TheoremProved

Euclidean Distance Formula in the Plane

Statement

For any two points $A(x_1,y_1)$ and $B(x_2,y_2)$ in the Cartesian plane, the distance between them is $AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ .

Why is it true?

Dropping lines parallel to the axes from $A$ and $B$ forms a right triangle whose legs have lengths $|x_2-x_1|$ and $|y_2-y_1|$ , so the hypotenuse $AB$ follows directly from the Pythagorean theorem.

Proof sketch

Introduce the auxiliary point $C(x_2,y_1)$ , which shares the second coordinate $y_1$ with $A(x_1,y_1)$ and the first coordinate $x_2$ with $B(x_2,y_2)$ .

The segment $AC$ is parallel to the first axis with length $AC=|x_2-x_1|$ , while the segment $CB$ is parallel to the second axis with length $CB=|y_2-y_1|$ . Since the two coordinate axes are perpendicular, $\angle ACB=90^\circ$ .

Applying the Pythagorean theorem to the right triangle $\triangle ACB$ gives $AB^2=AC^2+CB^2=|x_2-x_1|^2+|y_2-y_1|^2=(x_2-x_1)^2+(y_2-y_1)^2$ . Taking the nonnegative square root yields $AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ (which also holds when $x_1=x_2$ or $y_1=y_2$ ).

Topics that use this theorem

Coordinate methods in the plane

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

René Descartes (trans. David Eugene Smith, Marcia L. Latham) (1954). The Geometry of René Descartes
H. S. M. Coxeter (1969). Introduction to Geometry