Coordinate Geometry | Math Learning Guide

📋 Prerequisites

Coordinate geometry (or analytic geometry) turns shapes into algebra. Put a figure on the coordinate plane, assign coordinates to every point, and suddenly distances, midpoints, areas, and intersections become calculations you can just compute. When a geometry problem feels stuck, try coordinates — brute force works.

📚 Key Concepts

Distance Formula

Between two points $ (x_1, y_1) $ and $ (x_2, y_2) $:

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

This is just the Pythagorean theorem: the difference in x is one leg, difference in y is the other leg, and the distance is the hypotenuse.

Midpoint Formula

The midpoint M of segment with endpoints $ (x_1, y_1) $ and $ (x_2, y_2) $:

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

Average the x-coordinates, average the y-coordinates.

Section Formula (Point Dividing a Segment)

If point P divides segment AB in ratio $ m:n $ (AP : PB = m : n):

\[ P = \left( \frac{nx_1 + mx_2}{m+n}, \frac{ny_1 + my_2}{m+n} \right) \]

Midpoint is the special case where m:n = 1:1.

Weighted average intuition: closer to A means more weight on A's coordinates.

Equations of Lines

Slope-intercept form: $ y = mx + b $

Point-slope form: $ y - y_1 = m(x - x_1) $

Standard form: $ Ax + By + C = 0 $

Slope between two points: $ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\Delta y}{\Delta x} $

Parallel & Perpendicular Lines

Parallel lines: same slope ($ m_1 = m_2 $)
Perpendicular lines: slopes are negative reciprocals ($ m_1 \cdot m_2 = -1 $)
Vertical line (undefined slope) is perpendicular to horizontal line (slope = 0)

Distance from Point to Line

Distance from point $ (x_0, y_0) $ to line $ Ax + By + C = 0 $:

\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

Useful for finding heights in triangles when you know coordinates and the equation of the base.

Equation of a Circle

Circle with center $ (h, k) $ and radius $ r $:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

Comes directly from the distance formula: every point on the circle is distance r from the center.

General form: $ x^2 + y^2 + Dx + Ey + F = 0 $ — complete the square to find center and radius.

Shoelace Formula (Area of Polygon)

For vertices $ (x_1,y_1), (x_2,y_2), ..., (x_n,y_n) $ in order:

\[ \text{Area} = \frac{1}{2} \left| \sum_{i=1}^{n} (x_i y_{i+1} - x_{i+1} y_i) \right| \]

Where $ (x_{n+1}, y_{n+1}) = (x_1, y_1) $.

One of the most powerful tools in coordinate geometry — plug in coordinates, get area.

"Coordinate Bashing" Strategy

When a synthetic geometry proof is elusive:

Place the figure on coordinate axes strategically (put a point at origin, a side on x-axis)
Assign coordinates to all known points
Find equations of lines and circles
Compute intersections, distances, slopes algebraically
Verify what you need to prove

It's called "bashing" because it's often messy but reliable. Competition problems are usually designed so the coordinates work out nicely.

⚠️ Common Mistake: In the distance formula, both differences get squared — so order doesn't matter ($x_1 - x_2$ gives the same result as $x_2 - x_1$). But for slope, sign matters! Also: the point-to-line distance formula requires the line in $Ax + By + C = 0$ form (everything on one side, zero on the other). Don't forget the absolute value — distance is always positive.

💡 Key Insight: Smart placement of coordinates makes everything easier. Put the origin at a key point, align sides with axes, use symmetry to your advantage. If there's a right angle, put it at (0,0) with legs on the axes. If there's a midpoint, center coordinates on it. Good coordinate choice can eliminate fractions and halve your algebra.

✏️ Example Problems

📝 Example 1 (Distance between two points)

What is the distance between points A(2, 3) and B(14, 8)?

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

Step 1: Find the differences.

$ \Delta x = 14 - 2 = 12 $

$ \Delta y = 8 - 3 = 5 $

Step 2: Plug into formula.

\[ d = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \]

5-12-13 Pythagorean triple! ✓

Answer: 13

📝 Example 2 (Equation of a circle)

A circle has center at (3, -2) and passes through point (7, 1).
What is the square of the radius ($r^2$)?

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

Center (h, k) = (3, -2). Point (7, 1) is on the circle.

Step 1: The radius is the distance from center to (7, 1).

$ \Delta x = 7 - 3 = 4 $

$ \Delta y = 1 - (-2) = 3 $

Step 2: Compute $r^2$ (no need to take the square root since we're asked for $r^2$).

\[ r^2 = 4^2 + 3^2 = 16 + 9 = 25 \]

Equation: $ (x-3)^2 + (y+2)^2 = 25 $

Answer: 25

📝 Example 3 (Shoelace formula — triangle area)

Find the area of a triangle with vertices at (0, 0), (6, 0), and (2, 5).

Shoelace formula: $ \text{Area} = \frac{1}{2} |x_1y_2 + x_2y_3 + x_3y_1 - y_1x_2 - y_2x_3 - y_3x_1| $

Step 1: List the points in order, repeat the first at the end.

$(0, 0) \\to (6, 0) \\to (2, 5) \\to (0, 0)$

Step 2: Sum of $x_i y_{i+1}$:

$ 0(0) + 6(5) + 2(0) = 0 + 30 + 0 = 30 $

Step 3: Sum of $y_i x_{i+1}$:

$ 0(6) + 0(2) + 5(0) = 0 + 0 + 0 = 0 $

Step 4: Take absolute difference and halve.

\[ \text{Area} = \frac{1}{2}|30 - 0| = \frac{1}{2}(30) = 15 \]

Check with base-height: Base = 6 (along x-axis from 0 to 6), height = 5 (y-coordinate of third point).

$Area = \\frac{1}{2} \\times 6 \\times 5 = 15.$ ✓

Answer: 15