For line $\ell_2$: $$m_2 = \frac{2 - 1}{6 - 4} = \frac{1}{2}$$

Check: $m_1 = m_2$? Is $2 = \frac{1}{2}$? No, so not parallel.

Check: $m_1 \cdot m_2 = -1$? Is $2 \times \frac{1}{2} = 1 = -1$? No, so not perpendicular.

The lines are neither parallel nor perpendicular.

Problem 4: Write the equation of the circle with center $(4, -1)$ that passes through the point $(7, 3)$.

Show Answer

First, find the radius by calculating the distance from the center to the point on the circle:

$$r = \sqrt{(7 - 4)^2 + (3 - (-1))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

Now write the equation with center $(4, -1)$ and radius $5$:

$$(x - 4)^2 + (y - (-1))^2 = 5^2$$

$$(x - 4)^2 + (y + 1)^2 = 25$$

Problem 5: The vertices of a triangle are $A = (0, 0)$, $B = (6, 0)$, and $C = (3, 4)$. Find the perimeter of the triangle.

Show Answer

Calculate the length of each side:

Side $\overline{AB}$: $$AB = \sqrt{(6-0)^2 + (0-0)^2} = \sqrt{36 + 0} = 6$$

Side $\overline{BC}$: $$BC = \sqrt{(3-6)^2 + (4-0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

Side $\overline{AC}$: $$AC = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

Perimeter: $6 + 5 + 5 = 16$ units

(This is an isosceles triangle with two sides of length 5.)

Problem 6: Show that the quadrilateral with vertices $P = (0, 0)$, $Q = (4, 3)$, $R = (7, -1)$, and $S = (3, -4)$ is a rhombus.

Show Answer

A rhombus is a quadrilateral with all four sides equal. Let us calculate all side lengths:

Side $\overline{PQ}$: $$PQ = \sqrt{(4-0)^2 + (3-0)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$

Side $\overline{QR}$: $$QR = \sqrt{(7-4)^2 + (-1-3)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

Side $\overline{RS}$: $$RS = \sqrt{(3-7)^2 + (-4-(-1))^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$

Side $\overline{SP}$: $$SP = \sqrt{(0-3)^2 + (0-(-4))^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

All four sides have length 5, so PQRS is a rhombus.

Problem 7: Write the equation of the line that passes through $(1, 4)$ and is parallel to the line $2x - 3y = 9$.

Show Answer

First, find the slope of the given line by converting to slope-intercept form:

$$2x - 3y = 9$$ $$-3y = -2x + 9$$ $$y = \frac{2}{3}x - 3$$

The slope is $\frac{2}{3}$. A parallel line has the same slope.

Using point-slope form with point $(1, 4)$ and slope $\frac{2}{3}$:

$$y - 4 = \frac{2}{3}(x - 1)$$ $$y - 4 = \frac{2}{3}x - \frac{2}{3}$$ $$y = \frac{2}{3}x - \frac{2}{3} + 4$$ $$y = \frac{2}{3}x + \frac{10}{3}$$

Or in standard form: $2x - 3y = -10$

Summary

  • The coordinate plane uses ordered pairs $(x, y)$ to specify exact locations, turning geometric problems into algebraic ones.

  • The distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ comes directly from the Pythagorean theorem. It tells you exactly how far apart two points are.

  • The midpoint formula $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ finds the point exactly halfway between two given points by averaging coordinates.

  • Slope $m = \frac{y_2-y_1}{x_2-x_1}$ measures the steepness of a line. Positive slopes rise, negative slopes fall, zero means horizontal, and undefined means vertical.

  • Parallel lines have equal slopes. Perpendicular lines have slopes that are negative reciprocals (they multiply to $-1$).

  • Lines can be written in slope-intercept form ($y = mx + b$), point-slope form ($y - y_1 = m(x - x_1)$), or standard form ($Ax + By = C$).

  • The equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.

  • Coordinate proofs use algebra to prove geometric facts. By placing figures on a coordinate plane and using distance, midpoint, and slope formulas, you can prove that shapes have specific properties.

  • To classify quadrilaterals, check slopes for parallel and perpendicular relationships, use the distance formula to compare side lengths, and find midpoints to check if diagonals bisect each other.

  • Coordinate geometry is everywhere in real life: GPS navigation, computer graphics, urban planning, video games, and robotics all rely on these fundamental concepts.