How to Find the Third Point of a Right Triangle

If you know two vertices of a right triangle, there is generally no unique formula for the third point until you know where the right angle is and have another measurement or condition. For example, a side length or an angle can narrow the possibilities to one or two points.

Let the known vertices be $A=(x_1,y_1)$ and $B=(x_2,y_2)$. The formulas below show what the right-angle condition tells you in the two main cases.

Right angle at a known point

Suppose the right angle is at $A$. Then the sides $AB$ and $AC$ are perpendicular, so the vector from $A$ to the missing point $C$ must be perpendicular to the vector from $A$ to $B$.

Set $d=B-A=(x_2-x_1,y_2-y_1)$. A perpendicular vector is $(-d_y,d_x)$, so every possible point $C$ has the form

$$C=A+t(-d_y,d_x),$$

where $t$ is any nonzero real number. The value $t=0$ would give $C=A$, which is not a triangle. Without another condition, there are infinitely many possible values of $t$.

If you know the length $AC=L$, then $|AC|=|t|\,|AB|$, giving

$$t=\pm\frac{L}{|AB|}.$$

The two signs put $C$ on opposite sides of line $AB$. If the triangle must lie on a particular side of that line, choose the corresponding sign.

Right angle at the missing point

If $C$ is the right-angle vertex, then $CA$ and $CB$ must be perpendicular. In coordinate form, this means

$$(C-A)\mathbin{\cdot}(C-B)=0.$$

Geometrically, every such point $C$ lies on the circle whose diameter is $AB$ (except the endpoints, which would make a degenerate triangle). The circle has center $(A+B)/2$ and radius $|AB|/2$, so you can write

$$C=\frac{A+B}{2}+\frac{|B-A|}{2}(\cos\theta,\sin\theta),$$

for any angle $\theta$, excluding values that give $C=A$ or $C=B$. This still gives infinitely many points unless you also know a side length, a coordinate, or another constraint.

Worked example

Take $A=(0,0)$ and $B=(4,0)$. The segment $AB$ is horizontal and has length $4$.

If the right angle is at $A$ and $AC=3$, then $d=(4,0)$ and a perpendicular direction is $(0,4)$. The formula gives

$$C=(0,0)+t(0,4).$$

Since $|AC|=4|t|=3$, we have $t=\pm 3/4$. Therefore the two possible points are

$$C=(0,3) \quad\text{or}\quad C=(0,-3).$$

Each makes a right triangle with legs of lengths $4$ and $3$. If the hypotenuse were specified as $5$ instead, the Pythagorean theorem would give $AC=\sqrt{5^2-4^2}=3$, producing the same two choices.

If the right angle is at $C$ instead, the circle has center $(2,0)$ and radius $2$. One possible point is $C=(2,2)$. Indeed,

$$(C-A)\mathbin{\cdot}(C-B)=(2,2)\mathbin{\cdot}(-2,2)=-4+4=0.$$

So the angle at $C$ is $90^\circ$. There are many other choices on the same circle, which shows why the right-angle condition alone does not give a unique answer.

What extra information do you need?

Common useful information includes:

• The length of one of the other sides.
• The hypotenuse length, if the known segment is a leg.
• One acute angle.
• A coordinate or a line on which the missing point must lie.
• Which side of the known segment the triangle must occupy.

For instance, when the right angle is at a known point, a side length fixes the distance along the perpendicular line, but usually leaves two mirror-image choices. A condition such as “above the segment” selects one of them. If the right angle is at the missing point, knowing just the hypotenuse does not help when $AB$ is already the hypotenuse: its length fixes the circle, not the location on it. You need another condition to select a point on that circle.

Common mistakes

• Assuming two points determine a triangle. They determine a segment, not the third vertex. The right-angle condition still allows infinitely many locations in either case.
• Using the wrong perpendicular. If the right angle is at $A$, use a direction perpendicular to $B-A$. If it is at $C$, use the dot-product condition involving both known points.
• Forgetting the two sides of a line. A perpendicular direction has two orientations, and a fixed side length commonly yields two reflected points.
• Mixing up the hypotenuse and a leg. The hypotenuse is opposite the right angle. Apply the Pythagorean theorem only with the correct side roles.

To find a particular third point, label the right-angle vertex first, then use the corresponding formula and the additional measurement or constraint to determine the remaining choice.