Finding the fourth-quadrant vertex of an equilateral triangle

Key takeaway: Equilateral triangle coordinates are determined by combining distance geometry with a 60-degree rotation.

Key idea

The points $A(0,0)$ and $B(4,0)$ form a horizontal base of length 4. In an equilateral triangle, all three sides are equal, so the third vertex must be exactly 4 units from both $A$ and $B$. Because the base lies on the $x$-axis and the third vertex is in the fourth quadrant, the point must be below the axis.

For a segment with endpoints $(0,0)$ and $(4,0)$, the midpoint is $(2,0)$. The third vertex of an equilateral triangle lies directly above or below the midpoint on the perpendicular bisector. Since we need the fourth quadrant, we choose the lower point.

Method

The height of an equilateral triangle with side length $s$ is $\frac{\sqrt{3}}{2}s$. Here, $s=4$, so the height is

$h=\frac{\sqrt{3}}{2}\cdot 4=2\sqrt{3}.$

That means the third vertex is 2 units left and right balanced around the midpoint, and $2\sqrt{3}$ units vertically from the base line. The lower vertex is therefore

$C=(2,-2\sqrt{3}).$

You can also verify this by checking distances:

$CA=\sqrt{(2-0)^2+(-2\sqrt{3}-0)^2}=\sqrt{4+12}=4,$

and

$CB=\sqrt{(2-4)^2+(-2\sqrt{3}-0)^2}=\sqrt{4+12}=4.$

Both sides match the base length, so the triangle is equilateral.

Why the quadrant matters

The same base produces two possible equilateral vertices, one above the $x$-axis and one below it. The point in the first quadrant would be $(2,2\sqrt{3})$, while the point in the fourth quadrant is $(2,-2\sqrt{3})$. The quadrant condition removes the ambiguity.

Final answer

The coordinates of the third vertex are $\boxed{(2,-2\sqrt{3})}$.

Common mistake

A frequent error is to place the third vertex at $(2,-\sqrt{3})$ or $(2,-4\sqrt{3})$ by mixing up the height formula for an equilateral triangle. Another mistake is to forget that the point must be centered over the midpoint of the base, not over one endpoint. The base length is 4, so the altitude is $2\sqrt{3}$, not 4. The side lengths must also be checked; if the distances to $A$ and $B$ are not both 4, the point is not correct.

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Pitfalls the pros know 👇 A common trap with $A(0,0)$ and $B(4,0)$ is to treat the third vertex as if only the height matters. That can lead to answers like $(0,-2\sqrt{3})$ or $(4,-2\sqrt{3})$, which are not centered over the base and do not give equal side lengths. Another mistake is forgetting that there are two equilateral positions, one on each side of the segment. The phrase “fourth quadrant” is the key filter: it tells you to choose the vertex below the $x$-axis. If you compute the height correctly but place it at the wrong horizontal location, the point will fail the distance check from one endpoint.

What if the problem changes? If the same base had third vertex $C(x,y)$ in the first quadrant instead of the fourth quadrant, the geometry would be almost identical, but the sign of the $y$-coordinate would change. With $A(0,0)$ and $B(4,0)$, the midpoint is still $(2,0)$ and the altitude is still $2\sqrt{3}$. The only valid first-quadrant vertex would be $C=(2,2\sqrt{3})$. If the base were changed to $A(0,0)$ and $B(6,0)$, then the side length would be 6 and the altitude would become $3\sqrt{3}$, giving vertices $(3,\pm 3\sqrt{3})$. This shows how the midpoint and equilateral-triangle altitude control the coordinates.

Tags: equilateral triangle coordinates, perpendicular bisector, triangle altitude formula