Unit 7 · Lines in the Plane　平面直线

By the end of this chapter you can:

1. Compute the distance between two points and find the midpoint of a segment
2. Define slope using the inclination angle and use $k = \tan\alpha$ correctly
3. Write the equation of a line in any of the four standard forms and convert between them
4. Determine whether two lines are parallel or perpendicular, and calculate the distance from a point to a line

Exam weight on past CSCA papers: ~7% (3 of 48 MCQs).

7.1　Distance Between Points　两点间距离

Given two points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ in the Cartesian plane, the distance between them is derived directly from the Pythagorean theorem:

$$|P_1P_2| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

The midpoint $M$ of the segment $P_1P_2$ is:

$$M = \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2}\right)$$

⚠️ Common mistake: Students sometimes write $\sqrt{(x_2 - x_1)^2 - (y_2 - y_1)^2}$ (a minus sign inside). The formula always uses addition under the radical.

🔑 Order doesn't matter: $(x_2 - x_1)^2 = (x_1 - x_2)^2$, so you can subtract in either direction — the square kills the sign.

Worked Example 7.1.A

Find the distance and midpoint between $A(-1, 3)$ and $B(5, -1)$.

Solution.

$$|AB| = \sqrt{(5-(-1))^2 + (-1-3)^2} = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$$

$$M = \left(\frac{-1+5}{2},\ \frac{3+(-1)}{2}\right) = (2,\ 1)$$

$$\boxed{|AB| = 2\sqrt{13},\quad M = (2,\ 1)}$$

7.2　Slope and Inclination　斜率与倾斜角

Inclination angle　倾斜角

Every non-vertical line makes an angle $\alpha$ with the positive $x$-axis, measured counter-clockwise. This angle satisfies:

$$0° \le \alpha < 180° \qquad (\text{i.e. } 0 \le \alpha < \pi)$$

A horizontal line has $\alpha = 0°$. A vertical line has $\alpha = 90°$ but its slope is undefined.

Slope　斜率

For a non-vertical line with inclination angle $\alpha$:

$$k = \tan\alpha$$

For a line through two distinct points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ with $x_1 \ne x_2$:

$$k = \frac{y_2 - y_1}{x_2 - x_1}$$

Restrictions to memorise

⚠️ Vertical lines have undefined slope. If you are asked for the slope of $x = 3$, the answer is "undefined" — not zero. Zero slope belongs to horizontal lines like $y = 3$.

🔑 $k = \tan\alpha$ works because: think of "rise over run" on a right triangle formed by the line and the axes — that ratio is exactly $\tan\alpha$.

Worked Example 7.2.A

A line passes through $P(1, 2)$ and $Q(4, -1)$. Find its inclination angle.

Solution.

$$k = \frac{-1-2}{4-1} = \frac{-3}{3} = -1$$

Since $k = \tan\alpha = -1$ and $\alpha \in [0°, 180°)$, we get $\alpha = 135°$.

$$\boxed{\alpha = 135°}$$

7.3　Line Equations　直线方程

There are four standard forms. Learn all four — CSCA questions often give you information in one form and ask for another.

The four forms — comparison table

🔑 Point-slope is the most flexible form. Given any point and a slope — or given two points (compute $k$ first) — point-slope immediately gives you the equation without having to solve for the intercept.

Converting between forms

To convert point-slope → slope-intercept: expand and solve for $y$.

To convert any form → general form: move all terms to one side so the equation equals zero.

Worked Example 7.3.A

Find the equation of the line through $(2, -3)$ with slope $k = \tfrac{1}{2}$. Express in slope-intercept and general form.

Solution.

Point-slope: $y - (-3) = \tfrac{1}{2}(x - 2)$, i.e. $y + 3 = \tfrac{1}{2}(x-2)$.

Slope-intercept: $y = \tfrac{1}{2}x - 1 - 3 = \tfrac{1}{2}x - 4$.

General form (multiply through by 2): $x - 2y - 8 = 0$.

$$\boxed{y = \tfrac{1}{2}x - 4 \quad\text{or equivalently}\quad x - 2y - 8 = 0}$$

Worked Example 7.3.B

Find the equation of the line through $A(1, 3)$ and $B(4, -3)$.

Solution.

Slope: $k = \dfrac{-3-3}{4-1} = \dfrac{-6}{3} = -2$.

Point-slope (using $A$): $y - 3 = -2(x - 1) \implies y = -2x + 5$.

$$\boxed{y = -2x + 5 \quad\text{or}\quad 2x + y - 5 = 0}$$

7.4　Parallel and Perpendicular Lines　平行与垂直

Parallel lines　平行线

Two non-vertical lines $\ell_1$ (slope $k_1$) and $\ell_2$ (slope $k_2$) are parallel if and only if:

$$k_1 = k_2 \quad \text{and the lines are distinct}$$

In general form: $A_1x + B_1y + C_1 = 0$ and $A_2x + B_2y + C_2 = 0$ are parallel iff $A_1B_2 - A_2B_1 = 0$ but $A_1C_2 - A_2C_1 \ne 0$.

⚠️ Parallel ≠ identical. Two lines with the same slope are parallel only if they are distinct. If they also share a point, they are the same line — not parallel. Always verify the $y$-intercepts are different.

Perpendicular lines　垂直线

Two non-vertical, non-horizontal lines are perpendicular if and only if:

$$k_1 \cdot k_2 = -1 \qquad \left(\text{equivalently, } k_2 = -\frac{1}{k_1}\right)$$

Special cases: a vertical line ($x = a$) is perpendicular to any horizontal line ($y = b$).

Worked Example 7.4.A

Line $\ell_1$: $2x - y + 3 = 0$. Find lines through $(1, -1)$ that are (a) parallel and (b) perpendicular to $\ell_1$.

Solution.

Rewrite $\ell_1$: $y = 2x + 3$, so $k_1 = 2$.

(a) Parallel — same slope $k = 2$:
$$y - (-1) = 2(x - 1) \implies y = 2x - 3 \implies \boxed{2x - y - 3 = 0}$$

(b) Perpendicular — slope $k = -\tfrac{1}{2}$:
$$y + 1 = -\tfrac{1}{2}(x - 1) \implies y = -\tfrac{1}{2}x - \tfrac{1}{2} \implies \boxed{x + 2y + 1 = 0}$$

7.5　Distance from a Point to a Line　点到直线距离

Given a line in general form $Ax + By + C = 0$ and a point $P_0(x_0, y_0)$, the perpendicular distance from $P_0$ to the line is:

$$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$$

The absolute value in the numerator ensures the distance is always non-negative regardless of which side of the line $P_0$ lies on.

Distance between two parallel lines

For parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ (same $A, B$, different constant), the distance between them is:

$$d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$$

Derivation: pick any point on the first line and apply the point-to-line formula for the second.

🔑 Get general form first. The formula requires $Ax + By + C = 0$. If the line is given as $y = kx + b$, convert it to $kx - y + b = 0$ before substituting.

Worked Example 7.5.A

Find the distance from $P(3, -2)$ to the line $4x - 3y + 1 = 0$.

Solution.

Here $A = 4$, $B = -3$, $C = 1$, and $(x_0, y_0) = (3, -2)$.

$$d = \frac{|4(3) + (-3)(-2) + 1|}{\sqrt{4^2 + (-3)^2}} = \frac{|12 + 6 + 1|}{\sqrt{16 + 9}} = \frac{|19|}{\sqrt{25}} = \frac{19}{5}$$

$$\boxed{d = \frac{19}{5}}$$

Worked Example 7.5.B

Find the distance between the parallel lines $2x + y - 3 = 0$ and $2x + y + 7 = 0$.

Solution.

Same $A = 2$, $B = 1$; constants $C_1 = -3$ and $C_2 = 7$.

$$d = \frac{|-3 - 7|}{\sqrt{4 + 1}} = \frac{10}{\sqrt{5}} = \frac{10\sqrt{5}}{5} = 2\sqrt{5}$$

$$\boxed{d = 2\sqrt{5}}$$

7.6　Intersection of Lines　直线交点

Two distinct non-parallel lines intersect at exactly one point. To find it, solve the system formed by their two equations simultaneously.

Method — elimination or substitution

1. Write both lines in general form (or slope-intercept).
2. Eliminate one variable by adding/subtracting multiples of the two equations.
3. Solve for the remaining variable, then back-substitute.

If the system has no solution, the lines are parallel (including possibly identical). If it has infinitely many solutions, the equations represent the same line.

Worked Example 7.6.A

Find the intersection of $\ell_1: 2x + y - 5 = 0$ and $\ell_2: x - 2y + 5 = 0$.

Solution.

From $\ell_1$: $y = 5 - 2x$. Substitute into $\ell_2$:

$$x - 2(5 - 2x) + 5 = 0 \implies x - 10 + 4x + 5 = 0 \implies 5x = 5 \implies x = 1$$

Back-substitute: $y = 5 - 2(1) = 3$.

$$\boxed{\text{Intersection: } (1,\ 3)}$$

Check: $2(1) + 3 - 5 = 0$ ✓ and $1 - 2(3) + 5 = 0$ ✓

Worked Example 7.6.B — Concurrent lines

Three lines are concurrent (all pass through one point) if and only if their intersection point (found from any two of the lines) also satisfies the third.

Lines: $\ell_1: x + y = 4$, $\ell_2: 2x - y = 2$, $\ell_3: x - 2y + 2 = 0$.

Intersection of $\ell_1$ and $\ell_2$: add them → $3x = 6 \implies x = 2, y = 2$.

Check $\ell_3$: $2 - 2(2) + 2 = 0$ ✓. The three lines are concurrent at $(2, 2)$.

Try it!　自测练习

Q1. Find the distance between $A(-2, 1)$ and $B(4, 9)$.

Q2. A line has inclination angle $\alpha = 120°$. What is its slope?

Q3. Find the equation of the line with $x$-intercept $3$ and $y$-intercept $-2$. Write in general form.

Q4. Are the lines $3x - 2y + 1 = 0$ and $6x - 4y + 7 = 0$ parallel, perpendicular, or neither?

Q5. Find the distance from the origin $(0, 0)$ to the line $5x - 12y + 26 = 0$.

📌 Chapter summary

Topic	Key formula / rule
Distance between points	$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
Midpoint	$\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$
Slope from angle	$k = \tan\alpha$, with $\alpha \in [0°, 180°)$; vertical line: $k$ undefined
Point-slope form	$y - y_0 = k(x - x_0)$ — most flexible
General form	$Ax + By + C = 0$ — universal, works for vertical lines
Parallel	$k_1 = k_2$ (and lines are distinct)
Perpendicular	$k_1 \cdot k_2 = -1$
Point-to-line distance	$d = \dfrac{|Ax_0 + By_0 + C|}{\sqrt{A^2+B^2}}$
Parallel-line distance	$d = \dfrac{|C_1 - C_2|}{\sqrt{A^2+B^2}}$
Intersection	Solve the two-equation system simultaneously

What's next → Unit 8 extends these ideas to circles: the standard equation $(x-a)^2 + (y-b)^2 = r^2$ is essentially the distance formula applied to all points equidistant from the centre.