Q1. Find the slope of the line joining (1, 2) and (4, 8).
m = (y₂ − y₁)/(x₂ − x₁)
m = (8 − 2)/(4 − 1) = 6/3
m = 2
A line is fixed by a slope and a point. Compute the slope and distance between two points, then learn every form of the line equation.
Enter two points to get the slope of the line joining them and the distance between them.
A vertical line (x₁ = x₂) has an undefined slope; a horizontal line has slope 0.
Q1. Find the slope of the line joining (1, 2) and (4, 8).
m = (y₂ − y₁)/(x₂ − x₁)
m = (8 − 2)/(4 − 1) = 6/3
m = 2
| Quantity | Formula |
|---|---|
| Slope | m = (y₂−y₁)/(x₂−x₁) |
| Slope-intercept form | y = mx + c |
| Distance of a point from a line | |ax₁+by₁+c| / √(a²+b²) |
| Perpendicular lines | m₁·m₂ = −1 |
m = (y₂ − y₁)/(x₂ − x₁); it equals the tangent of the angle the line makes with the positive x-axis.
The product of their slopes is −1 (m₁·m₂ = −1).
y = mx + c, where m is the slope and c the y-intercept.
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