{"id":63363,"date":"2026-06-09T17:57:32","date_gmt":"2026-06-09T09:57:32","guid":{"rendered":"https:\/\/blog-admin.thethinkacademy.com\/?p=63363"},"modified":"2026-06-09T17:57:35","modified_gmt":"2026-06-09T09:57:35","slug":"linear-interpolation-formula","status":"publish","type":"post","link":"https:\/\/blog-admin.thethinkacademy.com\/blog\/2026\/06\/09\/linear-interpolation-formula\/","title":{"rendered":"Linear Interpolation Formula: Definition, Steps and Examples (Grade 9\u201310 and Pascal Contest)"},"content":{"rendered":"\n

The linear interpolation formula is one of those techniques that sounds more complicated than it is. At its core, it answers one question: given two known points on a line, what is the value at a point between them? The formula is a single equation, and once you understand where it comes from, you can apply it to tables of data, graph readings, and the kind of multi-step reasoning problems that appear in the Pascal Contest<\/a> and senior Ontario math.<\/p>\n\n\n\n

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What is linear interpolation?<\/h2>\n\n\n\n

Linear interpolation is a method for estimating an unknown value that sits between two known values, by assuming the relationship between them is linear \u2014 that is, that the values change at a constant rate.<\/p>\n\n\n\n

If you have two data points and you want to find the value at an x-coordinate that falls between them, linear interpolation draws a straight line connecting those two points and reads off the y-value at the x-coordinate you want.<\/p>\n\n\n\n

It is called ‘linear’ because it assumes a straight-line relationship. It is called ‘interpolation’ because you are estimating within a range of known values, not beyond them (that would be extrapolation).<\/p>\n\n\n\n

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The linear interpolation formula<\/h2>\n\n\n\n

The standard linear interpolation formula is:<\/p>\n\n\n\n

y = y\u2081 + (x \u2212 x\u2081) \u00d7 (y\u2082 \u2212 y\u2081) \/ (x\u2082 \u2212 x\u2081)<\/strong><\/p>\n\n\n\n

Where:<\/p>\n\n\n\n

Symbol<\/th>Meaning<\/th><\/tr><\/thead>
(x\u2081, y\u2081)<\/td>The first known point<\/td><\/tr>
(x\u2082, y\u2082)<\/td>The second known point<\/td><\/tr>
x<\/td>The x-value at which you want to estimate y<\/td><\/tr>
y<\/td>The estimated (interpolated) value<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

This formula is the equation of the straight line connecting (x\u2081, y\u2081) and (x\u2082, y\u2082), evaluated at x.<\/p>\n\n\n\n

Where the formula comes from<\/h3>\n\n\n\n

The fraction (y\u2082 \u2212 y\u2081) \/ (x\u2082 \u2212 x\u2081) is simply the slope of the line between the two known points \u2014 the same slope formula used throughout Grade 9 and 10 Ontario math<\/a>. Multiplying that slope by (x \u2212 x\u2081) gives the vertical distance you have travelled from y\u2081. Adding y\u2081 gives the full estimated value.<\/p>\n\n\n\n

In other words, linear interpolation is slope reasoning applied to estimation. Students who are confident with slope-intercept form<\/a> and linear relations will find the formula straightforward.<\/p>\n\n\n

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\"Linear<\/figure>\n<\/div>\n\n\n
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Step-by-step examples<\/h2>\n\n\n\n

Example 1: Basic interpolation<\/h3>\n\n\n\n

Given two points (2, 4) and (6, 12), estimate y when x = 4.<\/p>\n\n\n\n

Step 1:<\/strong> Identify the values. x\u2081 = 2, y\u2081 = 4, x\u2082 = 6, y\u2082 = 12, x = 4<\/p>\n\n\n\n

Step 2:<\/strong> Apply the formula. y = 4 + (4 \u2212 2) \u00d7 (12 \u2212 4) \/ (6 \u2212 2) y = 4 + 2 \u00d7 8 \/ 4 y = 4 + 4 y = 8<\/strong><\/p>\n\n\n\n

Check:<\/strong> The midpoint of x = 2 and x = 6 is x = 4. The midpoint of y = 4 and y = 12 is y = 8. The answer is consistent.<\/p>\n\n\n\n

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Example 2: Non-midpoint interpolation<\/h3>\n\n\n\n

Given two points (10, 20) and (15, 35), estimate y when x = 12.<\/p>\n\n\n\n

Step 1:<\/strong> Identify the values. x\u2081 = 10, y\u2081 = 20, x\u2082 = 15, y\u2082 = 35, x = 12<\/p>\n\n\n\n

Step 2:<\/strong> Apply the formula. y = 20 + (12 \u2212 10) \u00d7 (35 \u2212 20) \/ (15 \u2212 10) y = 20 + 2 \u00d7 15 \/ 5 y = 20 + 6 y = 26<\/strong><\/p>\n\n\n\n

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Example 3: Reading from a data table<\/h3>\n\n\n\n

A scientist records the temperature of a substance at two times:<\/p>\n\n\n\n

Time (minutes)<\/th>Temperature (\u00b0C)<\/th><\/tr><\/thead>
3<\/td>45<\/td><\/tr>
7<\/td>65<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Estimate the temperature at t = 5 minutes.<\/p>\n\n\n\n

x\u2081 = 3, y\u2081 = 45, x\u2082 = 7, y\u2082 = 65, x = 5<\/p>\n\n\n\n

y = 45 + (5 \u2212 3) \u00d7 (65 \u2212 45) \/ (7 \u2212 3) y = 45 + 2 \u00d7 20 \/ 4 y = 45 + 10 y = 55\u00b0C<\/strong><\/p>\n\n\n\n

This type of problem \u2014 reading a value from a table where the exact point is not listed \u2014 appears frequently in data-heavy math problems at the Grade 9\u201310 level.<\/p>\n\n\n

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\"Data<\/figure>\n<\/div>\n\n\n
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Example 4: Pascal-style reasoning problem<\/h3>\n\n\n\n

A sequence of values is recorded as follows:<\/p>\n\n\n\n

x<\/th>y<\/th><\/tr><\/thead>
0<\/td>3<\/td><\/tr>
4<\/td>11<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

If the relationship is linear, what is the value of y when x = 1?<\/p>\n\n\n\n

x\u2081 = 0, y\u2081 = 3, x\u2082 = 4, y\u2082 = 11, x = 1<\/p>\n\n\n\n

y = 3 + (1 \u2212 0) \u00d7 (11 \u2212 3) \/ (4 \u2212 0) y = 3 + 1 \u00d7 8 \/ 4 y = 3 + 2 y = 5<\/strong><\/p>\n\n\n\n

This can also be solved by identifying the slope (8\/4 = 2) and the y-intercept (3), giving y = 2x + 3. Linear interpolation and slope-intercept form produce the same answer when the relationship is truly linear \u2014 they are two ways of expressing the same calculation.<\/p>\n\n\n\n

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Linear interpolation vs extrapolation<\/h2>\n\n\n\n
<\/th>Interpolation<\/th>Extrapolation<\/th><\/tr><\/thead>
Where<\/td>Between two known points<\/td>Beyond the range of known data<\/td><\/tr>
Reliability<\/td>More reliable<\/td>Less reliable (assumes the pattern continues)<\/td><\/tr>
Example<\/td>Estimating y at x = 4 when you know x = 2 and x = 6<\/td>Estimating y at x = 10 when you only know x = 2 and x = 6<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Linear interpolation is only accurate when the relationship between the two points is genuinely linear, or approximately so. For curves or data that changes rapidly, interpolation can give a misleading estimate.<\/p>\n\n\n\n

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How linear interpolation connects to Grade 9\u201310 Ontario math<\/h2>\n\n\n\n

Linear interpolation draws directly on skills from the Ontario Grade 9 MTH1W<\/a> and Grade 10 MPM2D<\/a> curriculum:<\/p>\n\n\n\n

Slope and rate of change (MTH1W Algebra strand):<\/strong> The (y\u2082 \u2212 y\u2081) \/ (x\u2082 \u2212 x\u2081) component of the interpolation formula is the slope formula. Students who understand slope as a rate of change \u2014 not just a formula to memorise \u2014 can derive linear interpolation without needing to be taught it separately.<\/p>\n\n\n\n

Linear relations and equations (MTH1W):<\/strong> The interpolation formula is the point-slope form of a line rearranged. Students who are comfortable with y = mx + b and point-slope form will recognise the structure immediately.<\/p>\n\n\n\n

Data and statistics:<\/strong> Linear interpolation is used to estimate medians, quartiles, and percentiles from grouped frequency tables \u2014 a technique that appears in statistics units from Grade 9 upward.<\/p>\n\n\n\n

Problem-solving with tables:<\/strong> Many data problems present values at specific intervals and ask students to estimate values between them. Linear interpolation is the formal method for doing this accurately.<\/p>\n\n\n\n

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Linear interpolation and the Pascal Contest<\/h2>\n\n\n
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\"Grade<\/figure>\n<\/div>\n\n\n

The Pascal Contest<\/strong>, run by the University of Waterloo’s Centre for Education in Mathematics and Computing (CEMC), is written by Grade 10 students (and strong Grade 9 students). It tests logical thinking and mathematical problem-solving rather than content recall directly.<\/p>\n\n\n\n

Linear interpolation does not appear by name on the Pascal Contest. However, the reasoning it embodies \u2014 slope, linear relationships, reading data at non-listed values, and working between known quantities \u2014 is central to how Pascal problems are structured.<\/p>\n\n\n\n

A student who can confidently:<\/p>\n\n\n\n