Find the exact midpoint $M$ of a line segment connecting two endpoints, $P_1$ and $P_2$. The midpoint is calculated by finding the average of the coordinates for each dimension ($x, y$, and optionally $z$).
Endpoints $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$
Point
X-Coordinate
Y-Coordinate
Z-Coordinate (Optional)
P₁
P₂
Results
Midpoint Coordinates (M)
Segment Length (Distance)
Step-by-Step Calculation
The Midpoint Formula (2D and 3D)
The midpoint formula is a direct application of the arithmetic mean. For each dimension, the midpoint coordinate is simply the average of the two endpoints' coordinates.
Midpoint Formula (2 Dimensions)
Given endpoints $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$, the midpoint $M$ is:
The calculation principle remains identical: find the average position along each axis.
Related Formula: Distance Between Two Points
While the midpoint finds the location of the center of the segment, the distance formula finds the length of the segment ($D$). The 2D distance formula is:
$$D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Frequently Asked Questions (FAQ)
What is the midpoint formula?
The midpoint formula is the method used to find the coordinates of the point that lies exactly halfway between two given points, $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$. The formula is $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.
Is the midpoint formula just the average?
Yes, the midpoint formula is essentially finding the arithmetic mean (average) of the x-coordinates and the y-coordinates separately. Since the midpoint is equidistant from both endpoints, finding the average position gives you the exact middle point.
Can the midpoint formula be used in 3D (three dimensions)?
Yes. In three dimensions, you simply extend the formula to include the z-coordinate: $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right)$
How do I find a missing endpoint if I know the midpoint?
You can rearrange the formula to solve for a missing coordinate. For the x-coordinate, the formula is: $x_2 = 2M_x - x_1$. This works because the midpoint ($M_x$) is halfway between $x_1$ and $x_2$.