Linear Interpolation- Practice Worksheet and Solutions

What Linear Interpolation Actually Is

Linear interpolation is a method to estimate values that fall between two known data points. That's it. Nothing fancy. You're drawing a straight line between two points and finding where your unknown value sits on that line.

You'll encounter this in engineering, statistics, computer graphics, and anywhere else numbers need estimating. It's a foundational skill that many people fumble because they skip the basics.

The Formula You Need to Know

Before touching any worksheet, memorize this:

y = y₁ + (x - x₁) × (y₂ - y₁) / (x₂ - x₁)

Where:

• y₁ and y₂ are your known y-values
• x₁ and x₂ are your known x-values
• x is the value you want to find the corresponding y for

Write it on a sticky note. Put it on your monitor. This formula appears in every linear interpolation problem you'll ever solve.

Practice Problems

Work through these in order. Don't skip ahead to the solutions—struggle builds understanding.

Problem 1: Temperature Estimation

A weather station recorded 68°F at 2 PM and 74°F at 4 PM. What temperature would you estimate at 3 PM?

Problem 2: Salary Interpolation

A company pays employees based on years of experience. Someone with 3 years earns $52,000. Someone with 7 years earns $68,000. What should someone with 5 years of experience earn?

Problem 3: Material Stress Test

Steel beam deflection measures 2.3 mm at 500 kg load and 8.7 mm at 2000 kg load. Estimate the deflection at 1200 kg load.

Problem 4: Financial Projection

An investment was worth $10,500 after 2 years and $14,200 after 6 years. What was the estimated value after 4 years?

Step-by-Step Solutions

Solution 1: Temperature Estimation

Given data: (2, 68) and (4, 74). Find y when x = 3.

Step 1: Identify your points. x₁ = 2, y₁ = 68, x₂ = 4, y₂ = 74

Step 2: Plug into the formula:

y = 68 + (3 - 2) × (74 - 68) / (4 - 2)

y = 68 + (1) × 6 / 2

y = 68 + 3

Answer: 71°F

Solution 2: Salary Interpolation

Given data: (3, 52000) and (7, 68000). Find y when x = 5.

Step 1: Label your values. x₁ = 3, y₁ = 52000, x₂ = 7, y₂ = 68000

Step 2: Apply the formula:

y = 52000 + (5 - 3) × (68000 - 52000) / (7 - 3)

y = 52000 + (2) × 16000 / 4

y = 52000 + 8000

Answer: $60,000

Solution 3: Material Stress Test

Given data: (500, 2.3) and (2000, 8.7). Find y when x = 1200.

Step 1: Extract your coordinates. x₁ = 500, y₁ = 2.3, x₂ = 2000, y₂ = 8.7

Step 2: Calculate:

y = 2.3 + (1200 - 500) × (8.7 - 2.3) / (2000 - 500)

y = 2.3 + (700) × 6.4 / 1500

y = 2.3 + 2.9867

Answer: 5.29 mm (rounded to two decimal places)

Solution 4: Financial Projection

Given data: (2, 10500) and (6, 14200). Find y when x = 4.

Step 1: Assign variables. x₁ = 2, y₁ = 10500, x₂ = 6, y₂ = 14200

Step 2: Solve:

y = 10500 + (4 - 2) × (14200 - 10500) / (6 - 2)

y = 10500 + (2) × 3700 / 4

y = 10500 + 1850

Answer: $12,350

Quick Reference: Linear Interpolation vs Alternatives

Common Mistakes That Kill Your Answers

• Swapping x and y values — Always check which axis your target value sits on
• Forgetting to divide the slope fraction — The (y₂ - y₁) / (x₂ - x₁) part is one fraction, not two separate operations
• Using extrapolation instead of interpolation — If x is outside your [x₁, x₂] range, you're extrapolating, which is much less reliable
• Rounding too early — Keep full decimal precision until your final answer
• Mixing up which point is first — Consistency matters. Label everything before you start calculating

How to Get Started with Your Own Problems

When you encounter a real interpolation problem:

1. Write down your two known points as (x₁, y₁) and (x₂, y₂)
2. Identify your target x-value — the one between x₁ and x₂
3. Substitute into the formula exactly as written
4. Calculate step by step — don't try to do it in your head
5. Check your answer — it should fall between y₁ and y₂

If your answer falls outside that range, something went wrong. Go back and check your arithmetic.

When Linear Interpolation Breaks Down

This method assumes a straight-line relationship between points. That assumption fails when:

• Your data shows obvious curvature
• Points are clustered unevenly
• Physical constraints make a linear model unrealistic (like population growth or compound interest)

In those cases, you need polynomial or spline methods. Linear interpolation is a starting point, not a universal solution.

Bottom Line

Linear interpolation is straightforward once you internalize the formula and stop overthinking it. Practice with the problems above until the process becomes automatic. The goal isn't to understand the theory deeply—it's to solve these problems quickly and correctly.

Bookmark this page. Come back when you need a refresher. That's why it's here.