Pre-Algebra
Geometry
Distance between two points (Pythagorean theorem application)
20 practice questions
0 video lessons
Theory + worked examples
Distance Between Two Points
Texas Pre-Algebra (TEKS) • Standard 8.7(D) • Geometry
Distance Between Two Points is a topic in Geometry in the Texas Essential Knowledge and Skills. It is aligned to Standard 8.7(D), which requires students to find the distance between two points in the coordinate plane.
The distance between two points comes from the Pythagorean theorem using the horizontal and vertical gaps as legs.
Theory
The distance between \((x_1,y_1)\) and \((x_2,y_2)\) is found with the Pythagorean theorem on the horizontal and vertical gaps.
\(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\).
Distance as a hypotenuse.
The distance formula.
Distance:
\[d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]
\(\Delta x\) and \(\Delta y\) are the legs.
How to find distance
- Find the horizontal gap \(\Delta x\).
- Find the vertical gap \(\Delta y\).
- Square, add, and take the root.
- That is the straight-line distance.
Example 1 — Compute
Find the distance from \((1,1)\) to \((5,4)\).
Solution
Legs \(4\) and \(3\).
| \(\sqrt{4^2+3^2}\) | \(=\) | \(\sqrt{25}=5\) |
Example 2 — Horizontal
Distance from \((2,3)\) to \((7,3)\)?
Solution
Same \(y\): just the \(x\) gap.
| \(7-2\) | \(=\) | \(5\) |
Example 3 — Vertical
Distance from \((4,1)\) to \((4,9)\)?
Solution
Same \(x\): the \(y\) gap.
| \(9-1\) | \(=\) | \(8\) |
Example 4 — General
Distance from \((0,0)\) to \((6,8)\)?
Solution
Legs \(6\) and \(8\).
| \(\sqrt{36+64}\) | \(=\) | \(10\) |
Common pitfalls
Square the differences before adding.
It is the Pythagorean theorem in disguise.
Take the square root at the end.
Frequently asked questions
What is the distance formula?
\(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\).
Where does it come from?
The Pythagorean theorem.
Distance from \((1,1)\) to \((5,4)\)?
\(5\).
Distance from \((0,0)\) to \((6,8)\)?
\(10\).
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