📚 Points of Intersection | 交点问题
In Edexcel A-Level Mathematics, intersection problems ask you to find where two curves or lines meet. A point of intersection lies on both graphs, so its coordinates satisfy both equations. This topic connects algebra, coordinate geometry, graphs and numerical methods, and it appears frequently in exam questions.
在 Edexcel A-Level 数学中，交点问题要求你找出两条曲线或直线的相交位置。交点同时位于两个图像上，因此其坐标同时满足两个方程。该主题联系代数、坐标几何、图像与数值方法，在考试题中出现频率很高。
1. Core Idea: Simultaneous Equations | 核心思想：联立方程
A point of intersection lies on both graphs, so its coordinates satisfy both equations at the same time. To find it, solve the two equations simultaneously. The solutions give the x-coordinates, and substituting into either original equation gives the corresponding y-coordinates.
交点同时位于两个图像上，因此其坐标同时满足两个方程。求交点就是联立求解这两个方程。解出的值给出 x 坐标，再代回任意一个原方程即可求出对应的 y 坐标。
y = f(x) and y = g(x) → f(x) = g(x)
find the intersections of y = 2x + 1 and y = x² − 3x + 2, equate the two right-hand sides.
例如：求 y = 2x + 1 与 y = x² − 3x + 2 的交点，可将两个等号右侧相等。
2x + 1 = x² − 3x + 2 → x² − 5x + 1 = 0
Using the quadratic formula gives x = (5 ± √21) / 2, so the points are ((5 + √21)/2, 6 + √21) and ((5 − √21)/2, 6 − √21).
使用二次公式得到 x = (5 ± √21) / 2，因此交点为 ((5 + √21)/2, 6 + √21) 和 ((5 − √21)/2, 6 − √21)。
2. Intersection of Two Lines | 两条直线的交点
ressions and solve for x. The number of solutions depends on the gradients and intercepts.
对于两条直线 y = m₁x + c₁ 与 y = m₂x + c₂，令两个表达式相等并解出 x。解的个数取决于斜率和截距。
m₁x + c₁ = m₂x + c₂ → x = (c₂ − c₁) / (m₁ − m₂)
| Case | Gradient condition | Intersections |
|---|---|---|
| Distinct non-parallel lines | m₁ ≠ m₂ | One point |
| Parallel lines | m₁ = m₂, c₁ ≠ c₂ | None |
| Coincident lines | m₁ = m₂, c₁ = c₂ | Infinitely many |
Remember that parallel lines never meet, while coincident lines overlap completely and therefore share every point.
记住，平行线永不相交，而重合直线完全重叠，因此它们有无数个公共点。
3. Line and Quadratic: Discriminant | 直线与二次曲线：判别式
the quadratic. The result is a quadratic equation in x, and its discriminant tells you how many intersections exist.
当直线 y = mx + k 与二次曲线 y = ax² + bx + c 相交时，将直线代入二次曲线。结果得到关于 x 的二次方程，其判别式可以判断交点的个数。
ax² + bx + c = mx + k → ax² + (b − m)x + (c − k) = 0
D = (b − m)² − 4a(c − k)
angent and there is exactly one intersection. If D < 0, the line and curve do not meet.
如果 D > 0，则有两个不同交点。如果 D = 0，则直线是切线，恰好有一个交点。如果 D < 0，则直线与曲线不相交。
4. Circle and Line | 圆与直线
A line can cut a circle twice, touch it once as a tangent, or miss it entirely. Substitute the line equation into the circle equation; the resulting quadratic can be analysed using the discriminant.
直线与圆可以相交于两点、相切于一点，或者完全不相交。将直线方程代入圆的方程，得到的二次方程可用判别式分析。
Circle: (x − a)² + (y − b)² = r², Line: y = mx + k
² + x − 12 = 0. The solutions are x = 3 and x = −4, giving the points (3, 4) and (−4, −3).
例如：圆 x² + y² = 25 与直线 y = x + 1，代入后得到 x² + (x + 1)² = 25，化简为 x² + x − 12 = 0。解为 x = 3 和 x = −4，对应交点为 (3, 4) 和 (−4, −3)。
5. Intersection of Two Circles | 两圆的交点
Two circles can intersect at two points, touch at one point, or have no common point. A useful method is to subtract one circle equation from the other; this removes the squared terms and leaves a linear equation for the common chord, sometimes called the radical axis.
两个圆可以相交于两点、相切于一点，或者没有公共点。一个常用方法是将两个圆的方程相减；这样可以消去平方项，得到公共弦所在直线的方程，有时称为根轴。
ing into C₁ gives 16 + y² = 25, so y = ±3. The intersection points are (4, 3) and (4, −3).
例如：设 C₁ 为 x² + y² = 25，C₂ 为 (x − 4)² + y² = 9。相减得到 8x − 16 = 16，所以 x = 4。代入 C₁ 得到 16 + y² = 25，因此 y = ±3。交点为 (4, 3) 和 (4, −3)。
6. Parametric Curves | 参数曲线
When two curves are given parametrically, their parameters are usually independent. Use different letters, such as t and s, then set the x-coordinates equal and the y-coordinates equal. Solve the resulting simultaneous equations for the two parameters.
当两条曲线以参数形式给出时，它们的参数通常是相互独立的。使用不同的字母，例如 t 和 s，然后令 x 坐标相等、y 坐标相等，联立解出两个参数。
3 and t + 1 = 2s − 1. This gives s = 7/3 and t = 8/3, so the common point is (16/3, 11/3).
例如：曲线 A 为 x = 2t, y = t + 1；曲线 B 为 x = s + 3, y = 2s − 1。联立 2t = s + 3 和 t + 1 = 2s − 1，得到 s = 7/3，t = 8/3，因此公共点为 (16/3, 11/3)。
If two moving particles are involved, a collision requires the same point at the same time, so use the same parameter. If only their paths cross, different parameter values are allowed.
如果涉及两个运动质点，碰撞要求在同一时刻到达同一点，因此应使用同一个参数。如果只是路径相交，则允许参数取值不同。
7. Modulus and Other Functions | 绝对值函数与其他函数
When finding intersections involving modulus functions, split the problem into cases according to the sign of the expression inside the modulus. Always check that each solution satisfies the original equation and any domain restrictions.
求解涉及绝对值函数的交点时，应根据绝对值内部表达式的符号分情况讨论。务必检查每个解是否满足原方程以及任何定义域限制。
Example: Solve |2x − 1| 1 < 0, then −(2x − 1) = x + 3, giving x = −2/3. The points are (4, 7) and (−2/3, 7/3).
例如：解 |2x − 1| = x + 3。若 2x − 1 ≥ 0，则 2x − 1 = x + 3，得到 x = 4。若 2x − 1 < 0，则 −(2x − 1) = x + 3，得到 x = −2/3。交点为 (4, 7) 和 (−2/3, 7/3)。
8. Inequalities from Intersections | 由交点解不等式
he critical values, then test intervals between them to decide where the inequality holds.
交点常被用来解形如 f(x) < g(x) 的不等式。先解 f(x) = g(x) 得到临界值，然后在临界值之间的区间内检验，以确定不等式成立的区域。
= 0, so x = −1 or x = 2. Testing intervals shows that x² < x + 2 for −1 < x < 2.
例如：解 x² < x + 2，先解 x² = x + 2。得到 x² − x − 2 = 0，所以 x = −1 或 x = 2。检验区间可知，当 −1 < x < 2 时，x² < x + 2。
9. Numerical Methods for Hard Intersections | 复杂交点的数值方法
ethod, iteration, or the Newton-Raphson method to approximate the root of f(x) − g(x) = 0.
有些交点无法用代数方法精确求出。此时可以使用符号变化法、迭代法或牛顿-拉夫森方法来近似求解 f(x) − g(x) = 0 的根。
xₙ₊₁ = xₙ − f(xₙ) / f'(xₙ)
least one root between 0 and 1. A numerical method can then be used to improve the approximation.
例如，方程 x³ − 3x + 1 = 0 满足 f(0) = 1 和 f(1) = −1，因此在 0 和 1 之间至少存在一个根。随后可以用数值方法提高近似精度。
10. Exam Tips and Common Mistakes | 考试技巧与
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