Distance, midpoint and slope between two points, the line's equation in slope-intercept, standard and point-slope form, and where two lines meet.
By AdminDrafted with AI writing toolsUpdated Checked examples: 6
P₁(1, 2) and P₂(4, 6) are 5 apart with midpoint (2.5, 4); the line through them has slope 4/3 (inclination 53.13°).
Angle of inclination: 53.1301°, measured counterclockwise from the positive x-axis in [0°, 180°).
Slope-intercept: y = (4/3)x + 2/3 · Standard: 4x − 3y = −2 · Point-slope: y − 2 = (4/3)(x − 1)
Through M(2.5, 4) at right angles to P₁P₂: y = −0.75x + 5.875.
Enter two points to get the straight-line distance d = √((x₂ − x₁)² + (y₂ − y₁)²), the midpoint ((x₁ + x₂)/2, (y₁ + y₂)/2) and the slope m = (y₂ − y₁)/(x₂ − x₁). The line through them is written in slope-intercept form y = mx + b, in standard form Ax + By = C with whole-number coefficients and in point-slope form, alongside its perpendicular bisector. Fractions stay exact, so a slope of 4/3 is not rounded to 1.3333.
The default points (1, 2) and (4, 6) are 5 apart, a 3-4-5 triangle, with midpoint (2.5, 4) and line y = (4/3)x + 2/3. Switching on a second line adds whether the two are parallel, perpendicular or intersecting, the angle between them and the crossing point.
A vertical line has no slope, so only its standard form, such as x = 3, is given. Coordinates carry no unit: the distance is in whatever unit they use.
Slope is the change in y divided by the change in x: m = (y₂ − y₁)/(x₂ − x₁) (OpenStax, College Algebra 2e, §4.1). From (2, 3) to (6, 11), y rises 8 while x runs 4, so m = 8/4 = 2, a climb of 2 units per unit to the right. Any two points with different x-values have one.
The distance between two points is Pythagoras' theorem applied to the horizontal and vertical gaps: d = √((x₂ − x₁)² + (y₂ − y₁)²). For (2, 3) and (6, 11), d = √(4² + 8²) = √80 = 4√5 ≈ 8.94427191.
The midpoint averages each coordinate: M = ((x₁ + x₂)/2, (y₁ + y₂)/2). Here M = ((2 + 6)/2, (3 + 11)/2) = (4, 7).
Put the slope into y = mx + b and find the intercept from either point: b = y₁ − mx₁ = 3 − 2 × 2 = −1, so the line is y = 2x − 1. Substituting the other point checks it: 2 × 6 − 1 = 11.
The same line in standard form is 2x − y = 1, and in point-slope form y − 3 = 2(x − 2). Its angle of inclination, measured counterclockwise from the positive x-axis, is arctan 2 ≈ 63.43°.
| Points | Distance | Slope | Midpoint x, y |
|---|---|---|---|
| (1, 2) to (4, 6) | 5 | 1.33333333 | 2.5, 4 |
| (0, 0) to (3, 4) | 5 | 1.33333333 | 1.5, 2 |
| (0, 0) to (2, 2) | 2.82842712 | 1 | 1, 1 |
| (0, 4) to (4, 0) | 5.65685425 | -1 | 2, 2 |
| (3, 1) to (3, 7) | 6 | Undefined (vertical) | 3, 4 |
| (-2, 5) to (4, 5) | 6 | 0 | 1, 5 |
| (-3, -4) to (0, 0) | 5 | 1.33333333 | -1.5, -2 |
| (1/2, 1) to (5/2, 4) | 3.60555128 | 1.5 | 1.5, 2.5 |
From (0, 4) to (4, 0), y falls 4 while x rises 4, so the slope is −1. The first two rows share the slope 4/3 (1.33333333 rounded), and both are 3-4-5 triangles, so their distance is 5.
The vertical segment from (3, 1) to (3, 7) still has a length and a midpoint; only its slope is missing. Fractions work like any number: from (1/2, 1) to (5/2, 4), m = 3/2.
Parallel lines have equal slopes. Perpendicular lines have slopes that multiply to −1, so a line perpendicular to slope 2 has slope −1/2; a vertical and a horizontal line are perpendicular too.
The line through (0, 0) and (4, −2) has slope −1/2, so it crosses y = 2x − 1 at a right angle, at (0.4, −0.2). For other pairs, the angle between them satisfies tan θ = |(m₂ − m₁)/(1 + m₁m₂)|.
The perpendicular bisector of a segment passes through its midpoint at a right angle, and every point on it is the same distance from both ends. For (2, 3) and (6, 11) it is y = −0.5x + 9.
To see the lines plotted with other functions, use the graphing calculator. For a slope given as rise and run, angle or roof pitch, use the slope and roof pitch calculator.
Checked against: dx = 3, dy = 4 (3-4-5), m = 4/3, b = 2 − 4/3 = 2/3; bisector through (5/2, 4) with slope −3/4 has b = 47/8 = 5.875 (terminating fractions print as decimals)
Checked against: Equal x-coordinates: the line x = 3 is vertical; the bisector is the horizontal line through (3, 4)
Checked against: dx = 5, dy = −3, √34; −3x − 5y = −27/4 scaled by −4
Checked against: intersection (16/7, 26/7) by with exact fractions
Divide the change in y by the change in x: m = (y₂ − y₁)/(x₂ − x₁). From (1, 2) to (4, 6) the rise is 4 and the run is 3, so m = 4/3 ≈ 1.3333. A positive slope rises to the right, a negative one falls, 0 is horizontal, and when x₂ = x₁ the line is vertical and the slope is undefined.
d = √((x₂ − x₁)² + (y₂ − y₁)²), which is Pythagoras' theorem applied to the horizontal and vertical gaps. Between (1, 2) and (4, 6) the gaps are 3 and 4, so d = √25 = 5. For points given as latitude and longitude use a great-circle formula instead, because the flat formula ignores the Earth's curvature.
Find the slope m, then the intercept b = y₁ − m·x₁, and write y = mx + b. Through (1, 2) and (4, 6), m = 4/3 and b = 2 − 4/3 = 2/3, so y = (4/3)x + 2/3. Multiplying by 3 and rearranging gives the standard form 4x − 3y = −2, with whole-number coefficients.
For nonvertical lines, equal slopes mean the lines are parallel or coincident, and slopes whose product is −1 mean they are perpendicular, such as 1/2 and −2. A vertical and a horizontal line are also perpendicular. Distinct parallel lines do not meet. For other pairs with finite slopes, tan θ = |(m₂ − m₁)/(1 + m₁m₂)|: slopes of 4/3 and −1 give tan θ = 7, so the smaller angle between the lines is 81.87°.
The line through the midpoint of a segment at right angles to it; every point on it is equally far from both endpoints. For (1, 2) and (4, 6) it passes through (2.5, 4) with slope −3/4, the negative reciprocal of 4/3, giving y = −0.75x + 5.875. The bisectors of a triangle's three sides meet at the centre of its circumscribed circle.
This calculator includes 6 worked examples with independently sourced answers. They run in the test suite, and you can run them here too.