Line: Distance / Midpoint / Slope

Overview

The Line Geometry tool takes two points in the plane and reports their distance, midpoint, slope and the equation of the line through them in slope-intercept form. Enter two (x, y) pairs and you get every basic property you might need for an analytic-geometry problem.

It is built for students working through coordinate geometry, GIS analysts measuring straight-line distances between markers and game developers checking projectile paths. Each value falls out of a one-line formula but they're easy to mix up under time pressure.

How it works

For points P1 = (x1, y1) and P2 = (x2, y2):

• Distance: d = sqrt((x2 - x1)^2 + (y2 - y1)^2) (Pythagoras).
• Midpoint: ((x1 + x2) / 2, (y1 + y2) / 2).
• Slope: m = (y2 - y1) / (x2 - x1), undefined when the line is vertical.
• Equation: y = m * x + b where b = y1 - m * x1. For vertical lines, the form is x = x1.

Examples

P1 = (0, 0), P2 = (3, 4)
   →  distance 5, midpoint (1.5, 2), slope 4/3
   →  equation y = (4/3)x

P1 = (1, 2), P2 = (5, 10)
   →  distance ≈ 8.944, midpoint (3, 6), slope 2
   →  equation y = 2x

P1 = (2, 3), P2 = (2, 9)
   →  vertical line, equation x = 2

P1 = (-1, 4), P2 = (3, -4)
   →  slope -2, equation y = -2x + 2

FAQ

What if the two points are the same?

Distance is zero, slope is undefined and no unique line passes through a single point.

Why is the slope undefined for vertical lines?

The denominator x2 - x1 is zero. The line still exists but in the form x = constant, which has no slope-intercept representation.

Does the order of the two points matter?

For distance and midpoint, no. For slope, swapping the points gives the same value because both differences flip sign.

Can I get the perpendicular line?

Yes — once you have slope m, the perpendicular passing through the midpoint has slope -1 / m. Build the equation with the new slope and the same intercept formula.

Is the distance formula tied to the Pythagorean theorem?

It's a direct application. The horizontal and vertical legs are x2 - x1 and y2 - y1, and the hypotenuse is the distance.

Try Line: Distance / Midpoint / Slope