How this instrument works
An altitude runs from one vertex of a triangle straight into the opposite side, striking it at a right angle. A triangle has three altitudes, one dropped from each corner, and no matter how the triangle is stretched or skewed, the three always cross at one shared point: the orthocenter. That three lines drawn from three different vertices, by three different rules, should still meet at a single spot is the surprising part — it is a genuine theorem, provable from Ceva's concurrency condition, not a coincidence of any particular shape.
Intersecting three altitude lines directly is messy algebra, so this sheet takes a shortcut discovered by Euler: the orthocenter, the centroid, and the circumcenter all sit on one straight line, the Euler line, with the centroid always exactly one third of the way from the circumcenter to the orthocenter. That gives H = 3G − 2O — compute the centroid G as the plain average of the three vertices, borrow the circumcenter O from the same formula this site's circumcenter calculator uses, and the orthocenter falls out of a single subtraction.
One case collapses to something clean: in a right triangle, the two legs meeting at the right angle are already perpendicular to each other, so each leg is itself one of the triangle's altitudes, and their intersection — the right-angle vertex — is automatically where two of the three altitudes cross. The orthocenter of any right triangle therefore sits exactly on the right-angle corner, not merely near it. In an obtuse triangle the point does the opposite and wanders outside the triangle's own boundary entirely, past the vertex opposite the widest angle.
- Enter the first vertex in Vertex A: x and Vertex A: y, using any coordinate system you like.
- Enter the second and third vertices in Vertex B: x, Vertex B: y, Vertex C: x, and Vertex C: y — any labeling order works as long as the points aren't collinear.
- Centroid: x and Centroid: y show the plain average of the three vertices, computed on the way to the answer.
- Circumcenter: x and Circumcenter: y show the point this sheet borrows from the circumcenter formula before combining it with the centroid.
- Read the result in Orthocenter: x and Orthocenter: y — the point where all three altitudes meet.
Worked example — the 3-4-5 right triangle's orthocenter
Take the same right triangle used on the circumcenter page: A(0, 0), B(4, 0), C(0, 3), legs 4 and 3 along the axes. The centroid is G = ((0+4+0) ⁄ 3, (0+0+3) ⁄ 3) = (1.333333, 1.0), and the circumcenter — reusing that page's own result for this triangle — is O = (2.0, 1.5). The Euler-line relation gives hx = 3(1.333333) − 2(2.0) = 4.0 − 4.0 = 0.0 and hy = 3(1.0) − 2(1.5) = 3.0 − 3.0 = 0.0.
The orthocenter lands at (0.0, 0.0) — vertex A itself, exactly where the right angle sits. That is not a rounding coincidence: leg AB runs along the x-axis and leg AC runs along the y-axis, so each leg is already perpendicular to the other, which makes each leg one of the triangle's own altitudes. Their intersection is vertex A, and no separate altitude computation is needed to confirm it.
Questions
What is the orthocenter of a triangle?
It's the single point where a triangle's three altitudes cross — each altitude running from one vertex perpendicular into the line containing the opposite side. Every triangle, however skewed, has exactly one orthocenter, even though it can land inside, on an edge, or well outside the triangle itself.
How does the orthocenter relate to the centroid and circumcenter?
All three sit on a single straight line called the Euler line, with the centroid G always exactly one third of the way from the circumcenter O to the orthocenter H. That fixed ratio is what lets this sheet compute H = 3G − 2O instead of intersecting three altitude lines directly.
Why does a right triangle's orthocenter land exactly on its right-angle vertex?
Because the two legs meeting at that vertex are already perpendicular to one another, which makes each leg one of the triangle's own altitudes. Two altitudes intersecting at the right-angle vertex is enough to fix the orthocenter there — the third altitude, computed separately, passes through the same point.
Why does an obtuse triangle's orthocenter fall outside the triangle?
Once one interior angle passes 90 degrees, at least two of the altitude lines have to be extended beyond the triangle's own sides before they can meet, and the point where all three cross moves out past the vertex opposite the widest angle. Acute triangles keep their orthocenter inside; right triangles put it exactly on a vertex.
Does the order I enter the three vertices in change the result?
No. Labeling the corners A, B, and C in a different order describes the same triangle and produces the same orthocenter, since the underlying geometry depends only on the three points, not on which label is attached to which corner.
What happens if the three points I enter are collinear?
There's no triangle and no orthocenter — three points on a single line don't enclose any area, so d, twice the signed area, comes out to zero and the circumcenter step this sheet relies on has nothing to divide by. Move any one point off the line to get a real answer.