Linear Functions
and Systems一次函数与方程组

Suppose $L$ is a non-vertical line in the plane, and pick any four points $P_{1} = (x_{1}, y_{1})$, $P_{2} = (x_{2}, y_{2})$, $P_{3} = (x_{3}, y_{3})$, $P_{4} = (x_{4}, y_{4})$ on $L$ with $x_{1} \ne x_{2}$ and $x_{3} \ne x_{4}$. Drop perpendiculars to the $x$-axis from $P_{1}, P_{2}$ and from $P_{3}, P_{4}$. The two resulting right triangles share the angle that $L$ makes with the horizontal, so they are similar by AA. Hence corresponding sides are proportional:设 $L$ 是平面内一条非竖直直线，在 $L$ 上任取四点 $P_{1} = (x_{1}, y_{1})$、$P_{2} = (x_{2}, y_{2})$、$P_{3} = (x_{3}, y_{3})$、$P_{4} = (x_{4}, y_{4})$，且 $x_{1} \ne x_{2}$、$x_{3} \ne x_{4}$。分别从 $P_{1}, P_{2}$ 与 $P_{3}, P_{4}$ 向 $x$ 轴作垂线，得到两个直角三角形，它们共享 $L$ 与水平方向的夹角，因此由 AA 相似。故对应边成比例：

That is, the slope formula returns the same value whichever pair of points we pick. This is why $m$ is well-defined as a property of the line, not of the chosen points. The argument also extends to negative differences and to the four sign cases; vertical lines fail because the limiting triangle has zero base and the ratio is undefined.即斜率公式对任意两点取相同的值。这正是 $m$ 作为直线属性（而非所选点的属性）良定义的原因。该论证同样适用于差为负的情形与四种符号情形；竖直线则因极限三角形底为零、比值不存在而失效。

Slope-intercept form $y = mx + b$ cannot represent a vertical line because a vertical line has undefined slope. Standard form $A x + B y = C$ avoids the problem by parametrising the line through its direction vector $(B, -A)$ rather than through a slope: any non-zero choice of $(A, B)$ picks out a unique line direction. When $B = 0$ and $A \ne 0$, we recover the vertical line $x = C/A$; when $A = 0$ and $B \ne 0$, the horizontal line $y = C/B$. This unified treatment is why textbooks introduce standard form even though slope-intercept is computationally lighter.斜截式 $y = mx + b$ 无法表示竖直线，因为竖直线的斜率不存在。标准式 $A x + B y = C$ 改用方向向量 $(B, -A)$ 参数化直线而非用斜率：只要 $(A, B)$ 不全为零，就唯一确定一条直线方向。当 $B = 0$、$A \ne 0$ 时即得竖直线 $x = C/A$；当 $A = 0$、$B \ne 0$ 时得水平线 $y = C/B$。这种统一处理正是教科书引入标准式的原因，尽管斜截式在计算上更轻巧。

Suppose a line has slope $m \ne 0$ and $y$-intercept $b \ne 0$. The $y$-intercept is $(0, b)$ by definition. To find the $x$-intercept, set $y = 0$ in $y = mx + b$: $0 = mx + b$, so $x = -b/m$. The slope between $(0, b)$ and $(-b/m, 0)$ is设直线斜率 $m \ne 0$、$y$ 截距 $b \ne 0$。按定义 $y$ 截距为 $(0, b)$。令 $y = mx + b$ 中 $y = 0$，得 $x = -b/m$，即 $x$ 截距。$(0, b)$ 与 $(-b/m, 0)$ 间的斜率为

which is the same $m$ the slope-intercept form encodes. The two graphing methods produce the same line because the intercepts are derivable from $(m, b)$ and vice versa.即与斜截式中的 $m$ 相同。因为截距可由 $(m, b)$ 推出，反之亦然，所以两种作图法画出同一条直线。

Let $L_{1}$ and $L_{2}$ be two non-vertical lines making angles $\theta_{1}$ and $\theta_{2}$ with the positive $x$-axis, so $m_{i} = \tan \theta_{i}$. The lines are perpendicular if and only if $\theta_{2} = \theta_{1} + \pi/2$. Then设两条非竖直直线 $L_{1}$、$L_{2}$ 与 $x$ 轴正方向的夹角分别为 $\theta_{1}$、$\theta_{2}$，故 $m_{i} = \tan \theta_{i}$。两直线垂直当且仅当 $\theta_{2} = \theta_{1} + \pi/2$。于是

so $m_{1} m_{2} = -1$. The argument breaks if either line is vertical (one slope is undefined), in which case the perpendicular partner is horizontal and the rule is replaced with "horizontal and vertical lines are always perpendicular."故 $m_{1} m_{2} = -1$。若任一直线竖直（斜率不存在），论证失效；此时其垂线为水平线，规则改述为"水平线与竖直线恒垂直"。

Let the system be $E_{1}: a_{1} x + b_{1} y = c_{1}$ and $E_{2}: a_{2} x + b_{2} y = c_{2}$. Replace $E_{2}$ by $E_{2}' = E_{2} + k E_{1}$ for any constant $k$. We claim $\{E_{1}, E_{2}\}$ and $\{E_{1}, E_{2}'\}$ have identical solution sets.设方程组 $E_{1}: a_{1} x + b_{1} y = c_{1}$、$E_{2}: a_{2} x + b_{2} y = c_{2}$。对任意常数 $k$，把 $E_{2}$ 替换为 $E_{2}' = E_{2} + k E_{1}$。我们证明 $\{E_{1}, E_{2}\}$ 与 $\{E_{1}, E_{2}'\}$ 解集相同。

Forward direction: if $(x, y)$ satisfies $E_{1}$ and $E_{2}$, then it satisfies $k E_{1}$ (since both sides are scaled by $k$) and so it satisfies $E_{2}' = E_{2} + k E_{1}$.正向：若 $(x, y)$ 同时满足 $E_{1}$ 与 $E_{2}$，则它满足 $k E_{1}$（两边同乘 $k$），从而满足 $E_{2}' = E_{2} + k E_{1}$。

Reverse direction: if $(x, y)$ satisfies $E_{1}$ and $E_{2}'$, then it satisfies $E_{2}' - k E_{1} = E_{2}$.反向：若 $(x, y)$ 同时满足 $E_{1}$ 与 $E_{2}'$，则它满足 $E_{2}' - k E_{1} = E_{2}$。

The replacement is reversible (the same $k$ undoes it with the opposite sign), so the two systems have exactly the same solutions. This justifies HSA-REI.C.5: every elimination step is just a controlled replacement that preserves the solution set.该替换可逆（用相反符号的同一 $k$ 即可还原），故两方程组解集完全相同。这便证明了 HSA-REI.C.5：消元的每一步都是保持解集不变的可控替换。

A chemist needs $10$ litres of a $40\%$ acid solution. She has $20\%$ acid and $70\%$ acid stocks on hand. How many litres of each should she combine?一名化学家需要 $10$ 升 $40\%$ 浓度的酸溶液。手头有 $20\%$ 与 $70\%$ 两种原液。各需取多少升？

Let $x$ be litres of the $20\%$ stock and $y$ litres of the $70\%$ stock. The two equations:设取 $20\%$ 原液 $x$ 升、$70\%$ 原液 $y$ 升，得两方程：

From the first equation, $x = 10 - y$. Substitute:由第一式 $x = 10 - y$，代入：

Then $x = 6$. The chemist mixes $6$ litres of $20\%$ stock with $4$ litres of $70\%$ stock. A quick sanity check: $0.20 \cdot 6 + 0.70 \cdot 4 = 1.2 + 2.8 = 4.0$ litres of pure acid, which is $40\%$ of $10$ litres.再得 $x = 6$。即取 $6$ 升 $20\%$ 原液与 $4$ 升 $70\%$ 原液。快速验证：$0.20 \cdot 6 + 0.70 \cdot 4 = 1.2 + 2.8 = 4.0$ 升纯酸，恰为 $10$ 升的 $40\%$。

After row-reduction, an augmented matrix falls into one of three shapes that mirror the three outcomes of 1.5:行简化后，增广矩阵呈现三种形态，对应 §1.5 的三种结果：

• A pivot in every variable column.每个变量列都有主元。 The system has exactly one solution.方程组有唯一解。
• A row of the form $[0, 0, \ldots, 0 \,|\, c]$ with $c \ne 0$.存在形如 $[0, 0, \ldots, 0 \,|\, c]$（$c \ne 0$）的行。 The system has no solution; this row reads $0 = c$.方程组无解，因为该行读作 $0 = c$。
• A free variable (no pivot in some variable column) and no contradiction row.某变量列无主元（自由变量）且无矛盾行。 The system has infinitely many solutions parametrised by the free variable.方程组有无穷多解，由自由变量参数化。

For $2 \times 2$ systems this matches "one intersection / parallel lines / coincident lines"; for $3 \times 3$ systems it matches "single point / parallel planes / line or plane of solutions." The same trichotomy generalises to every dimension.对 $2 \times 2$ 方程组对应"一个交点 / 平行 / 重合"；对 $3 \times 3$ 对应"一点 / 平行平面 / 解为直线或平面"。该三分法可推广到任意维。