How to use the Linear interpolation calculator
Enter two known points
Enter the two known pairs (x₁, y₁) and (x₂, y₂), then enter the target x. The two x-values must differ, while negative values and decimals are accepted.
Select Calculate to estimate the y-value on the straight line through the two points. Editing any input immediately removes the old answer and disables copying.
Formula and interpretation
The calculator uses y = y₁ + (x − x₁)(y₂ − y₁)/(x₂ − x₁). The fraction t = (x − x₁)/(x₂ − x₁) reports where the target lies relative to the first and second x-values.
A fraction from 0 to 1 is interpolation. A fraction below 0 or above 1 is extrapolation, so the result is clearly marked because it extends the line beyond the known interval.
Reading the result
Results show at most two decimal places for ordinary values while internal arithmetic keeps number precision. Very small or very large supported values use compact scientific notation.
Use this for a straight-line estimate between table readings. It does not fit a regression, spline, polynomial, or uploaded data series.
Worked examples
Between two readings
For (10, 20), (30, 60), and x = 15, t = 0.25 and y = 30.
Negative and decimal values
For (−2, 5.5), (2, −2.5), and x = 0, t = 0.5 and y = 1.5.
Outside the interval
For (0, 10), (5, 20), and x = 8, t = 1.6 and y = 26. This is extrapolation, not interpolation.
Uses and limits
Good uses
Use it for lookup-table values, calibration readings, or any quantity reasonably assumed to change in a straight line between two known points.
Important limits
The estimate is only as good as the linear assumption. Equal x-values cannot define y as a function of x, and values outside the interval are flagged as extrapolation rather than rejected.
Frequently asked questions
Can x₂ be less than x₁?
Yes. The two x-values may be in either order, but they cannot be equal.
Are endpoints allowed?
Yes. At x₁ the result is y₁, and at x₂ it is y₂.
Why is my result marked extrapolation?
Your target x lies outside the interval between the two known x-values.
Does this find a best-fit line?
No. It uses exactly two points and does not perform regression.
How is rounding handled?
The display uses at most two decimals; calculation keeps the available internal number precision.