Determining the Plane Region of a System of Inequalities | 不等式组表示的平面区域判定

📚 Determining the Plane Region of a System of Inequalities | 不等式组表示的平面区域判定

In coordinate geometry, a single linear inequality in two variables describes a half-plane. When several inequalities are combined into a system, the set of points that satisfies every inequality at the same time is called the feasible region, or the solution region. This article explains, step by step, how to determine, sketch, and verify that region.

在平面解析几何中，一个含两个变量的线性不等式表示一个半平面。当多个不等式组成不等式组时，同时满足每一个不等式的点集称为可行区域，也叫解区域。本文逐步讲解如何判定、绘制并验证这个区域。

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1. Linear Inequalities in Two Variables | 二元一次不等式

A linear inequality in two variables can be written in one of four general forms:

ax + by < c,   ax + by > c,   ax + by ≤ c,   ax + by ≥ c

where a and b are not both zero. The solution set of such an inequality consists of all points (x, y) whose coordinates make the statement true. For example, the inequality y ≥ x − 1 is satisfied by points such as (2, 1), (0, 0), and (5, 4), but not by (0, −2).

二元一次不等式可以写成以下四种一般形式之一：

ax + by < c，   ax + by > c，   ax + by ≤ c，   ax + by ≥ c

其中 a 和 b 不全为零。该不等式的解集由所有能使不等式成立的坐标点 (x, y) 组成。例如，不等式 y ≥ x − 1 满足 (2, 1)、(0, 0)、(5, 4) 等点，但不满足 (0, −2)。

It is often convenient to rearrange a linear inequality into the slope-intercept form:

y > mx + c,   y < mx + c,   y ≥ mx + c,   y ≤ mx + c

This form makes it easy to see whether the required region lies above or below the line y = mx + c. However, not every line can be written this way — vertical lines such as x ≥ 2 must be handled separately.

为了方便判定，我们常把线性不等式改写成斜截式：

y > mx + c，   y < mx + c，   y ≥ mx + c，   y ≤ mx + c

这种形式能直观看出所需区域在直线 y = mx + c 的上方还是下方。但并非所有直线都能写成这种形式，例如竖直线 x ≥ 2 需要单独处理。

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2. Drawing the Boundary Line | 画出边界线

The first step in solving any linear inequality is to draw its boundary line. This is done by replacing the inequality symbol with an equals sign and plotting the resulting straight line.

求解任何线性不等式的第一步都是画出其边界线。做法是把不等号换成等号，然后画出所得直线。

• Procedure: For ax + by ≤ c, first draw the line ax + by = c.
• 一般步骤： 对于 ax + by ≤ c，先画出直线 ax + by = c。
• Find two points: The easiest points are usually the intercepts. Put x = 0 to find the y-intercept and y = 0 to find the x-intercept.
• 找两个点： 最方便的是截距点。令 x = 0 求 y 截距，令 y = 0 求 x 截距。
• Check the origin: If the line passes through (0, 0), choose another easy point such as (1, 0) or (0, 1) instead.
• 检查原点： 如果直线过 (0, 0)，则另选简单点如 (1, 0) 或 (0, 1)。

The boundary line itself may be solid or dashed, depending on whether the inequality is strict. This distinction is critical and is summarised below.

边界线本身是实线还是虚线，取决于不等式是否严格。这一区分至关重要，总结如下。

Inequality | 不等式	Line style | 线型	Boundary included? | 边界是否包含
y > mx + c or y < mx + c	dashed | 虚线	No | 不包含
y ≥ mx + c or y ≤ mx + c	solid | 实线	Yes | 包含

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3. The Test-Point Method | 测试点法

Once the boundary line is drawn, we must decide which side of the line is the solution half-plane. The most reliable method is the test-point method.

画好边界线后，必须判断直线哪一侧是解的半平面。最可靠的方法是测试点法。

Step 1: Choose a test point not on the boundary line.
Step 2: Substitute it into the original inequality.
Step 3: If the result is true, shade the side containing the test point. If false, shade the opposite side.

第一步：选一个不在边界线上的测试点。
第二步：将其代入原不等式。
第三步：若不等式成立，则涂画包含测试点的一侧；若不成立，则涂画另一侧。

The origin (0, 0) is usually the best test point because substituting zero is quick. The only exception is when the boundary line passes through the origin; in that case use a point such as (1, 0) or (0, 1).

原点 (0, 0) 通常是最佳测试点，因为代入零计算最快。唯一的例外是边界线过原点，此时应改用 (1, 0) 或 (0, 1) 等点。

Why does this method work? The expression ax + by − c cannot change sign without crossing zero, so its sign is constant on each side of the line ax + by = c. One test point therefore tells us the sign of the whole half-plane.

为什么这种方法有效？表达式 ax + by − c 只有在经过零时才会变号，因此在直线 ax + by = c 的每一侧符号恒定。所以一个测试点就可以判断整个半平面的符号。

Substitution result | 代入结果	Conclusion | 结论	Action | 操作
Inequality is true | 不等式成立	Test point lies in the solution region | 测试点在解区域内	Shade the side containing the test point | 涂画测试点所在侧
Inequality is false | 不等式不成立	Test point lies outside the solution region | 测试点在解区域外	Shade the opposite side | 涂画另一侧

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4. Strict vs Inclusive Inequalities | 严格不等式与包含不等式

It is essential to distinguish between strict inequalities and inclusive inequalities, because this determines both the line style and whether boundary points are valid solutions.

区分严格不等式与包含不等式至关重要，因为它既决定了线型，也决定了边界点是否为有效解。

• Strict inequalities using > or < exclude the boundary line. The line is drawn dashed, and points exactly on the line are not solutions.
• 严格不等式使用 > 或 <，不包含边界线。直线画成虚线，直线上的点不是解。
• Inclusive inequalities using ≥ or ≤ include the boundary line. The line is drawn solid, and points on the boundary are valid solutions.
• 包含不等式使用 ≥ 或 ≤，包含边界线。直线画成实线，边界上的点也是有效解。

For example, the inequality y > 2x + 1 gives the half-plane strictly above the dashed line y = 2x + 1. The point (0, 1) lies on the line, so it is not a solution. But for y ≥ 2x + 1 the same point is included.

例如，不等式 y > 2x + 1 表示虚线 y = 2x + 1 上方的半平面（不含直线）。点 (0, 1) 在这条直线上，因此不是解；而对于 y ≥ 2x + 1，该点则是解。

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5. Systems of Inequalities: The Feasible Region | 不等式组：可行区域

When a problem gives a system of two or more linear inequalities, the solution is the intersection of all the corresponding half-planes. A point must satisfy every inequality in the system simultaneously.

当题目给出两个或更多线性不等式组成的不等式组时，解就是所有对应半平面的交集。一个点必须同时满足组内的每一个不等式。

Feasible region = Half-plane 1 ∩ Half-plane 2 ∩ … ∩ Half-plane n

可行区域 = 半平面 1 ∩ 半平面 2 ∩ … ∩ 半平面 n

The practical sketching method is as follows:

具体的作图方法如下：

• Draw all boundary lines on the same axes, using solid or dashed lines as appropriate.
• 在同一坐标系中画出所有边界线，根据情况使用实线或虚线。
• Shade each half-plane lightly using the test-point method. Use different shading directions for different inequalities to avoid confusion.
• 用测试点法轻涂每个半平面。对不同不等式使用不同方向的斜线，避免混淆。
• The feasible region is the area where all the shadings overlap.
• 可行区域是所有阴影重叠的区域。

The feasible region may be bounded, like a triangle or quadrilateral, or unbounded, extending to infinity. In either case the same method applies.

可行区域可能是有界的，如三角形或四边形；也可能是无界的，向无穷远处延伸。两种情况方法相同。

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6. How to Verify a Point in the Region | 如何验证点是否在区域内

Sometimes a question asks whether a given point belongs to the feasible region. This is a purely algebraic check: substitute the coordinates into every inequality.

有时题目会问某个给定点是否属于可行区域。这是一个纯代数检验：把坐标代入每一个不等式即可。

A point (x₀, y₀) is in the feasible region if and only if it satisfies every inequality in the system.

点 (x₀, y₀) 属于可行区域，当且仅当它满足不等式组中的每一个不等式。

Consider the system:

y ≥ x − 1,   x + y ≤ 4,   y ≥ 0

Check the point (2, 2):

• 2 ≥ 2 − 1 → 2 ≥ 1 ✓, true | 成立
• 2 + 2 = 4 ≤ 4 ✓, true | 成立
• 2 ≥ 0 ✓, true | 成立

All three inequalities hold, so (2, 2) is inside the region. Now check (3, 2): the first inequality gives 2 ≥ 2 ✓ and the third gives 2 ≥ 0 ✓, but x + y = 5 which is greater than 4, so the second inequality fails. The point (3, 2) is therefore outside the region.

三个不等式全部成立，所以 (2, 2) 在区域内。再检查 (3, 2)：第一个不等式 2 ≥ 2 ✓，第三个 2 ≥ 0 ✓，但 x + y = 5 大于 4，所以第二个不等式不成立。因此 (3, 2) 在区域外。

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7. Finding the Vertices of the Region | 求区域的顶点

The vertices of a feasible region are the corner points where two boundary lines intersect. These points are especially important in linear programming, where the maximum or minimum of an objective function always occurs at a vertex.

可行区域的顶点是两条边界直线相交的角点。这些点在线性规划中尤为重要，因为目标函数的最大值或最小值总是在顶点处取得。

To find the vertices, follow these steps:

求顶点的步骤如下：

• Identify pairs of boundary lines that actually form the boundary of the feasible region.
• 找出真正构成可行区域边界的边界线对。
• Solve each pair simultaneously as linear equations.
• 联立求解每一对线性方程。
• Check that the intersection point satisfies all the other inequalities; otherwise it is not a true vertex of the region.
• 检验交点是否满足其余所有不等式；否则它不是该区域真正的顶点。

For example, for the system y ≥ x − 1, x + y ≤ 4 and y ≥ 0, the boundary lines are L₁: y = x − 1, L₂: x + y = 4 and L₃: y = 0. Solving them in pairs gives three vertices:

例如，对于不等式组 y ≥ x − 1、x + y ≤ 4、y ≥ 0，边界线为 L₁: y = x − 1，L₂: x + y = 4，L₃: y = 0。两两联立得到三个顶点：

• L₁ ∩ L₂: x + (x − 1) = 4 → 2x = 5 → x = 5/2, y = 3/2 → (5/2, 3/2)
• L₁ ∩ L₃: x − 1 = 0 → x = 1 → (1, 0)
• L₂ ∩ L₃: x + 0 = 4 → x = 4 → (4, 0)

All three

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