Distance Between Two Points
Aligned to 8.G.B.8 — Common Core State Standards for Mathematics.
What this lesson teaches
The distance between two points on the coordinate plane is found with the Pythagorean Theorem: the horizontal change and vertical change are the legs, and the distance is the hypotenuse. Distance = √((x2 − x1)^2 + (y2 − y1)^2).
Worked examples
Example 1
Find the distance between (1, 2) and (4, 6). Horizontal change = 4 − 1 = 3; vertical change = 6 − 2 = 4. Distance = √(3^2 + 4^2) = √(9 + 16) = √25 = 5.
Example 2
Distance from (−2, 1) to (3, 13): the changes are 5 and 12, so d = √(25 + 144) = √169 = 13.
Example 3
The distance formula is just the Pythagorean theorem with the horizontal and vertical changes as the legs. Drawing the right triangle makes it obvious.
Practice questions
- Find the distance between (0, 0) and (6, 8).
- Find the distance between (2, 1) and (5, 5).
- Find the distance between (−3, −2) and (1, 1).
Watch the lesson
Every lesson comes with a video taught in English and Spanish — the same video the QR code in the printed workbook opens.
▶ Watch this lessonEn español
Distancia entre Dos Puntos
La distancia entre dos puntos en el plano cartesiano se halla con el Teorema de Pitágoras: el cambio horizontal y el cambio vertical son los catetos, y la distancia es la hipotenusa. Distancia = √((x2 − x1)^2 + (y2 − y1)^2).
Ejemplo 1: Halla la distancia entre (1, 2) y (4, 6). Cambio horizontal = 4 − 1 = 3; cambio vertical = 6 − 2 = 4. Distancia = √(3^2 + 4^2) = √(9 + 16) = √25 = 5.
Ejemplo 2: Distancia de (−2, 1) a (3, 13): los cambios son 5 y 12, así que d = √(25 + 144) = √169 = 13.
Ejemplo 3: La fórmula de distancia es solo el teorema de Pitágoras con los cambios horizontal y vertical como catetos. Dibujar el triángulo rectángulo lo hace evidente.
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