Enter triangle vertices to compute the Euler line and verify center collinearity
Vertex A x₁
Vertex A y₁
Vertex B x₂
Vertex B y₂
Vertex C x₃
Vertex C y₃
Result
Centroid G
-
Circumcenter O
-
Orthocenter H
-
Nine-Point Center N (midpoint of OH): -
Detailed Derivation
Euler Line Properties
Centroid G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)
G divides OH in ratio OG:GH = 1:2
Nine-point center N = midpoint of OH
O, G, H are collinear (Euler Line)
The Euler line is one of the most famous results in triangle geometry. It demonstrates the elegant relationship between the centroid, circumcenter, orthocenter, and nine-point center. All lie on a single line.
⚠For equilateral triangles, all centers coincide at one point and the Euler line is undefined. For right triangles, the circumcenter is at the hypotenuse midpoint.
What Is the Euler Line?
The Euler line is a line through the centroid, circumcenter, orthocenter, and nine-point center of a triangle. Leonhard Euler discovered this collinearity in 1765. The ratios between these points are fixed: the centroid divides the line between O and H in a 1:2 ratio.
Centroid G
Average of vertices. Always lies between O and H on Euler line with OG:GH = 1:2.
Circumcenter O
Center of circumscribed circle. Intersection of perpendicular bisectors. For acute triangles it lies inside.
Orthocenter H
Intersection of altitudes. For acute triangles it lies inside. Together with O defines the Euler line direction.
Nine-Point Center
Midpoint of OH. Center of the nine-point circle which passes through nine significant triangle points.
Teaching Example: Triangle A(0,0), B(6,0), C(4,6). Centroid G = (3.33,2). Circumcenter O = (3,2.25). Orthocenter H = (4,3). Verify: slope OG = (2-2.25)/(3.33-3) = -0.75, slope GH = (3-2)/(4-3.33) = 1.49. Nine-point center N = ((3+4)/2, (2.25+3)/2) = (3.5,2.625).
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