📚 Regions Defined by Inequalities | 不等式定义的区域
pins down a single straight line, but an inequality such as x + y < 4 selects an entire half-plane on one side of that line. The topic appears early in Pure Year 1, sits inside the coordinate geometry and inequalities chapters, and returns later as the geometric foundation of linear programming problems.
在爱德思 A Level 数学中，区域是坐标平面内满足一个或多个不等式的点的集合。像 x + y = 4 这样的方程只确定一条直线，而不等式 x + y < 4 则选取该直线一侧的整个半平面。这一主题出现在 Pure Year 1 的早期，属于坐标几何与不等式章节，并在之后作为线性规划问题的几何基础再次出现。
1. What Is a Region? | 什么是区域
e inequality is not about that line alone - it describes every point on one particular side of it. The point (0, 0) gives 0 + 0 = 0 < 4, so it lies inside the region. The point (3, 3) gives 3 + 3 = 6, which is not less than 4, so it lies outside the region.
区域是所有满足某个不等式的点 (x, y) 的完整集合。以 x + y < 4 为例，与之对应的方程 x + y = 4 是一条直线，而不等式描述的不是这条直线本身，而是直线某一侧的所有点。点 (0, 0) 满足 0 + 0 = 0 < 4，因此落在区域内；点 (3, 3) 得到 3 + 3 = 6，不小于 4，因此落在区域外。
counts as part of the region depends on the inequality symbol: ≤ and ≥ include the boundary, while < and > exclude it.
为了表示区域，首先画出边界线 x + y = 4，然后判断直线的哪一侧满足不等式。边界线本身是否属于区域，取决于不等号：≤ 和 ≥ 把边界包含在内，而 < 和 > 则把它排除在外。
x + y = 4 → boundary line ; x + y < 4 → open half-plane
The word "region" is deliberately broad. It can be a half-plane bounded by one straight line, a polygon bounded by several straight lines, the inside or outside of a circle, or the area above a parabola. What all of these have in common is that the answer is an area of the plane, not a single curve.
"区域" 这个词的含义刻意保持宽泛。它可以是单条直线界定的半平面，可以是多条直线围成的多边形，也可以是圆的内部或外部，还可以是抛物线之上的区域。它们的共同点是：答案都是平面上的一片面积，而不是一条曲线。
2. Boundary Lines: Solid vs Dashed | 边界线：实线与虚线
ry line is included in the region and is drawn as a solid line. If the inequality uses < or >, the boundary line is not included and is drawn as a dashed (broken) line. This single detail is worth a mark in most Edexcel questions, so it should never be rushed.
处理每一道区域题目的第一步，都是决定边界线怎么画。如果不等式使用 ≤ 或 ≥，边界线包含在区域内，应画成实线；如果使用 < 或 >，边界线不包含在区域内，应画成虚线（断续线）。这个细节在大多数爱德思题目中都占一分，因此绝不能草率处理。
| Inequality / 不等式 | Boundary line / 边界线 | Included? / 是否包含 |
|---|---|---|
| y ≤ 2x + 1 | Solid 实线 | Yes 是 |
| y < 2x + 1 | Dashed 虚线 | No 否 |
| y ≥ 2x + 1 | Solid 实线 | Yes 是 |
| y > 2x + 1 | Dashed 虚线 | No 否 |
| y = 2x + 1 | Solid 实线 | Line only 仅直线本身 |
A dashed line is not a decorative choice: it carries mathematical meaning. If a vertex of a feasible region lies on a dashed boundary, that point is excluded and cannot be the exact optimum in a linear programming problem, although it may still be the limiting value that the objective approaches.
虚线不是装饰性的选择，它承载着数学含义。如果可行区域的一个顶点落在虚线上，该点是被排除的，在线性规划问题中不能作为精确的最优解，尽管它仍可能是目标函数不断逼近的极限值。
3. The Test Point Method | 检验点法
Once the boundary is drawn, you must decide which side of it to shade. The reliable method is to choose a test point that is definitely not on the line, substitute its coordinates into the inequality, and see whether the statement is true or false.
边界线画好之后，必须判断该给哪一侧上阴影。可靠的方法是选取一个明确不在直线上的检验点，把它的坐标代入不等式，看这个陈述是真还是假。
- The origin (0, 0) is usually the fastest choice, because both coordinates are zero and the arithmetic almost disappears. / 原点 (0, 0) 通常是最快的选择，因为两个坐标都是零，运算几乎为零。
- If the line passes through the origin, pick (1, 0) or (0, 1) instead. / 如果直线经过原点，改用 (1, 0) 或 (0, 1)。
- If the test point makes the inequality true, the whole side containing that point is the required region. / 如果检验点使不等式成立，那么包含该点的整个一侧就是所求区域。
- If it makes the inequality false, shade the opposite side. / 如果使不等式不成立，就给相反的一侧上阴影。
y ≤ 2x + 1 gives 0 ≤ 1, which is true, so the region is the side containing the origin, and the boundary is solid.
例如，把 (0, 0) 代入 y > 2x + 1 得到 0 > 1，不成立，因此区域位于虚线 y = 2x + 1 的另一侧。把 (0, 0) 代入 y ≤ 2x + 1 得到 0 ≤ 1，成立，因此区域是包含原点的那一侧，并且边界为实线。
The test point method works unchanged for curves. For x² + y² < 9, testing (0, 0) gives 0 < 9, which is true, so the region is the interior of the circle of radius 3 centred on the origin - and the circle itself is dashed.
检验点法对曲线同样适用。对于 x² + y² < 9，代入 (0, 0) 得到 0 < 9，成立，因此区域是以原点为圆心、半径为 3 的圆的内部，而圆周本身画成虚线。
4. Worked Example: One Linear Inequality | 例题：单个线性不等式
ves y = 4, so the line crosses the y-axis at (0, 4). Setting y = 0 gives x = 6, so it crosses the x-axis at (6, 0). Joining these two intercepts produces the boundary.
考虑不等式 2x + 3y ≥ 12。第一步是把不等号换成等号，求出边界线 2x + 3y = 12。令 x = 0 得 y = 4，直线与 y 轴交于 (0, 4)；令 y = 0 得 x = 6，直线与 x 轴交于 (6, 0)。连接这两个截距即得边界线。
and the statement 0 ≥ 12 is false. The origin therefore lies outside the region, so the required region is everything on the far side of the line, away from the origin.
第二步：由于符号是 ≥，边界线包含在内，因此画成实线。第三步：检验原点。代入 (0, 0) 得 2(0) + 3(0) = 0，而 0 ≥ 12 不成立，说明原点在区域之外，因此所求区域是直线远离原点的那一侧。
tside, such as (1, 1): 2 + 3 = 5, and 5 ≥ 12 is false. Both checks agree with the shading.
第四步是快速自检。选一个应落在区域内的点，例如 (0, 5)：2(0) + 3(5) = 15，而 15 ≥ 12 成立。再选一个应落在区域外的点，例如 (1, 1)：2 + 3 = 5，而 5 ≥ 12 不成立。两次检验都与阴影一致。
2x + 3y = 12 passes through (6, 0) and (0, 4) ; region: 2x + 3y ≥ 12
5. Regions Defined by Two or More Inequalities | 两个或多个不等式定义的区域
When a question gives several inequalities at once, the answer is the intersection of the individual regions - the set of points that satisfy every inequality simultaneously. Each inequality contributes one boundary line and one half-plane; the final region is the overlap of all of them.
当题目一次给出若干个不等式时，答案是各个单独区域的交集，即同时满足所有不等式的点的集合。每个不等式贡献一条边界线和一侧半平面，最终区域是它们的重叠部分。
The practical procedure is always the same:
实际操作步骤始终相同：
- Draw every boundary line accurately, using intercepts or the gradient-intercept form. / 准确画出每一条边界线，可用截距法或斜截式。
- Use the correct line style for each inequality (solid for ≤, ≥; dashed for <, >). / 为每个不等式选用正确的线型（≤、≥ 用实线；<、> 用虚线）。
- Test a convenient point for each inequality separately, or shade out the unwanted side of each line. / 对每个不等式分别用方便的检验点判断，或把每条线不想要的一侧划掉。
- Identify the region where all the shadings agree, then label it, usually as R. / 找出所有阴影共同覆盖的区域并加以标注，通常记作 R。
Edexcel accepts two conventions. You may shade the region that is not wanted and leave the required region clear, or you may shade the required region directly. Either is fine provided you clearly label the final answer, normally with the letter R written inside the region itself.
爱德思接受两种约定：你可以把不要的区域涂上阴影、把所求区域留白，也可以直接把所求区域涂上阴影。只要清楚标注最终答案（通常是在区域内部写上字母 R），两种做法都可以。
6. Worked Example: A Triangular Feasible Region | 例题：三角形可行区域
1. Begin by drawing each boundary as a solid line, because all three inequalities include their boundaries.
设某区域由三个不等式定义：x + y ≤ 8、y ≥ 2、y ≤ 2x - 1。先把三条边界线都画成实线，因为这三个不等式都包含边界。
oint (3, 2) satisfies all three: 5 ≤ 8, 2 ≥ 2 and 2 ≤ 5. So the region is the small triangle containing (3, 2).
直线 x + y = 8 经过 (8, 0) 与 (0, 8)；直线 y = 2 是水平线；直线 y = 2x - 1 的斜率为 2，经过 (0, -1) 与 (0.5, 0)。点 (3, 2) 同时满足三个不等式：5 ≤ 8、2 ≥ 2、2 ≤ 5，因此所求区域是包含 (3, 2) 的那个小三角形。
es (6, 2). Solving y = 2x - 1 with x + y = 8 gives x = 3 and y = 5, so the vertex is (3, 5).
顶点由边界线两两联立求得。联立 y = 2 与 y = 2x - 1 得 x = 1.5，顶点为 (1.5, 2)；联立 y = 2 与 x + y = 8 得 (6, 2)；联立 y = 2x - 1 与 x + y = 8 得 x = 3、y = 5，顶点为 (3, 5)。
eet lines that lie outside the region, and those crossing points must be discarded because they fail at least one of the other inequalities. Always substitute a candidate vertex back into all the inequalities before accepting it.
注意，并不是每两条边界线的交点都是区域的顶点。有些交点落在区域之外，因为这些点至少不满足其余不等式中的一个，必须舍去。接受任何一个候选顶点之前，都应把它代回全部不等式进行检验。
Vertices: (1.5, 2) , (6, 2) , (3, 5) → base 4.5, height 3, area = ½ × 4.5 × 3 = 6.75
7. Shading Conventions and Exam Marking | 阴影约定与考试评分
In Edexcel papers the marks for a region question are normally split between the boundary lines and the final region. Typically one mark is available for each correctly drawn boundary line, including the correct solid or dashed style, and one mark for correctly identifying and labelling the region.
在爱德思试卷中，区域题目的分值通常分配给边界线与最终区域两部分。一般而言，每一条画得正确的边界线（包括正确的实线或虚线线型）可得一分，正确识别并标注区域再得一分。
Common marking requirements include the following.
常见的评分要求包括以下几点。
- The boundary lines must be drawn with a ruler, or the curve sketched with a clear shape. / 直线须用直尺画出，曲线须画出清晰的形状。
