直线方程zhíxiàn fāngchéng

Core Concept

The equation of a line (直线方程) is an algebraic sentence that is true for exactly the points lying on that line. The CSCA exam uses four standard forms, and much of the difficulty is not solving — it is choosing the right form for the information you are given.

The Four Standard Forms

Form	Equation	Given	Cannot represent
Point-slope 点斜式	$y - y_0 = k(x - x_0)$	a point + slope	vertical lines
Slope-intercept 斜截式	$y = kx + b$	slope + $y$-intercept	vertical lines
Two-point 两点式	$\frac{y-y_1}{y_2-y_1} = \frac{x-x_1}{x_2-x_1}$	two points	vertical and horizontal lines
General 一般式	$Ax + By + C = 0$	anything	nothing — fully general

The rightmost column is the exam's favourite trap: three of the four forms have blind spots, and only the general form can describe every line.

Point-Slope Form

$y - y_0 = k(x - x_0)$

This is the workhorse. Given any point $(x_0, y_0)$ on the line and the slope $k$, you can write the equation immediately without solving anything.

Slope-Intercept Form

$y = kx + b$

Here $b$ is the $y$-intercept — the $y$-coordinate where the line crosses the $y$-axis. This form makes both the slope and the intercept readable at a glance, which is why it is the standard target when a question says "convert to..." .

General Form

$Ax + By + C = 0 \quad (A, B \text{ not both } 0)$

From the general form you can read off:

• Slope: $k = -\dfrac{A}{B}$ (when $B \neq 0$)
• $y$-intercept: $-\dfrac{C}{B}$ (when $B \neq 0$)
• $x$-intercept: $-\dfrac{C}{A}$ (when $A \neq 0$)

When $B = 0$, the equation reduces to $x = -\frac{C}{A}$ — a vertical line, whose slope does not exist.

Parallel and Perpendicular Conditions

Written in general form, with $l_1: A_1x + B_1y + C_1 = 0$ and $l_2: A_2x + B_2y + C_2 = 0$:

$l_1 \parallel l_2 \iff A_1B_2 - A_2B_1 = 0$

$l_1 \perp l_2 \iff A_1A_2 + B_1B_2 = 0$

These versions are strictly better than the slope versions, because they stay valid when a line is vertical. Memorising them removes the need to case-split.

Worked Applications

Application 1: Line Through Two Points

Find the equation of the line through $A(2, 1)$ and $B(4, 7)$.

1. Slope: $k = \frac{7-1}{4-2} = 3$
2. Point-slope with $A$: $y - 1 = 3(x - 2)$
3. Simplify: $y = 3x - 5$, or in general form $3x - y - 5 = 0$

Application 2: Intersection of Two Lines

Find where $l_1: 2x + y - 5 = 0$ and $l_2: x - y - 1 = 0$ meet.

Solve the system by addition: $(2x + y) + (x - y) = 5 + 1$, so $3x = 6$ and $x = 2$. Substituting into $l_2$: $y = 1$. The intersection is $(2, 1)$.

Geometric meaning of the algebra: one solution means the lines intersect; no solution means they are parallel; infinitely many means they coincide.

CSCA Practice Problems

💡 Note: These problems follow the CSCA exam syllabus and Chinese standardized test formats.

Example 1: Basic (Difficulty ★★☆☆☆)

The line through $(1, -2)$ with slope $3$ has equation:

Options:

• A. $3x - y - 5 = 0$
• B. $3x - y + 5 = 0$
• C. $3x + y - 1 = 0$
• D. $x - 3y - 7 = 0$

Detailed Solution:

Start from point-slope form:

$y - (-2) = 3(x - 1)$

$y + 2 = 3x - 3$

$3x - y - 5 = 0$

The sign trap here is $y - (-2) = y + 2$. Substituting a negative coordinate without keeping the double negative gives option B.

Answer: A

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Example 2: Intermediate (Difficulty ★★★☆☆)

Line $l$ passes through $(2, 3)$ and is perpendicular to $x + 2y - 4 = 0$. Find its equation.

Options:

• A. $2x - y - 1 = 0$
• B. $x + 2y - 8 = 0$
• C. $2x + y - 7 = 0$
• D. $x - 2y + 4 = 0$

Detailed Solution:

Method 1 (via slopes):

1. Given line: $2y = -x + 4$, so $y = -\frac{1}{2}x + 2$ and $k_1 = -\frac{1}{2}$.
2. Perpendicular slope: $k = -\frac{1}{k_1} = 2$.
3. Point-slope: $y - 3 = 2(x - 2) \implies y = 2x - 1 \implies 2x - y - 1 = 0$.

Method 2 (faster, using the general-form condition):

For $l_1: x + 2y - 4 = 0$, a perpendicular line has the form $2x - y + C = 0$ (swap the coefficients and negate one — this automatically satisfies $A_1A_2 + B_1B_2 = 1\cdot2 + 2\cdot(-1) = 0$).

Substitute $(2, 3)$: $2(2) - 3 + C = 0 \implies C = -1$.

So $2x - y - 1 = 0$.

Answer: A

Common Mistakes

❌ Mistake 1: Using point-slope form when the line may be vertical

If a question says "line $l$ passes through $P$ and is tangent to circle $C$", the tangent may be vertical. Writing only $y - y_0 = k(x-x_0)$ silently discards that solution, and you lose half the marks. Always check the vertical case separately, or use the general form.

❌ Mistake 2: Confusing the two intercepts

In $y = kx + b$, the value $b$ is the $y$-intercept. The $x$-intercept is $-\frac{b}{k}$, which is a different number.

❌ Mistake 3: Sign errors converting to general form

Moving terms across the equals sign flips their signs. After converting, spot-check by substituting a known point back into your general form — it should give exactly $0$.

Study Tips

1. ✅ Pick the form that matches your data — point + slope → point-slope; two points → find slope first, then point-slope
2. ✅ Answer in general form unless the question asks otherwise; it is the conventional final answer
3. ✅ Learn the coefficient-swap trick for perpendiculars: $Ax + By + C = 0 \perp Bx - Ay + C' = 0$
4. ✅ Always ask about the vertical case before committing to a slope-based form
5. ✅ Verify by substitution — plug the given point back in and confirm you get $0$

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💡 Exam Tip: 直线的点斜式方程及辨析 and 求直线交点坐标 are both high-frequency CSCA knowledge points. Intersection questions reduce to solving a two-variable linear system, so practice that until it is mechanical.