What Is The Orthocenter of A Triangle?
The orthocenter of a triangle is the point where its three altitudes intersect, usually written as $H$. An altitude is a line segment drawn from a vertex straight down to the opposite side, meeting that side at a right angle (90°). Every triangle has exactly three altitudes, one from each vertex - and all three pass through the same single point.
That "same single point" is not obvious. There is no reason three separately-drawn perpendiculars should agree on where to cross. But they always do, for every triangle. That guaranteed meeting is called concurrency, and the orthocenter is one of a triangle's four classic centres.
How Do You Find The Orthocenter?
You never actually need all three altitudes. Because the three altitudes are guaranteed to be concurrent, finding where any two of them cross gives the orthocenter - the third altitude will pass through the same point automatically.
By hand, the steps are:
Draw one altitude - from a vertex, drop a perpendicular to the opposite side.
Draw a second altitude - from a different vertex, drop a perpendicular to its opposite side.
Mark where they cross. That intersection is the orthocenter $H$.
On the coordinate plane, you turn "perpendicular" into slopes. The altitude from a vertex is perpendicular to the opposite side, so its slope is the negative reciprocal of that side's slope. If a side has slope $m$, the altitude to it has slope $-\frac{1}{m}$. Find the equations of two altitudes and solve them together - the solution is $H$.
Where does the orthocenter sit?
Its position depends entirely on the triangle's shape:
Acute triangle (all angles < 90°) - orthocenter lies inside.
Right triangle (one 90° angle) - orthocenter sits exactly on the right-angle vertex, because the two legs are themselves altitudes.
Obtuse triangle (one angle > 90°) - orthocenter lies outside the triangle, and two of the altitudes must be extended beyond the sides to reach it.
How Is the Orthocenter Different From the Circumcenter, Incenter, and Centroid?
Every triangle has four famous centres, and each comes from a different set of concurrent lines. Mixing them up is the most common source of error, so hold the distinctions clearly.
Centre | Built from | Special property |
|---|---|---|
Orthocenter ($H$) | The three altitudes | Inside/on/outside by triangle type |
Circumcenter ($O$) | The three perpendicular bisectors of the sides | Centre of the circle through all three vertices |
Incenter ($I$) | The three angle bisectors | Centre of the circle inside, touching all three sides |
Centroid ($G$) | The three medians (vertex to midpoint of opposite side) | Balance point; divides each median $2:1$ |
Two facts make the set easier to hold together. In an equilateral triangle, all four centres land on the exact same point. And in any triangle, the orthocenter $H$, centroid $G$, and circumcenter $O$ always lie on one straight line called the Euler line, with the centroid sitting between the other two in a fixed $HG:GO = 2:1$ ratio.
Examples of the Orthocenter
Each example is worked step by step. The problem statement is in bold; the working is not.
Example 1
In which region of an acute triangle does the orthocenter lie?
An acute triangle has all three angles less than 90°.
By the position rule, the orthocenter of an acute triangle lies inside the triangle.
Final answer: inside the triangle.
Example 2
A student is asked for the orthocenter of a right triangle with the right angle at C. They start solving two altitude equations from scratch. Is there a faster way, and is the setup even right?
The instinct is to treat every triangle the same: write slopes, find negative reciprocals, solve simultaneously. That works, but for a right triangle it hides something.
Watch what the general method gives. Take the right triangle with vertices $A(0, 0)$, $B(4, 0)$, $C(0, 3)$, right angle at $C$... except the right angle is actually at $A$ here, since $AB$ lies along the x-axis and $AC$ along the y-axis, meeting at $A(0,0)$ at 90°.
That mislabelling is the trap. Before grinding through algebra, look at the shape. The two legs of a right triangle are altitudes: leg $AB$ is perpendicular to leg $AC$, so each leg is the altitude from its far vertex. Both legs meet at the right-angle vertex.
So the orthocenter sits exactly on the right-angle vertex - here, $A(0, 0)$. No simultaneous equations needed.
Final answer: The orthocenter of a right triangle is the right-angle vertex itself. Read the triangle type first; the algebra confirms what the shape already tells you.
Example 3
Find the orthocenter of the triangle with vertices A(0, 0), B(6, 0), and C(2, 4).
Find two altitudes.
Altitude from $C$ to side $AB$: side $AB$ lies along the x-axis (slope $0$), so its altitude is vertical through $C(2, 4)$: $$x = 2$$
Altitude from $A$ to side $BC$: side $BC$ goes from $B(6, 0)$ to $C(2, 4)$, slope $= \frac{4 - 0}{2 - 6} = \frac{4}{-4} = -1$. The altitude is perpendicular, slope $= +1$, through $A(0, 0)$: $$y = x$$
Solve the two together: substitute $x = 2$ into $y = x$: $$y = 2$$
Final answer: The orthocenter is at $(2, 2)$.
Example 4
A triangle has one angle of 120°. Where is its orthocenter?
A 120° angle is greater than 90°, so this is an obtuse triangle.
The orthocenter of an obtuse triangle lies outside the triangle; two of its altitudes must be extended beyond the sides to reach the meeting point.
Final answer: outside the triangle.
Example 5
Find the orthocenter of the triangle with vertices A(1, 1), B(5, 1), C(3, 5).
Altitude from $C$ to $AB$: $AB$ is horizontal (slope $0$), altitude is vertical through $C(3, 5)$: $$x = 3$$
Altitude from $A$ to $BC$: $BC$ runs from $B(5, 1)$ to $C(3, 5)$, slope $= \frac{5 - 1}{3 - 5} = \frac{4}{-2} = -2$. Perpendicular slope $= \frac{1}{2}$, through $A(1, 1)$: $$y - 1 = \tfrac{1}{2}(x - 1)$$
Substitute $x = 3$: $$y - 1 = \tfrac{1}{2}(3 - 1) = 1$$ $$y = 2$$
Final answer: The orthocenter is at $(3, 2)$.
Example 6
Verify that the third altitude of the triangle in Example 5 also passes through (3, 2).
The third altitude is from $B(5, 1)$ to side $AC$. Side $AC$ runs from $A(1, 1)$ to $C(3, 5)$, slope $= \frac{5 - 1}{3 - 1} = 2$. Perpendicular slope $= -\frac{1}{2}$, through $B(5, 1)$: $$y - 1 = -\tfrac{1}{2}(x - 5)$$
Test the point $(3, 2)$: $$2 - 1 = -\tfrac{1}{2}(3 - 5)$$ $$1 = -\tfrac{1}{2}(-2) = 1 \checkmark$$
Final answer: The third altitude passes through $(3, 2)$, confirming the concurrency of all three altitudes.
Why the Orthocenter Matters — "Three lines forced to agree"
The orthocenter earns its keep as a first, clean lesson in concurrency - the idea that certain constructions in a triangle are guaranteed to meet at one point, not merely likely to. That guarantee is a theorem, not luck, and it is the reason triangle geometry is so tightly interlocked.
Where that idea does real work:
The Euler line - the discovery by the Swiss mathematician Leonhard Euler in 1765 that the orthocenter, centroid, and circumcenter always line up is one of the surprises of classical geometry. Three centres built from three completely different rules, yet always collinear.
Coordinate geometry and computer graphics - turning "the altitudes meet here" into an $(x, y)$ solution is exactly the kind of intersection-of-perpendiculars computation that rendering engines and CAD systems run constantly.
The nine-point circle - the orthocenter is one of the anchor points for a remarkable circle that passes through nine special points of any triangle, a result worth meeting later.
Carry this forward: the orthocenter is less about the point itself and more about why three altitudes can never disagree. That question opens the door to the deeper structure of the triangle.
What Are the Most Common Mistakes With the Orthocenter?
Mistake 1: Confusing the orthocenter with the centroid
Where it slips in: Whenever a question names a triangle centre without spelling out which lines build it.
Don't do this: Using medians (vertex to midpoint) when the question asks for the orthocenter.
The correct way: The orthocenter comes from altitudes (perpendicular to the opposite side); the centroid comes from medians (to the midpoint of the opposite side). The point of confusion between two close ideas is exactly here - "line from a vertex to the opposite side" describes both, and the difference is perpendicular foot versus midpoint. Read which one the construction actually uses before computing.
Mistake 2: Expecting the orthocenter to be inside every triangle
Where it slips in: Obtuse-triangle problems, where a student solves the algebra correctly but then rejects a valid answer for landing outside the triangle.
Don't do this: Assuming a coordinate answer is wrong because $H$ falls outside the shape.
The correct way: For an obtuse triangle the orthocenter genuinely lies outside - the second-guesser who "corrects" a right answer here loses marks for a point that was correct all along. Check the triangle type first: obtuse means expect an exterior orthocenter.
Mistake 3: Using the side's slope instead of the perpendicular slope
Where it slips in: Setting up altitude equations on the coordinate plane.
Don't do this: Giving the altitude the same slope as the side it drops to.
The correct way: An altitude is perpendicular to the opposite side, so its slope is the negative reciprocal of the side's slope. If the side has slope $2$, the altitude has slope $-\frac{1}{2}$. The rusher who copies the side's slope straight into the altitude equation gets a line parallel to the side - which can never be an altitude. This is the coordinate-geometry cousin of a real navigation error: a ship taking a bearing perpendicular to a coastline must turn 90° from the shore's direction, and copying the coastline's own heading (as dead-reckoning mistakes have historically shown) sends the plotted fix badly off course.
Conclusion
The orthocenter is the point where a triangle's three altitudes meet.
Find it by solving any two altitude equations; the third passes through the same point.
On coordinates, an altitude's slope is the negative reciprocal of the opposite side's slope.
Its position depends on the triangle: inside (acute), on the right-angle vertex (right), outside (obtuse).
It differs from the circumcenter (perpendicular bisectors), incenter (angle bisectors), and centroid (medians), and lies with $G$ and $O$ on the Euler line.
To take the orthocenter and triangle centres further with a teacher, explore Bhanzu's geometry tutor or high school math tutor programs, or browse math classes online.
A Practical Next Step
Work through the exercises below to solidify your understanding. Find the orthocenter of the triangle with vertices $A(0, 0)$, $B(4, 0)$, $C(1, 3)$ by writing two altitude equations. Then classify the triangle as acute, right, or obtuse and predict where $H$ should sit before you solve. If the negative-reciprocal step trips you up, return to the coordinate-method section. Want a live Bhanzu trainer to walk through more triangle-centre problems? Book a free demo class.
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