The Linear Interpolation Formula is a fundamental mathematical tool used to estimate an unknown value that lies between two known data points on a straight line. Expressed as y = y₁ + [(x − x₁) / (x₂ − x₁)] × (y₂ − y₁), this formula appears in NCERT Mathematics for Class 11 and Class 12, and is equally important for students preparing for JEE Main, JEE Advanced, and NEET. This article covers the complete definition, derivation, variable breakdown, a formula cheat sheet, three progressive solved examples, CBSE exam tips for 2025-26, common mistakes, and JEE/NEET applications.
Key Linear Interpolation Formulas at a Glance
Quick reference for the most important linear interpolation expressions.
- Standard form: \( y = y_1 + \dfrac{x – x_1}{x_2 – x_1} \times (y_2 – y_1) \)
- Slope form: \( y = y_1 + m(x – x_1) \) where \( m = \dfrac{y_2 – y_1}{x_2 – x_1} \)
- Fraction form: \( \dfrac{y – y_1}{y_2 – y_1} = \dfrac{x – x_1}{x_2 – x_1} \)
- Interpolation weight: \( t = \dfrac{x – x_1}{x_2 – x_1} \), then \( y = (1-t)y_1 + t\,y_2 \)
- Finding unknown x: \( x = x_1 + \dfrac{y – y_1}{y_2 – y_1} \times (x_2 – x_1) \)
- Percentage interpolation: \( y = y_1 + \dfrac{(x – x_1)}{(x_2 – x_1)} \times 100\% \) of \( (y_2 – y_1) \)
What is the Linear Interpolation Formula?
The Linear Interpolation Formula provides a method to find an unknown value between two known data points. It assumes that the change between the two points follows a straight-line (linear) pattern. The word “interpolation” comes from the Latin interpolare, meaning to refurbish or alter. In mathematics, it means estimating a value within a known range.
This concept is introduced in NCERT Mathematics at the Class 11 level, particularly in the context of coordinate geometry and straight lines (Chapter 10). It is also applied in statistics (Class 11, Chapter 15) when estimating median or mode from grouped data. The formula essentially uses the equation of a straight line passing through two points \( (x_1, y_1) \) and \( (x_2, y_2) \) to find the y-value corresponding to any x-value between \( x_1 \) and \( x_2 \).
In simple terms, if you know two data points and need to estimate a value between them, the Linear Interpolation Formula gives you the best linear estimate. It is widely used in numerical methods, statistics, physics, and engineering calculations. Students encounter this formula in logarithm tables, trigonometric tables, and statistical interpolation problems in CBSE board examinations.
Linear Interpolation Formula — Expression and Variables
The standard Linear Interpolation Formula is written as:
\[ y = y_1 + \frac{x – x_1}{x_2 – x_1} \times (y_2 – y_1) \]
This can also be written in the equivalent fraction form:
\[ \frac{y – y_1}{y_2 – y_1} = \frac{x – x_1}{x_2 – x_1} \]
Here, \( (x_1, y_1) \) and \( (x_2, y_2) \) are the two known data points, \( x \) is the known input value for which we want to find \( y \).
| Symbol | Quantity | Description |
|---|---|---|
| \( x \) | Input value | The known x-value at which we want to estimate y |
| \( y \) | Output value (unknown) | The estimated y-value corresponding to x |
| \( x_1 \) | Lower known x-value | The x-coordinate of the first known data point |
| \( y_1 \) | Lower known y-value | The y-coordinate of the first known data point |
| \( x_2 \) | Upper known x-value | The x-coordinate of the second known data point |
| \( y_2 \) | Upper known y-value | The y-coordinate of the second known data point |
| \( t \) | Interpolation parameter | \( t = (x – x_1)/(x_2 – x_1) \), ranges from 0 to 1 |
Derivation of the Linear Interpolation Formula
The derivation uses the concept of the slope of a straight line. Consider two known points \( (x_1, y_1) \) and \( (x_2, y_2) \). The slope of the line joining them is:
\[ m = \frac{y_2 – y_1}{x_2 – x_1} \]
Using the point-slope form of a line through \( (x_1, y_1) \):
\[ y – y_1 = m(x – x_1) \]
Substituting the value of \( m \):
\[ y – y_1 = \frac{y_2 – y_1}{x_2 – x_1} \times (x – x_1) \]
Rearranging gives the standard Linear Interpolation Formula:
\[ y = y_1 + \frac{x – x_1}{x_2 – x_1} \times (y_2 – y_1) \]
This derivation confirms that linear interpolation is simply the equation of the straight line through two given points, evaluated at a specific x-value.
Complete Algebra & Interpolation Formula Sheet
| Formula Name | Expression | Variables | Use Case | NCERT Chapter |
|---|---|---|---|---|
| Linear Interpolation (Standard) | \( y = y_1 + \dfrac{x – x_1}{x_2 – x_1}(y_2 – y_1) \) | x, y, x₁, y₁, x₂, y₂ | Estimate y for a given x between two points | Class 11, Ch 10 |
| Linear Interpolation (Fraction Form) | \( \dfrac{y – y_1}{y_2 – y_1} = \dfrac{x – x_1}{x_2 – x_1} \) | x, y, x₁, y₁, x₂, y₂ | Cross-multiply to find any unknown | Class 11, Ch 10 |
| Slope of a Line | \( m = \dfrac{y_2 – y_1}{x_2 – x_1} \) | m = slope, (x₁,y₁), (x₂,y₂) | Find gradient between two points | Class 11, Ch 10 |
| Point-Slope Form | \( y – y_1 = m(x – x_1) \) | m = slope, (x₁,y₁) = known point | Equation of line through a point | Class 11, Ch 10 |
| Two-Point Form | \( \dfrac{y – y_1}{x – x_1} = \dfrac{y_2 – y_1}{x_2 – x_1} \) | (x₁,y₁), (x₂,y₂) | Line through two known points | Class 11, Ch 10 |
| Weighted Average (Interpolation) | \( y = (1-t)y_1 + t\,y_2 \) | t = interpolation parameter (0 to 1) | Blend between two values | Class 11, Ch 15 |
| Median (Grouped Data) | \( M = l + \dfrac{n/2 – cf}{f} \times h \) | l=lower boundary, cf=cumulative freq, f=freq, h=class width | Statistical interpolation for median | Class 10, Ch 14 |
| Mode (Grouped Data) | \( Z = l + \dfrac{f_1 – f_0}{2f_1 – f_0 – f_2} \times h \) | l=lower boundary, f₁=modal freq, f₀, f₂=adjacent freqs | Statistical interpolation for mode | Class 10, Ch 14 |
| Linear Equation (Two Variables) | \( ax + by + c = 0 \) | a, b, c = constants; x, y = variables | Represents a straight line | Class 9, Ch 4 |
| Distance Formula | \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \) | d = distance, (x₁,y₁), (x₂,y₂) | Distance between two points | Class 10, Ch 7 |
Linear Interpolation Formula — Solved Examples
Example 1 (Class 9-10 Level)
Problem: The temperature of a metal rod is 30°C at position x = 2 cm and 50°C at position x = 6 cm. Estimate the temperature at x = 4 cm using the Linear Interpolation Formula.
Given: \( x_1 = 2,\; y_1 = 30,\; x_2 = 6,\; y_2 = 50,\; x = 4 \)
Step 1: Write the Linear Interpolation Formula: \( y = y_1 + \dfrac{x – x_1}{x_2 – x_1} \times (y_2 – y_1) \)
Step 2: Substitute the known values: \( y = 30 + \dfrac{4 – 2}{6 – 2} \times (50 – 30) \)
Step 3: Simplify the fraction: \( \dfrac{4-2}{6-2} = \dfrac{2}{4} = 0.5 \)
Step 4: Calculate the result: \( y = 30 + 0.5 \times 20 = 30 + 10 = 40 \)
Answer
The estimated temperature at x = 4 cm is 40°C.
Example 2 (Class 11-12 Level)
Problem: From a statistical table, the cumulative frequency at class boundary 20 is 15 and at class boundary 30 is 38. Use linear interpolation to estimate the value of x corresponding to a cumulative frequency of 25.
Given: \( x_1 = 20,\; y_1 = 15,\; x_2 = 30,\; y_2 = 38,\; y = 25 \)
Step 1: We need to find x, so rearrange the formula: \( x = x_1 + \dfrac{y – y_1}{y_2 – y_1} \times (x_2 – x_1) \)
Step 2: Substitute values: \( x = 20 + \dfrac{25 – 15}{38 – 15} \times (30 – 20) \)
Step 3: Simplify: \( x = 20 + \dfrac{10}{23} \times 10 \)
Step 4: Calculate: \( x = 20 + \dfrac{100}{23} = 20 + 4.35 \approx 24.35 \)
Step 5: Verify: The value 24.35 lies between 20 and 30, confirming the result is valid.
Answer
The estimated class boundary value is approximately 24.35.
Example 3 (JEE/NEET Level)
Problem: A scientist measures the pressure of a gas at two temperatures: at T = 300 K, pressure P = 1.20 × 10⁵ Pa; at T = 400 K, pressure P = 1.60 × 10⁵ Pa. Using linear interpolation, estimate the pressure at T = 350 K. Also, find the interpolation parameter t and interpret its physical meaning.
Given: \( x_1 = 300,\; y_1 = 1.20 \times 10^5,\; x_2 = 400,\; y_2 = 1.60 \times 10^5,\; x = 350 \)
Step 1: Calculate the interpolation parameter: \( t = \dfrac{x – x_1}{x_2 – x_1} = \dfrac{350 – 300}{400 – 300} = \dfrac{50}{100} = 0.5 \)
Step 2: Apply the weighted average form: \( y = (1-t)y_1 + t\,y_2 \)
Step 3: Substitute: \( y = (1 – 0.5)(1.20 \times 10^5) + (0.5)(1.60 \times 10^5) \)
Step 4: Calculate each term: \( y = 0.5 \times 1.20 \times 10^5 + 0.5 \times 1.60 \times 10^5 \)
Step 5: Add: \( y = 0.60 \times 10^5 + 0.80 \times 10^5 = 1.40 \times 10^5 \) Pa
Physical Interpretation: Since \( t = 0.5 \), the temperature 350 K is exactly halfway between 300 K and 400 K. The estimated pressure is exactly the arithmetic mean of the two known pressures, which is consistent with Gay-Lussac’s Law for an ideal gas at constant volume.
Answer
The estimated pressure at T = 350 K is 1.40 × 10⁵ Pa. The parameter t = 0.5 indicates the midpoint of the interval.
CBSE Exam Tips 2025-26
- Memorise both forms: We recommend learning both the standard form and the fraction form of the Linear Interpolation Formula. Either form can appear in CBSE 2025-26 papers.
- Check the direction: Always verify that your x-value lies between \( x_1 \) and \( x_2 \). If it does not, the method is called extrapolation, not interpolation.
- Apply to statistics: The median formula for grouped data is a direct application of linear interpolation. Recognising this connection helps you remember both formulas together.
- Show all steps: CBSE examiners award marks for each step. Write the formula first, then substitute, then simplify. Never skip the formula-writing step.
- Unit consistency: Ensure x and y values use consistent units throughout your calculation. A unit mismatch is a common source of errors in board exams.
- Verify your answer: After finding y, check that it lies between \( y_1 \) and \( y_2 \). If it does not, you have made an arithmetic error.
Common Mistakes to Avoid
- Swapping numerator and denominator: Many students write \( \dfrac{x_2 – x_1}{x – x_1} \) instead of \( \dfrac{x – x_1}{x_2 – x_1} \). Always place the difference involving the unknown x in the numerator.
- Inconsistent pairing: The subscripts must be consistent. If you use \( x_1 \) in the numerator, you must subtract \( y_1 \) (not \( y_2 \)) in the y-difference term. Mixing subscripts gives a completely wrong answer.
- Confusing interpolation with extrapolation: The Linear Interpolation Formula is valid only when the unknown x lies strictly between \( x_1 \) and \( x_2 \). Using it outside this range is extrapolation and may give large errors.
- Forgetting to add \( y_1 \): A frequent mistake is computing only the fractional part and forgetting to add \( y_1 \) at the end. The full formula is \( y_1 \) plus the correction term.
- Rounding too early: Students often round the intermediate fraction \( \dfrac{x – x_1}{x_2 – x_1} \) too early. Carry at least four decimal places until the final step to avoid accumulated rounding error.
JEE/NEET Application of the Linear Interpolation Formula
In our experience, JEE aspirants encounter the Linear Interpolation Formula most often in numerical methods, data interpretation, and physics problems involving proportional change. Here are the key application patterns you must know.
Pattern 1: Reading Mathematical Tables
JEE problems sometimes ask students to find logarithm or trigonometric values that fall between standard table entries. For instance, finding \( \log(2.35) \) when the table gives \( \log(2.3) \) and \( \log(2.4) \). The Linear Interpolation Formula provides the precise estimate. This is a direct plug-and-calculate application that tests conceptual clarity.
Pattern 2: Physics — Proportional Reasoning
In JEE Physics and NEET Biology/Chemistry, linear interpolation appears in problems involving uniform variation. Examples include estimating the refractive index of a medium at a wavelength between two known values, or estimating enzyme activity at a temperature between two measured data points. The formula \( y = y_1 + \dfrac{x – x_1}{x_2 – x_1}(y_2 – y_1) \) directly models these linear relationships.
Pattern 3: Numerical Methods in JEE Advanced
JEE Advanced occasionally tests the concept of linear interpolation as part of numerical methods. Students may be asked to find the root of an equation using the method of false position (Regula Falsi), which is a direct application of the Linear Interpolation Formula. The formula is applied iteratively: \( x = x_1 – f(x_1) \times \dfrac{x_2 – x_1}{f(x_2) – f(x_1)} \). Recognising this as linear interpolation saves time and prevents errors. Our experts suggest practising at least five such problems before your JEE Advanced attempt.
For further reading on related topics, visit the official NCERT website to access the Class 11 Mathematics textbook chapters on straight lines and statistics.
FAQs on Linear Interpolation Formula
We hope this comprehensive guide to the Linear Interpolation Formula has clarified every concept you need for CBSE board exams and competitive entrance tests. For more algebra formulas, explore our complete Algebra Formulas hub. You may also find our articles on the Linear Equations Formula and the Diagonal Formula useful for building a strong foundation. For factorial-based problems, visit our guide on the Factorial Formula.