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Instrument MI-01-337 · Mathematics

Linear Interpolation Calculator

Two known points and an x that falls between them: this sheet returns the y a straight line drawn through those points would give, with the fraction shown at every step.

Instrument MI-01-337
Sheet 1 OF 1
Rev A
Verified
Type 05 — Algebra SER. 2026-01337

Interpolated y

30.00000000

y = y₁ + (x−x₁)(y₂−y₁) ⁄ (x₂−x₁)

The working Every figure verified twice
  1. y = 0 + (3 − 0)·(100 − 0) ⁄ (10 − 0) = 30.00000000
Worksheet log
  1. No entries yet — change an input to log a scenario.

How this instrument works

Linear interpolation estimates an unknown y-value at some x sitting between two already-known points, (x₁, y₁) and (x₂, y₂), by assuming the straight line joining them tells the truth in between. The engine computes one fraction and applies it twice: how far x has moved from x₁ toward x₂, as a share of the whole gap, is the same share applied to the gap between y₁ and y₂. Move 30% of the way across in x and the formula hands back a y that has moved exactly 30% of the way across too, no more and no less.

The formula is the two-point slope rearranged to solve for a missing y instead of a rate: multiply the fractional position t = (x − x₁) ⁄ (x₂ − x₁) by the total rise (y₂ − y₁), then add that back onto y₁. Long before calculators existed, sailors, surveyors, and astronomers ran this exact arithmetic by hand to read values between the printed rows of logarithm and trigonometric tables, a practice old enough that the word 'interpolation' traces to the Latin for smoothing a value in among ones already known.

Two boundary checks sit built into the shape of the formula. Set x equal to x₁ and the fraction collapses to zero, returning y₁ untouched; set it equal to x₂ and the fraction becomes exactly 1, returning y₂ untouched in turn. Push x past either boundary and the arithmetic keeps running, the fraction simply exceeding 1 or dropping below 0, but the result is now an extrapolation riding on a straight-line assumption that may not hold once the true relationship curves away from the two points that justified it.

y=y1+(xx1)(y2y1)x2x1y = y_1 + \dfrac{(x - x_1)(y_2 - y_1)}{x_2 - x_1}t=xx1x2x1t = \dfrac{x - x_1}{x_2 - x_1}y=y1+t(y2y1)y = y_1 + t\,(y_2 - y_1)
x₁, y₁ — the first known point · x₂, y₂ — the second known point · x — the input to interpolate at · t — the fraction of the way from x₁ to x₂ that x has travelled · y — the interpolated result, the straight line's height at x.
  • Enter the first known point's coordinates into x₁ and y₁.
  • Enter the second known point into x₂ and y₂ — together the two points fix the straight line the formula assumes.
  • Enter the value you need a reading for into x (to interpolate at); it is most reliable sitting between x₁ and x₂.
  • Read Interpolated y — the height that same straight line reaches at your chosen x, with the fraction and rise shown in the working.
  • If x₁ and x₂ come out equal, the sheet flags it: a single x cannot fix a line's direction.

Worked example — a thermometer calibrated from 0 to 10

A shop thermometer is calibrated at two points: a raw sensor reading of x₁ = 0 lines up with y₁ = 0°C, the ice point, and a raw reading of x₂ = 10 lines up with y₂ = 100°C, the boiling point. A customer's unit comes back reading x = 3. The fraction travelled is t = (3 − 0) ⁄ (10 − 0) = 0.3, so the temperature is y = 0 + 0.3 × (100 − 0) = 30.0°C — three-tenths of the way from ice point to boiling point lands exactly three-tenths of the way up the scale.

The straight line drawn through those two calibration points is only ever a working assumption in between them; a real sensor can bow slightly off a straight response, which is why a lab thermometer gets checked against more than two fixed points before anyone trusts it fully. Here, though, that assumption is the calibration itself: reading a raw 3 back as 30.0°C is the entire point of running the two known anchors through this formula in the first place.

Questions

What assumption does linear interpolation make about the data?

It assumes the true value changes at a constant rate between the two known points, so a straight line through (x₁, y₁) and (x₂, y₂) tells the truth everywhere in between. That assumption is exact for genuinely linear data and only ever approximate for anything that curves, which is why interpolation works best when the two known points sit close together.

How is this different from just calculating a slope?

A slope calculation reports the rate of change itself, rise over run. Linear interpolation reuses that identical ratio internally but asks a different question: given the rate implied by two points, what y-value sits at some third x in between them? The slope is a byproduct of the arithmetic here, not the figure being asked for.

What happens if x falls outside the range between x₁ and x₂?

The formula keeps computing — the fraction t simply climbs above 1 or drops below 0 — but the result is now an extrapolation, not an interpolation. Extrapolating trusts the straight-line assumption well past the two points that justified it, and the error tends to grow the further outside that range x sits, especially once the true relationship curves.

Why is it called linear interpolation, and is there a curved version?

'Linear' names the straight-line assumption connecting the two points. Feed in three or more known points and fit a curve instead, and the result is quadratic or cubic interpolation, or a spline stitched from several curved pieces — methods built to catch curvature that two points alone cannot reveal, at the cost of needing more known data to start from.

What's the most common mistake when using this formula by hand?

Mismatching which y goes with which x: swapping y₁ and y₂ while leaving x₁ and x₂ in their original order flips the sign of the rise and silently produces the wrong answer. Keep each point's x and y paired exactly as measured, and check that x₁ is genuinely the point you mean to measure the fraction from.

References