Free · Mathematical relationships

Linear Interpolation through Two Points

Two points with distinct x coordinates define a line. Interpolation uses its value between the points; outside that interval the same calculation is extrapolation.

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How to use this calculator

  1. Enter the known values in the units shown. Results update as you type.
  2. Where results are editable, change one to solve backwards. Lock a value to hold it fixed.
  3. Use the worked example to check the method. Reset restores the starting fields.

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Formula and method

With x1 = 1, y1 = 3, x2 = 5, y2 = 11, x3 = 3, slope = 2, intercept = 1, y3 = 7.

m=(y₂−y₁)/(x₂−x₁); y=mx+b

Worked example

Enter these known values and leave the other values blank.

x₁
1
y₁
3
x₂
5
y₂
11
x₃
3
y₃
7
m
2
b
1

Assumptions and limitations

  • All x coordinates must be distinct in this workflow. A horizontal line evaluates forward at a supplied x but does not determine an unknown x.
  • Two points with distinct x coordinates define a line. Interpolation uses its value between the points; outside that interval the same calculation is extrapolation.
  • Enter enough independent values and clear the quantity being solved. Locks preserve known values; contradictory inputs can make the system inconsistent. Unit controls convert physical quantities rather than merely changing their labels.
  • Displayed precision is numerical rounding, not an estimate of measurement uncertainty.

Common questions

How do I solve for a missing quantity?

Clear the missing value, enter the independent known values and retain compatible units. A nonunique inverse may require another known value.

What relationship is used?

m=(y₂−y₁)/(x₂−x₁); y=mx+b

References

A horizontal line can be evaluated at a supplied x coordinate. Equal y values cannot determine an unknown x, but they do not invalidate consistent forward interpolation.

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