Algebra • Linear Inequalities

Linear Inequality Point Checker

Verify whether an ordered pair (x₀, y₀) satisfies a linear inequality in standard form Ax + By ≤ C or slope-intercept form y ≥ mx + b with step-by-step algebraic substitution.

Reviewed by Sanjay Samanta & Academic Review Board

Last Updated: September 2026

Verified Accurate: Convex Optimization & Linear Systems

Interactive Calculator

Linear Inequality Point Verifier

Presets:

Inequality: Ax + By [op] C

A (x coeff)

B (y coeff)

Operator

C (constant)

Coordinates of Test Point (x₀, y₀)

Point x-coordinate (x₀)

Point y-coordinate (y₀)

Shaded Half-Plane Visualizer Range: [-10, 10]

Boundary Line

Point (x₀, y₀)

Satisfaction Status

Satisfies Inequality

Point lies within the solution region

Inequality Balance

0 ≤ 6

True Mathematical Statement

Boundary & Location

Interior (Solid Line)

Strictly inside shaded region

Algebraic Step-by-Step Proof

Direct Answer & Overview

Verified Educational Guide

How to Determine if a Point Satisfies a Linear Inequality

To verify if a coordinate pair (x₀, y₀) satisfies a linear inequality, substitute x₀ for x and y₀ for y into the algebraic inequality. Simplify both sides. If the inequality statement is true (e.g., 0 ≤ 6), the point belongs to the solution set and falls within the shaded half-plane. If false (e.g., 8 < 3), the point is outside the solution region. Boundary points satisfy non-strict inequalities (≤, ≥) with solid lines, but violate strict inequalities (<, >) with dashed lines.

Primary Mathematical Formula Coordinate Substitution & Half-Plane Inclusion Test

Standard Equation

ƒ(x)

Q.E.D.

A(x_0) + B(y_0) \le C \quad \text{or} \quad y_0 \ge m(x_0) + b

Boundary lines are solid for non-strict inequalities (≤, ≥) and dashed for strict inequalities (<, >).

Exact Formula

Input Parameters

Required

Linear Inequality: Standard form Ax + By [op] C or slope-intercept form y [op] mx + b.

Ordered Pair (x₀, y₀): Coordinates of the point being evaluated.

Expected Outputs

Calculated

Satisfaction Status: True (inside solution set) vs. False (outside solution set).

Boundary Position: Interior, on solid boundary, on dashed boundary, or exterior.

Boundary Line Rendering: Solid line (points included) vs. Dashed line (points excluded).

Worked Numerical Example

Instant Verification

Check if (0, 0) satisfies 2x + 3y < 6

→ Substitute x = 0 and y = 0: LHS = 2(0) + 3(0) = 0. RHS = 6. 0 < 6 is True.

True: Point (0, 0) lies strictly inside the shaded solution region.

What Is a Linear Inequality in Two Variables?

While a linear equation Ax + By = C represents a 1-dimensional line across the Cartesian coordinate plane, a linear inequality represents a full 2-dimensional half-plane: an infinite geometric region bounded on one edge by the straight line.

Any straight line divides the entire 2D coordinate plane into three distinct, non-overlapping geometric sets:

1. Half-Plane A

All coordinates where Ax + By < C strictly holds true.

2. Boundary Line

All coordinates where Ax + By = C is exactly equal. Verify with our linear equation point checker.

3. Half-Plane B

All coordinates where Ax + By > C strictly holds true.

Half-Planes and Solution Sets

The solution set of an inequality consists of every point whose coordinates satisfy the inequality statement. Graphically, this entire region is shaded.

Because there are infinitely many points in a half-plane, testing individual coordinates allows you to determine whether a given point is an allowable solution for real-world constraints such as manufacturing budgets, calorie minimums, or structural tolerances.

Solid vs. Dashed Boundary Line Rules

The most critical algebraic detail in inequalities is how boundary points are treated:

Non-Strict: ≤ and ≥ (Solid Line)

The inequality includes "or equal to". The boundary line is drawn as a solid line, and any point falling directly on this line is an acceptable solution.

Ax₀ + By₀ = C ⇒ SATISFIED

Strict: < and > (Dashed Line)

The inequality is strict. The boundary line is drawn as a dashed line to indicate that it serves as an open fence: points on the boundary are NOT solutions.

Ax₀ + By₀ = C ⇒ NOT SATISFIED

The Test Point Method (Origin Test)

When graphing linear inequalities by hand, the standard algorithm is the Test Point Method:

1. Graph the Boundary Line: Plot Ax + By = C (solid for ≤, ≥; dashed for <, >).

2. Select a Test Point: Choose a point not on the line. The origin (0, 0) is ideal because A(0) + B(0) = 0.

3. Evaluate: If (0, 0) yields a true statement, shade the half-plane containing (0, 0). If false, shade the opposite half-plane.

Step-by-Step Verification Algorithm

Substitute Coordinates: Insert x₀ into x and y₀ into y in the inequality.

Evaluate LHS Arithmetic: Multiply coefficients and add/subtract terms to obtain a single real number.

Compare LHS with RHS: Check if the inequality relation (<, ≤, >, ≥) holds.

Check Boundary Inclusion: If LHS equals RHS, check if the symbol includes equality (≤ or ≥). If strict (< or >), the statement is FALSE.

Step-by-Step Graded Worked Examples

Example 1 • Interior Point Satisfied Basic Tier

Determine if P(1, 2) satisfies 3x - 2y < 5

1. Substitute x = 1, y = 2: 3(1) - 2(2)

2. LHS = 3 - 4 = -1

3. RHS = 5

4. -1 < 5 is TRUE

Conclusion: P(1, 2) satisfies the inequality and lies in the shaded half-plane.

Example 2 • Point on Strict Dashed Boundary Intermediate Tier

Determine if Q(3, -1) satisfies 2x + 4y > 2

1. Substitute x = 3, y = -1: 2(3) + 4(-1)

2. LHS = 6 - 4 = 2

3. RHS = 2

4. 2 > 2 is FALSE (2 is equal to 2, not strictly greater)

Conclusion: Q(3, -1) lies on the dashed boundary line and is NOT part of the solution set.

Example 3 • Slope-Intercept Non-Strict Inequality Advanced Tier

Check if R(-2, 5) satisfies y ≥ -3x - 1

1. LHS = y₀ = 5

2. RHS = -3(-2) - 1 = 6 - 1 = 5

3. Compare: 5 ≥ 5 is TRUE (equality is allowed by ≥)

Conclusion: R(-2, 5) lies on the solid boundary line and IS part of the solution set.

Common Pitfalls and Boundary Errors

Including Boundary Points for Strict Inequalities

Assuming that because a point is on the boundary line, it must satisfy the inequality. For < and >, equality is excluded, making boundary points invalid solutions.

Forgetting to Flip Inequality Signs

When converting standard inequalities with negative B to slope-intercept form (e.g. -2y ≤ 4), dividing by a negative number reverses the inequality direction (y ≥ -2).

Reversing the Test Point Shading

When testing (0, 0), if the statement is false, shading the side containing (0, 0) anyway. A false test result means the other side must be shaded.

Confusing x and y Coordinate Substitution

Swapping coordinates when plugging in (such as evaluating y as the first number). Always carefully substitute the first coordinate into x and the second into y.

Fact-Checked & Verified • Computational Accuracy Standards

Updated September 2026 • Editorial Policy

Authored By

Sanjay Samanta

Lead Developer & Founder of Basic Math Tools. Specializes in browser-native computational algorithms and applied mathematics.

Reviewed & Verified By

Academic Review Board

Mathematics & curriculum specialists. Audited against standard algebraic and arithmetic principles.

Found an error or have an improvement suggestion? Report a calculation issue

Frequently Asked Questions

How do you check if a point satisfies a linear inequality?

Substitute the point's x-coordinate into x and y-coordinate into y. Calculate the numerical value on both sides. If the inequality relation (<, ≤, >, ≥) is mathematically true (e.g. 0 ≤ 6), the point satisfies the inequality and belongs to the solution region.

What is the difference between a solid and dashed boundary line?

Non-strict inequalities (≤ and ≥) include the boundary line in the solution set, which is drawn as a solid line. Strict inequalities (< and >) do not include boundary points in the solution set, which is drawn as a dashed line.

Does a point on the boundary line satisfy a strict inequality like 2x + y < 5?

No. If substituting a point yields equality (such as 5 < 5), the statement is false because 5 is not strictly less than 5. Points on the boundary of a strict inequality do not satisfy it.

What is the "Origin Test" when graphing linear inequalities?

The origin (0, 0) is the easiest test point to check because 0 eliminates the variable terms. If substituting (0, 0) makes the inequality true, the side of the boundary line containing (0, 0) is shaded. If false, the opposite half-plane is shaded. If the line passes through (0, 0), choose another point like (1, 0) or (0, 1).

How does this tool handle slope-intercept inequalities like y ≥ mx + b?

In slope-intercept form, y ≥ mx + b represents the region on or above the boundary line, while y ≤ mx + b represents the region on or below the boundary line. You can switch between standard form and slope-intercept form anytime.

How does checking a single inequality extend to systems of inequalities?

In a system of inequalities, a point is a solution if and only if it satisfies every individual inequality in the system simultaneously, meaning it lies in the intersection of all shaded half-planes.