Linear Interpolation Formula with Worked Examples

Key Points

• The linear interpolation formula is $y = y_1 + \frac{(x - x_1)(y_2 - y_1)}{x_2 - x_1}$, estimating $y$ at an $x$ between two known points.

• The fraction $\frac{y_2-y_1}{x_2-x_1}$ is the slope of the line joining the two points, so the formula walks along that line.

• The method assumes the quantity changes at a steady rate between the points, which is an approximation rather than an exact result.

• Reading at an $x$ outside the two known points is extrapolation, not interpolation, and its error grows with distance.

• Interpolating at the midpoint gives exactly the average of the two $y$ values, which is a quick check on any implementation.

What Is the Linear Interpolation Formula?

The linear interpolation formula estimates a value lying between two known data points $(x_1,y_1)$ and $(x_2,y_2)$:

$$y = y_1 + \frac{(x - x_1)(y_2 - y_1)}{x_2 - x_1}$$

It assumes the straight line joining the two points is a reasonable stand-in for whatever happens between them.

Why the Formula Is a Slope Walk

Read the formula from left to right and it describes a journey. Start at the known height $y_1$, then add the extra height gained by moving sideways from $x_1$ to $x$.

The height gained per unit of sideways movement is the slope, $\frac{y_2-y_1}{x_2-x_1}$. Multiplying it by the sideways distance $(x-x_1)$ gives the rise over that stretch.

That is why the formula has the shape it does: a starting value plus a rate times a distance. It is the same structure as any straight-line equation, written from a point you already know.

Variable Key

Examples of the Linear Interpolation Formula

Example 1

Given $(10, 25)$ and $(20, 45)$, estimate $y$ at $x = 14$.

$$y = 25 + \frac{(14-10)(45-25)}{20-10} = 25 + \frac{4 \times 20}{10} = 25 + 8 = 33$$

Final answer: 33

Example 2

Given the same points, estimate $y$ at $x = 14$ by averaging the two $y$ values.

Averaging gives $\frac{25+45}{2} = 35$, which is at least in the right region.

But 35 is the value at the midpoint $x = 15$, not at $x = 14$. The target sits four tenths of the way along the interval rather than halfway, so the estimate must be lower than the average.

Weighting by position instead of averaging:

$$y = 25 + 0.4 \times 20 = 33$$

Final answer: 33. Averaging is only correct when the target is exactly midway, which is the check in the Key Points.

Example 3

A table gives $\sqrt{16} = 4$ and $\sqrt{25} = 5$. Estimate $\sqrt{20}$ by interpolation.

$$y = 4 + \frac{(20-16)(5-4)}{25-16} = 4 + \frac{4}{9} \approx 4.444$$

Final answer: about 4.444, against a true value of about 4.472.

The estimate is low by 0.028 because the square-root curve bends, and interpolation assumed a straight line. That gap is the method's error, not an arithmetic slip.

Example 4

Estimate $y$ at $x = 15$ for the points in Example 1, and check the midpoint property.

$$y = 25 + \frac{(15-10)(20)}{10} = 25 + 10 = 35$$

Final answer: 35, which equals $\frac{25+45}{2}$ exactly, confirming the midpoint check.

Example 5

Given $(10,25)$ and $(20,45)$, estimate $y$ at $x = 30$ and comment.

$$y = 25 + \frac{(30-10)(20)}{10} = 25 + 40 = 65$$

Final answer: 65, but this is extrapolation rather than interpolation, since 30 lies outside both known points.

The arithmetic is identical and the reliability is not. Nothing in the data supports the straight line continuing past $x = 20$.

Interpolation and Its Neighbours

The fourth row is what you reach for when linear interpolation is visibly too crude, as with the square-root example. Fitting a curve through three or more points reduces the error at the cost of more arithmetic.

Where Do Students Trip Up on Linear Interpolation?

Mistake 1: Averaging instead of weighting by position

Where it slips in: The target lies between two values, and averaging is the familiar way to land between two numbers.

Don't do this: $y = \frac{y_1+y_2}{2}$ for any $x$ in the interval.

The correct way: Weight by how far along the interval $x$ sits. Averaging is the special case where that fraction is exactly one half.

Students who first work out what fraction of the way along the target sits, as a decimal, before touching the $y$ values get this right consistently. That fraction is the whole of the interpolation, and the rest is multiplication.

Mistake 2: Extrapolating without saying so

Where it slips in: The formula accepts any $x$ and returns a number without complaint.

Don't do this: Report an estimate at $x = 30$ from data spanning 10 to 20 as though it were interpolated.

The correct way: Check that $x$ lies between $x_1$ and $x_2$. If it does not, the result is an extrapolation and should be labelled as one.

Mistake 3: Subtracting the coordinates in inconsistent order

Where it slips in: The formula contains three subtractions, and reversing one flips a sign.

Don't do this: Use $x - x_1$ in the numerator but $x_1 - x_2$ in the denominator.

The correct way: Keep point 1 first in every subtraction. A sign error makes the estimate fall outside the two known $y$ values, which is an impossible result for an interpolation.

Bottom Line

• The linear interpolation formula is $y = y_1 + \frac{(x-x_1)(y_2-y_1)}{x_2-x_1}$.

• It is a starting value plus a slope times a sideways distance, which is why it reads as a walk along a line.

• The method assumes a steady rate between the points, so a curving relationship leaves a predictable error.

• An $x$ outside the known points makes the result an extrapolation, which should always be labelled.

• To work through graphs and rates with a teacher, explore an algebra tutor or math classes online.

Where to Go From Here

Given $(2, 7)$ and $(8, 25)$, estimate $y$ at $x = 5$ and confirm it equals the average of the two $y$ values. Then estimate at $x = 4$ and check your answer lies between 7 and 25.

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