📚 Coordinates in the Plane | 平面坐标系
Coordinate geometry, also known as analytic geometry, allows us to describe the position of points, lines, and shapes using numbers and equations. This powerful branch of mathematics bridges algebra and geometry, enabling us to solve geometric problems with algebraic techniques. In the IB Mathematics curriculum, mastering the fundamentals of coordinates in the plane is essential for functions, calculus, and further study of vectors and transformations.
坐标几何，也称为解析几何，使我们能够用数字和方程来描述点、直线和图形的位置。这个强有力的数学分支将代数与几何连接起来，让我们能够用代数方法解决几何问题。在IB数学课程中，掌握平面坐标的基础知识对于函数、微积分以及进一步学习向量和变换至关重要。
1. The Cartesian Coordinate System | 笛卡尔坐标系
The Cartesian plane is formed by two perpendicular number lines: the horizontal x-axis and the vertical y-axis. Their intersection is called the origin, denoted by O(0, 0). The axes divide the plane into four regions called quadrants, numbered I, II, III, and IV in a counterclockwise direction starting from the upper right.
笛卡尔平面由两条互相垂直的数轴构成：水平的x轴和垂直的y轴。它们的交点称为原点，记作O(0, 0)。坐标轴将平面分成四个称为象限的区域，从右上角开始按逆时针方向分别编号为I、II、III、IV象限。
In quadrant I, both x and y coordinates are positive. In quadrant II, x is negative and y is positive. In quadrant III, both are negative. In quadrant IV, x is positive and y is negative. This sign convention is crucial for understanding the position of any point on the plane.
在第I象限，x坐标和y坐标均为正。在第II象限，x为负，y为正。在第III象限，两者均为负。在第IV象限，x为正，y为负。这种符号规则对于理解平面上任何点的位置至关重要。
2. Plotting Points and Ordered Pairs | 描点与有序数对
Every point in the plane is represented by an ordered pair (x, y), where the first number gives the horizontal displacement from the origin, and the second gives the vertical displacement. To plot the point P(3, 4), move 3 units to the right along the x-axis and then 4 units up parallel to the y-axis. The order is essential: (3, 4) is completely different from (4, 3).
平面上的每一个点都由一个有序数对(x, y)表示，第一个数表示从原点出发的水平位移，第二个数表示垂直位移。要描出点P(3, 4)，沿x轴向右移动3个单位，然后平行于y轴向上移动4个单位。顺序至关重要：(3, 4)和(4, 3)是完全不同的点。
Points lying directly on the x-axis have a y-coordinate of zero, such as (a, 0). Similarly, points on the y-axis have an x-coordinate of zero, (0, b). The origin itself is (0, 0). Coordinates allow us to uniquely identify any location on the flat plane.
位于x轴上的点其y坐标为零，例如(a, 0)。同样，位于y轴上的点其x坐标为零，如(0, b)。原点本身为(0, 0)。坐标使我们能够唯一地确定平面上任意位置。
3. Distance Between Two Points | 两点间的距离
The distance d between two points A(x₁, y₁) and B(x₂, y₂) is derived from the Pythagorean theorem. By forming a right triangle with the segment AB as the hypotenuse, the horizontal leg has length |x₂ − x₁| and the vertical leg has length |y₂ − y₁|. The distance formula is therefore:
两点A(x₁, y₁)和B(x₂, y₂)之间的距离d是从勾股定理推导出来的。以线段AB为斜边构造直角三角形，水平直角边的长度为|x₂ − x₁|，垂直直角边的长度为|y₂ − y₁|。因此距离公式为：
d = √[(x₂ − x₁)² + (y₂ − y₁)²]
r example, the distance between (1, 2) and (4, 6) is √[(4 − 1)² + (6 − 2)²] = √[3² + 4²] = √9 + 16 = √25 5. The order of the points does not matter because squaring eliminates any negative signs.
例如，(1, 2)和(4, 6)之间的距离为√[(4 − 1)² + (6 − 2)²] = √[3² + 4²] = √9 + 16 = √25 = 5。点的先后顺序无关紧要，因为平方消除了任何负号。
This formula is fundamental in coordinate geometry and is used extensively to calculate lengths, perimeters, and to verify properties such as whether a triangle is isosceles or right-angled.
这个公式是坐标几何的基础，广泛用于计算长度、周长以及验证三角形的性质，例如是否是等腰三角形或直角三角形。
4. Midpoint of a Segment | 线段的中点
The midpoint M of the line segment joining A(x₁, y₁) and B(x₂, y₂) is found by averaging the x-coordinates and averaging the y-coordinates separately. The midpoint formula is:
连接A(x₁, y₁)和B(x₂, y₂)的线段的中点M通过分别计算x坐标的平均值和y坐标的平均值来求得。中点公式为：
M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
(5, 3). The midpoint is often used to find the center of a circle given the endpoints of a diameter or to construct perpendicular bisectors.
这在直观上是合理的：中点在水平和垂直方向上都正好位于一半的位置。例如，P(2, 5)和Q(8, 1)的中点为((2+8)/2, (5+1)/2) = (5, 3)。中点常用于已知直径端点求圆心，或用于构造垂直平分线。
Note that the midpoint always lies on the segment and divides it into two equal lengths. It also serves as the point of symmetry for the two endpoints.
请注意，中点始终在线段上并将其分成两个相等的长度。它也是两个端点的对称中心。
5. Gradient (Slope) of a Line | 直线的斜率
The gradient or slope of a straight line measures its steepness and direction. Given two distinct points A(x₁, y₁) and B(x₂, y₂) on the line, the gradient m is defined as the ratio of the vertical change to the horizontal change:
直线的斜率衡量其倾斜程度和方向。给定直线上两个不同的点A(x₁, y₁)和B(x₂, y₂)，斜率m定义为垂直变化与水平变化之比：
m = (y₂ − y₁) / (x₂ − x₁)
e (y = constant). An undefined gradient occurs when x₁ = x₂, representing a vertical line.
正斜率表示直线随x增大而上升；负斜率表示随x增大而下降。斜率为零表示水平直线（y为常数）。当x₁ = x₂时，斜率无定义，代表一条垂直直线。
2. The line through (2, 5) and (2, −3) has undefined slope and is the vertical line x = 2.
例如，通过(1, 3)和(4, 9)的直线斜率为m = (9 − 3)/(4 − 1) = 6/3 = 2。通过(2, 5)和(2, −3)的直线斜率无定义，是垂直直线x = 2。
6. Parallel and Perpendicular Lines | 平行与垂直直线
is vertical. That is, m₁ × m₂ = −1, or equivalently, m₂ = −1/m₁ (the negative reciprocal).
两条非垂直直线平行当且仅当它们的斜率相等：m₁ = m₂。如果两条直线垂直，则它们斜率的乘积为−1，前提是两条直线都不垂直。即m₁ × m₂ = −1，或等价地，m₂ = −1/m₁（负倒数）。
he slope product rule does not apply because the vertical slope is undefined. These conditions allow us to determine unknown coefficients in linear equations or to prove geometric properties of quadrilaterals.
对于垂直和水平直线，垂直直线x = a与水平直线y = b垂直。此时斜率乘积规则不适用，因为垂直直线的斜率无定义。这些条件使我们能够确定线性方程中的未知系数，或证明四边形的几何性质。
− 5 (both have gradient 3). The line perpendicular to y = 3x + 2 will have gradient −1/3.
例如，直线y = 3x + 2与y = 3x − 5平行（两者的斜率均为3）。垂直于y = 3x + 2的直线其斜率将为−1/3。
7. Equation of a Straight Line: Slope-Intercept Form | 直线的斜截式方程
a straight line is written as y = mx + c, where m is the gradient and c is the y-intercept — for quickly identifying both the steepness and the starting height of a line on the graph.
直线的斜截式方程写作y = mx + c，其中m为斜率，c为y轴截距——即直线与y轴相交时的y坐标（当x = 0时）。这种形式非常便于快速识别直线的倾斜程度以及它在图上的起始高度。
, then use the point to find c: 7 = 2(3) + c, giving c = 1, so the equation is y = 2x + 1.
要写出斜率为2且经过点(3, 7)的直线方程，我们可以先将m = 2代入y = 2x + c，然后用该点求c：7 = 2(3) + c，得c = 1，因此方程为y = 2x + 1。
form; they are written as x = k. Horizontal lines have gradient 0 and take the form y = c.
请注意，垂直直线无法用这种形式表示；它们写作x = k。水平直线的斜率为0，形式为y = c。
8. Equation of a Straight Line: Point-Slope Form | 点斜式方程
When we know the gradient m and one point (x₁, y₁) on the line, we can use the point-slope form:
当我们知道直线的斜率m和直线上的一点(x₁, y₁)时，可以使用点斜式方程：
y − y₁ = m(x − x₁)
₁), rearranged algebraically. It is particularly useful when we are not given the y-intercept directly but have a known point and gradient.
这种形式直接来源于斜率定义m = (y − y₁)/(x − x₁)的代数变形。当我们没有直接给出y轴截距，但知道斜率和某一点时，这种形式尤其有用。
ept form. This flexibility makes point-slope form a powerful tool for building equations quickly.
例如，斜率为4并经过点(2, 3)的直线可以写成y − 3 = 4(x − 2)。展开得y − 3 = 4x − 8，即y = 4x − 5，这就是斜截式。这种灵活性使点斜式成为快速构建方程的有力工具。
9. General Form of a Linear Equation | 直线的一般式方程
n be found by rearranging: By = −Ax − C, so y = (−A/B)x − (C/B), giving gradient m = −A/B.
直线的一般式方程为Ax + By + C = 0，其中A、B和C为常数，且A和B不同时为零。这种形式可以表示平面上的任何直线，包括垂直直线（当B = 0时）。斜率可以通过整理得到：By = −Ax − C，所以y = (−A/B)x − (C/B)，得出斜率m = −A/B。
s form is often used when dealing with systems of linear equations or when points of intersection are required.
要求x轴截距，设y = 0并求解x：Ax + C = 0得x = −C/A（若A ≠ 0）。y轴截距通过设x = 0求得，即y = −C/B（若B ≠ 0）。在处理线性方程组或需要求交点时常用这种形式。
ent m = −3/(−4) = 3/4, x-intercept at x = −12/3 = −4, and y-intercept at y = −12/(−4) = 3.
例如，方程3x − 4y + 12 = 0的斜率为m = −3/(−4) = 3/4，x轴截距为x = −12/3 = −4，y轴截距为y = −12/(−4) = 3。
10. Intersection of Two Lines | 两直线的交点
The point where two lines intersect satisfies both linear equations simultaneously. To find the intersection, we solve the system of two equations, either by substitution or by elimination. The solution gives the coordinates (x, y) of the intersection point. If the lines are parallel and distinct, there is no solution. If the equations represent the same line, there are infinitely many intersection points.
两条直线的交点同时满足这两个线性方程。为了求出交点，我们通过代入法或消元法来解这个方程组。解出的坐标(x, y)即为交点。如果两条直线平行且不重合，则无解。如果两个方程表示同一条直线，则有无穷多个交点。
s (1, 3). In general form, we could solve 2x − y + 1 = 0 and x + y − 4 = 0 simultaneously.
例如，求y = 2x + 1与y = −x + 4的交点。令2x + 1 = −x + 4，得3x = 3，x = 1，然后y = 2(1) + 1 = 3。交点为(1, 3)。在一般式下，我们可以解2x − y + 1 = 0和x + y − 4 = 0的方程组。
This technique is essential for many coordinate geometry problems, including finding the vertices of triangles and quadrilaterals or the intersection of medians and altitudes.
这一技巧对于许多坐标几何问题至关重要，包括求三角形和四边形的顶点，或求中线与高线的交点。
11. Perpendicular Bisector | 垂直平分线
The perpendicular bisector of a segment AB is the line that is perpendicular to AB and passes through its midpoint. To find its equation, first calculate the midpoint M and the gradient mAB of the segment. The gradient of the perpendicular bisector mperp is the negative reciprocal −1/mAB. Then use the point-slope form with point M and gradient mperp.
线段AB的垂直平分线是垂直于AB并通过其中点的直线。要求它的方程，首先计算中点M和线段AB的斜率mAB。垂直平分线的斜率mperp是负倒数−1/mAB。然后利用点斜式，以点M和斜率mperp写出方程。
B(5, 4), the midpoint is M(3, 3). The gradient of AB is (4 − 2)/(5 − 1) = 2/4 = 1/2, so mperp ation of the perpendicular bisector is y − 3 = −2(x − 3), which simplifies to y = −2x + 9.
例如，给定A(1, 2)和B(5, 4)，中点为M(3, 3)。AB的斜率为(4 − 2)/(5 − 1) = 2/4 = 1/2，因此mperp = −2。垂直平分线的方程为y − 3 = −2(x − 3)，化简为y = −2x + 9。
The perpendicular bisector is used to find the circumcenter of a triangle, as it is the intersection point of the three perpendicular bisectors of the sides. It also helps solve locus problems where a point is equidistant from two given points.
垂直平分线用于求三角形的外心，因为它是三角形三条边的垂直平分线的交点。它还有助于解决动点到两定点等距的轨迹问题。
12. Area of a Triangle Using Coordinates | 利用坐标求三角形面积
Given the vertices of a triangle A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃), the area can be computed directly from the coordinates using the formula:
给定三角形顶点A(x₁, y₁)、B(x₂, y₂)和C(x₃, y₃)，面积可以直接由坐标用以下公式计算：
Area = ½ | x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂) |
The absolute value ensures the area is always positive. If the points are collinear, the expression inside the absolute value equals zero, so the area is zero. This formula is derived from the determinant of vectors and is very efficient for coordinate-based calculations.
绝对值确保面积始终为正。如果三点共线，绝对值内部的表达式等于零，因此面积为零。该公式源自向量的行列式，对于基于坐标的计算非常高效。
ple: Find the area of triangle with vertices (1, 1), (4, 3), (2, 6). Plugging in: Area = ½ | 1(3 − 6) + 4(6 − 1) + 2(1 − 3) | = ½ | 1(−3) + 4(5) + 2(−2) | = ½ | −3 + 20 − 4 | = ½ × 13 = 6.5 square units.
例如：求顶点为(1, 1)、(4, 3)、(2, 6)的三角形面积。代入得：面积 = ½ | 1(3 − 6) + 4(6 − 1) + 2(1 − 3) | = ½ | 1(−3) + 4(5) + 2(−2) | = ½ | −3 + 20 − 4 | = ½ × 13 = 6.5 平方单位。
This formula can also be extended to find the area of any polygon by breaking it into triangles or using the surveyor's formula. Understanding this approach reinforces the link between algebraic expressions and geometric areas.
这个公式还可以通过将多边形分割成三角形或使用测量员公式来求任意多边形的面积。理解这种方法能强化代数表达式与几何面积之间的联系。
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