Lesson 1 of 10, 11 min read
The coordinate plane, distance and midpoint
Coordinates and quadrants, the distance formula, the midpoint formula, and using them to classify triangles and find missing endpoints.
Turning geometry into algebra
René Descartes had a simple, powerful idea: describe every point with a pair of numbers, and every shape with an equation. Geometry problems then become algebra problems.
Two perpendicular number lines, the \(x\)-axis (horizontal) and the \(y\)-axis (vertical), cross at the origin \((0, 0)\). Every point is an ordered pair \((x, y)\): first how far to move horizontally, then vertically.
The axes split the plane into four quadrants, numbered counterclockwise starting from the upper right:
| Quadrant | Sign of \(x\) | Sign of \(y\) |
|---|---|---|
| I | \(+\) | \(+\) |
| II | \(-\) | \(+\) |
| III | \(-\) | \(-\) |
| IV | \(+\) | \(-\) |
Points on the axes don't belong to any quadrant.
The distance formula
Between \((x_1, y_1)\) and \((x_2, y_2)\), draw a right triangle with horizontal and vertical legs. The legs have lengths \(|x_2 - x_1|\) and \(|y_2 - y_1|\), and the segment you want is the hypotenuse. The Pythagorean theorem gives:
The midpoint formula
Classifying a triangle with coordinates
To check for a right triangle, compare \(c^2\) with \(a^2 + b^2\) using the squared distances, so you can skip the square roots.
Practice
Quick check
What is the distance between \((1, 1)\) and \((4, 5)\)?
\(\sqrt{3^2 + 4^2} = \sqrt{25} = 5\).
Quick check
What is the midpoint of the segment from \((-4, 6)\) to \((2, -2)\)?
\(\left(\frac{-4 + 2}{2}, \frac{6 - 2}{2}\right) = (-1, 2)\).
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