📚 G-1 Practice Workbook 200: Straight Line Graphs and Coordinate Geometry | G-1 练习册 200：直线图像与坐标几何
applying midpoint, distance, parallel and perpendicular line rules. Work through each section carefully and check your understanding with the exam-style examples.
在 IGCSE 数学中，坐标几何与直线图像是教学大纲的核心内容。本练习册文章将复习描点、求斜率与截距、使用 y = mx + c，以及应用中点、距离、平行线与垂直线规则等关键技能。请仔细学习每一节，并通过考试风格例题检查理解程度。
1. Understanding Coordinates and Plotting Points | 理解坐标与描点
A point in the Cartesian plane is written as (x, y). The x-coordinate is the horizontal position; the y-coordinate is the vertical position. Always plot x first, then y.
笛卡尔平面中的点写作 (x, y)。x 坐标表示水平位置，y 坐标表示垂直位置。描点时总是先确定 x，再确定 y。
For example, the point A(2, 3) is 2 units to the right of the origin and 3 units up. The point B(−1, 4) is 1 unit left and 4 units up. The origin is (0, 0).
例如，点 A(2, 3) 位于原点右侧 2 个单位、上方 3 个单位。点 B(−1, 4) 位于原点左侧 1 个单位、上方 4 个单位。原点为 (0, 0)。
You must be confident with positive and negative coordinates in all four quadrants. Quadrant I has positive x and positive y; Quadrant II has negative x and positive y; Quadrant III has negative x and negative y; Quadrant IV has positive x and negative y.
你必须熟练处理四个象限中的正负坐标。第一象限 x 和 y 均为正；第二象限 x 为负、y 为正；第三象限 x 和 y 均为负；第四象限 x 为正、y 为负。
2. The Equation of a Straight Line: y = mx + c | 直线方程 y = mx + c
radient of the line and c represents the y-intercept. The y-intercept is the point where the line crosses the y-axis, so its x-coordinate is always 0.
直线的一般方程为 y = mx + c，其中 m 表示直线的斜率，c 表示 y 截距。y 截距是直线与 y 轴相交的点，因此该点的 x 坐标总是为 0。
y = mx + c
the gradient is 3 and the y-intercept is 2. This means the line crosses the y-axis at (0, 2) and rises 3 units for every 1 unit it moves to the right.
例如，在方程 y = 3x + 2 中，斜率为 3，y 截距为 2。这意味着直线在 (0, 2) 处与 y 轴相交，并且每向右移动 1 个单位就上升 3 个单位。
3. Finding Gradient from Two Points | 由两点求斜率
The gradient of a straight line measures how steep the line is. To find the gradient between two points (x₁, y₁) and (x₂, y₂), use the formula below.
直线的斜率衡量直线的倾斜程度。要求两点 (x₁, y₁) 和 (x₂, y₂) 之间的斜率，请使用以下公式。
m = (y₂ − y₁) ÷ (x₂ − x₁)
example, if A(1, 2) and B(5, 10) are two points, then m = (10 − 2) ÷ (5 − 1) = 8 ÷ 4 = 2.
分子和分母中必须保持相同的相减顺序。例如，如果 A(1, 2) 和 B(5, 10) 是两点，那么 m = (10 − 2) ÷ (5 − 1) = 8 ÷ 4 = 2。
A positive gradient means the line slopes upward from left to right. A negative gradient means the line slopes downward. A zero gradient means the line is horizontal, and an undefined gradient means the line is vertical.
正斜率表示直线从左到右向上倾斜。负斜率表示直线向下倾斜。斜率为零表示直线是水平的，斜率不存在表示直线是垂直的。
4. Finding the y-intercept and x-intercept | 求 y 截距与 x 截距
the line crosses the y-axis. The x-intercept is found by setting y = 0 and solving for x.
求直线的 y 截距时，在方程中令 x = 0，就得到直线与 y 轴相交时的 y 值。求 x 截距时，令 y = 0 并解出 x。
tercept is found by putting y = 0: 2x = 6, so x = 3. The intercepts are (0, 2) and (3, 0).
对于直线 2x + 3y = 6，令 x = 0 求 y 截距：3y = 6，所以 y = 2。令 y = 0 求 x 截距：2x = 6，所以 x = 3。两个截距分别是 (0, 2) 和 (3, 0)。
c. The x-intercept is calculated by setting y = 0, so x = −c ÷ m, provided m is not zero.
在方程 y = mx + c 中，y 截距就是 c。x 截距通过令 y = 0 计算，因此 x = −c ÷ m，前提是 m 不为零。
5. Midpoint of a Line Segment | 线段中点
The midpoint of a line segment is the point that divides the segment into two equal lengths. To find the midpoint M of two points (x₁, y₁) and (x₂, y₂), use the average of the x-coordinates and the average of the y-coordinates.
线段的中点是将其分成两段相等长度的点。要求两点 (x₁, y₁) 和 (x₂, y₂) 的中点 M，请分别取 x 坐标的平均值和 y 坐标的平均值。
M = ((x₁ + x₂) ÷ 2, (y₁ + y₂) ÷ 2)
e midpoint of (2, 6) and (8, 4) is ((2 + 8) ÷ 2, (6 + 4) ÷ 2) = (10 ÷ 2, 10 ÷ 2) = (5, 5).
例如，(2, 6) 和 (8, 4) 的中点是 ((2 + 8) ÷ 2, (6 + 4) ÷ 2) = (10 ÷ 2, 10 ÷ 2) = (5, 5)。
Midpoint questions often appear in IGCSE exams because they test both coordinate skills and careful arithmetic with negative numbers.
中点问题常出现在 IGCSE 考试中，因为它们既考查坐标技能，也考查对负数进行仔细运算的能力。
6. Distance Between Two Points | 两点间距离
The distance between two points is found using Pythagoras' theorem in the coordinate plane. The formula comes from the horizontal and vertical differences between the points.
两点之间的距离可以利用坐标平面中的勾股定理求得。该公式来源于两点之间的水平差和垂直差。
d = √[(x₂ − x₁)² + (y₂ − y₁)²]
For example, the distance between (1, 2) and (4, 6) is √[(4 − 1)² + (6 − 2)²] = √3² + 4² = √9 + 16 = √25 = 5.
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。
Remember to subtract the coordinates in the same order for both x and y. Squaring removes negative signs, but working neatly reduces mistakes.
记住 x 和 y 都要按相同顺序相减。平方会消除负号，但书写工整可以减少错误。
7. Parallel and Perpendicular Lines | 平行线与垂直线
arallel to y = 2x + 1, then its gradient is also 2, so its equation must start y = 2x + c.
平行直线的斜率相等。这是 IGCSE 题目的关键规则：如果一条直线与 y = 2x + 1 平行，那么它的斜率也是 2，因此它的方程必须以 y = 2x + c 开头。
. If one line has gradient m₁, the gradient of a line perpendicular to it is m₂ = −1 ÷ m₁.
垂直直线的斜率乘积为 −1。如果一条直线的斜率为 m₁，那么与它垂直的直线斜率为 m₂ = −1 ÷ m₁。
m₁ × m₂ = −1
For example, if a line has gradient 3, a perpendicular line has gradient −1/3. If a line has gradient −2, a perpendicular line has gradient 1/2.
例如，如果一条直线的斜率为 3，那么垂直直线的斜率为 −1/3。如果一条直线的斜率为 −2，那么垂直直线的斜率为 1/2。
8. Drawing Straight Line Graphs from Equations | 由方程绘制直线图像
To draw a straight line graph, create a table of values for x and calculate the corresponding y values. Three points are enough because a straight line is fixed by two points, and the third acts as a check.
要绘制直线图像，请先列一个 x 的取值表，并计算对应的 y 值。三个点就足够了，因为两点确定一条直线，第三个点可用于检查。
= 1. When x = 1, y = 3. Plot (−1, −1), (0, 1), (1, 3) and join them with a straight line.
对于 y = 2x + 1，取 x = −1, 0, 1。当 x = −1 时，y = 2(−1) + 1 = −1。当 x = 0 时，y = 1。当 x = 1 时，y = 3。描出 (−1, −1), (0, 1), (1, 3)，并用直线连接。
Always label the axes and use a ruler. In exams, marks are often awarded for a correct straight line and for plotting at least two accurate points.
务必标记坐标轴并使用直尺。在考试中，正确画出直线并准确描出至少两个点通常可以得分。
9. Finding Equation of a Line from Graph or Conditions | 由图或条件求直线方程
find the gradient m by choosing two clear points and using rise over run: m = rise ÷ run.
如果题目给出图像，请直接从 y 轴读出 y 截距 c。然后选择两个清晰的点，用纵向变化量除以横向变化量求斜率：m = rise ÷ run。
nd m using the gradient formula, then substitute one point into y = mx + c to solve for c.
知道 m 和 c 后，将它们代入 y = mx + c。如果题目给出两点但没有图像，先用斜率公式求 m，再将其中一个点代入 y = mx + c 解出 c。
ion is y = 3x + c. Substitute (2, 5): 5 = 3(2) + c, so c = −1. The equation is y = 3x − 1.
例如，一条直线经过 (2, 5) 且斜率为 3。方程为 y = 3x + c。代入 (2, 5)：5 = 3(2) + c，所以 c = −1。方程为 y = 3x − 1。
10. Exam-Style Worked Examples | 考试风格例题
ion is y = 2x + c. Substitute A(1, 3): 3 = 2(1) + c, so c = 1. The equation is y = 2x + 1.
例题 1：求经过 A(1, 3) 和 B(5, 11) 的直线方程。先求斜率：m = (11 − 3) ÷ (5 − 1) = 8 ÷ 4 = 2。方程为 y = 2x + c。代入 A(1, 3)：3 = 2(1) + c，所以 c = 1。方程为 y = 2x + 1。
ute (8, 1): 1 = −1/4 (8) + c, so 1 = −2 + c, giving c = 3. The equation is y = −1/4 x + 3.
例题 2：求与 y = 4x − 3 垂直且经过 (8, 1) 的直线方程。已知斜率为 4，因此垂直斜率为 m = −1 ÷ 4 = −1/4。方程为 y = −1/4 x + c。代入 (8
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