PracticeVideosBlogFlashcards

9.1 Equations of Straight Lines and Positional Relationships

Key Takeaways

  • Every straight line in Cartesian coordinates has an angle of inclination α∈[0,π)\alpha \in [0, \pi) and, when non-vertical (α≠π2\alpha \neq \frac{\pi}{2}), a unique slope k=tan⁡αk = \tan \alpha.

  • The five standard representations of a straight line are Point-Slope, Slope-Intercept, Two-Point, Intercept, and General Form (Ax+By+C=0Ax + By + C = 0 with A2+B2≠0A^2 + B^2 \neq 0). General Form is the only form capable of representing every possible line without exception.

  • For two non-vertical lines l1:y=k1x+b1l_1: y = k_1 x + b_1 and l2:y=k2x+b2l_2: y = k_2 x + b_2, l1∥l2  ⟺  k1=k2l_1 \parallel l_2 \iff k_1 = k_2 and b1≠b2b_1 \neq b_2; l1⊥l2  ⟺  k1k2=−1l_1 \perp l_2 \iff k_1 k_2 = -1. In General Form, l1∥l2  ⟺  A1B2−A2B1=0l_1 \parallel l_2 \iff A_1 B_2 - A_2 B_1 = 0 and A1C2−A2C1≠0A_1 C_2 - A_2 C_1 \neq 0, while l1⊥l2  ⟺  A1A2+B1B2=0l_1 \perp l_2 \iff A_1 A_2 + B_1 B_2 = 0.

  • The perpendicular distance from point P(x0,y0)P(x_0, y_0) to line Ax+By+C=0Ax + By + C = 0 is d=∣Ax0+By0+C∣A2+B2d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}, and the distance between parallel lines Ax+By+C1=0Ax + By + C_1 = 0 and Ax+By+C2=0Ax + By + C_2 = 0 is d=∣C1−C2∣A2+B2d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}.

  • Symmetry problems across a mirror line l:Ax+By+C=0l: Ax + By + C = 0 are solved by setting up a two-equation linear system enforcing that the segment's midpoint lies on ll and the segment is perpendicular to ll.

Last updated: July 2026

9.1 Equations of Straight Lines and Positional Relationships

Analytic geometry bridges algebra and geometry by translating geometric figures into algebraic equations. In the Gaokao Mathematics examination, the straight line serves as the fundamental linear building block. Mastering the analytical properties of straight lines—their inclinations, slopes, structural equation forms, positional relationships, distance metrics, and reflection properties—is essential for solving both standalone coordinate geometry items and complex comprehensive conic section problems.


1. Inclination Angle and Slope of a Line

Definition of Inclination Angle

In a rectangular Cartesian coordinate system (xOy)(xOy), when a straight line ll intersects the xx-axis, the angle α\alpha formed by rotating the positive xx-axis counterclockwise around the intersection point to coincide with the line ll is called the inclination angle (倾斜角) of the line ll.

  • By convention, when a line ll is parallel to or coincides with the xx-axis, its inclination angle is defined as α=0∘\alpha = 0^\circ (00 radians).
  • The range of the inclination angle α\alpha for any line in the coordinate plane is strictly restricted to: α∈[0∘,180∘)orα∈[0,π)\alpha \in [0^\circ, 180^\circ) \quad \text{or} \quad \alpha \in [0, \pi)

Definition of Slope

When the inclination angle α≠90∘\alpha \neq 90^\circ (α≠π2\alpha \neq \frac{\pi}{2}), the tangent value of the inclination angle is defined as the slope (斜率) of the line ll, denoted by kk:

k=tan⁡αk = \tan \alpha
  • When α∈[0∘,90∘)\alpha \in [0^\circ, 90^\circ), k≥0k \ge 0, and kk increases monotonically from 00 to +∞+\infty as α\alpha increases from 0∘0^\circ toward 90∘90^\circ.
  • When α=90∘\alpha = 90^\circ, tan⁡90∘\tan 90^\circ is undefined; hence vertical lines have no defined slope.
  • When α∈(90∘,180∘)\alpha \in (90^\circ, 180^\circ), k<0k < 0, and kk increases monotonically from −∞-\infty to 0−0^- as α\alpha increases from 90∘90^\circ toward 180∘180^\circ.

Two-Point Slope Formula

If a non-vertical line ll passes through two distinct points P1(x1,y1)P_1(x_1, y_1) and P2(x2,y2)P_2(x_2, y_2) where x1≠x2x_1 \neq x_2, the slope kk is given by:

k=y2−y1x2−x1=y1−y2x1−x2k = \frac{y_2 - y_1}{x_2 - x_1} = \frac{y_1 - y_2}{x_1 - x_2}

Proof: Let the directional vector of line ll be v⃗=P1P2⃗=(x2−x1,y2−y1)\vec{v} = \vec{P_1 P_2} = (x_2 - x_1, y_2 - y_1). By definition of vector direction, tan⁡α=ΔyΔx=y2−y1x2−x1\tan \alpha = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}.


2. The Five Standard Forms of Straight Line Equations

Depending on the given geometric parameters (such as points, slopes, or intercepts), a line can be expressed in five distinct algebraic forms:

FormEquationParametersApplication & Constraints
Point-Slope Formy−y0=k(x−x0)y - y_0 = k(x - x_0)Point (x0,y0)(x_0, y_0), Slope kkCannot represent vertical lines (x=x0x = x_0).
Slope-Intercept Formy=kx+by = kx + bSlope kk, yy-intercept bbCannot represent vertical lines. Frequently used in functional analysis.
Two-Point Formy−y1y2−y1=x−x1x2−x1\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}Points (x1,y1),(x2,y2)(x_1, y_1), (x_2, y_2)Requires x1≠x2x_1 \neq x_2 and y1≠y2y_1 \neq y_2. Cannot represent lines parallel to axes.
Intercept Formxa+yb=1\frac{x}{a} + \frac{y}{b} = 1xx-intercept aa, yy-intercept bbRequires a≠0a \neq 0 and b≠0b \neq 0. Cannot represent lines through the origin or parallel to axes.
General FormAx+By+C=0Ax + By + C = 0Coefficients A,B,CA, B, C with A2+B2≠0A^2 + B^2 \neq 0Universal form. Represents every straight line in the 2D Cartesian plane without exception.

Conversion to General Form

In Gaokao final submissions, final line equations should always be simplified into General Form Ax+By+C=0Ax + By + C = 0 where A,B,CA, B, C are simplified coprime integers and A≥0A \ge 0.

  • Slope of Ax+By+C=0Ax + By + C = 0 (when B≠0B \neq 0): k=−ABk = -\frac{A}{B}.
  • xx-intercept (when A≠0A \neq 0): x=−CAx = -\frac{C}{A}; yy-intercept (when B≠0B \neq 0): y=−CBy = -\frac{C}{B}.
  • Normal vector of the line: n⃗=(A,B)\vec{n} = (A, B); Directional vector: v⃗=(−B,A)\vec{v} = (-B, A) or (B,−A)(B, -A).

3. Positional Relationships Between Two Lines

Let two lines be given by l1:A1x+B1y+C1=0l_1: A_1 x + B_1 y + C_1 = 0 and l2:A2x+B2y+C2=0l_2: A_2 x + B_2 y + C_2 = 0.

Parallelism Condition (l1∥l2l_1 \parallel l_2)

Two lines are parallel if and only if their directional vectors are collinear while they do not coincide:

l1∥l2  ⟺  A1B2−A2B1=0andA1C2−A2C1≠0l_1 \parallel l_2 \iff A_1 B_2 - A_2 B_1 = 0 \quad \text{and} \quad A_1 C_2 - A_2 C_1 \neq 0

If expressed in slope-intercept form y=k1x+b1y = k_1 x + b_1 and y=k2x+b2y = k_2 x + b_2:

l1∥l2  ⟺  k1=k2andb1≠b2l_1 \parallel l_2 \iff k_1 = k_2 \quad \text{and} \quad b_1 \neq b_2

Warning: If A1B2−A2B1=0A_1 B_2 - A_2 B_1 = 0 and A1C2−A2C1=0A_1 C_2 - A_2 C_1 = 0, the two lines are coincident (重合), not parallel.

Perpendicularity Condition (l1⊥l2l_1 \perp l_2)

Two lines are perpendicular if and only if their normal vectors n⃗1=(A1,B1)\vec{n}_1 = (A_1, B_1) and n⃗2=(A2,B2)\vec{n}_2 = (A_2, B_2) are orthogonal (n⃗1⋅n⃗2=0\vec{n}_1 \cdot \vec{n}_2 = 0):

l1⊥l2  ⟺  A1A2+B1B2=0l_1 \perp l_2 \iff A_1 A_2 + B_1 B_2 = 0

If both lines have defined slopes k1k_1 and k2k_2:

l1⊥l2  ⟺  k1k2=−1l_1 \perp l_2 \iff k_1 k_2 = -1

(Note: The general form A1A2+B1B2=0A_1 A_2 + B_1 B_2 = 0 handles vertical/horizontal line pairs naturally without encountering division by zero.)

Inclination Angle Between Two Intersecting Lines

Let θ\theta be the acute or right angle between l1l_1 and l2l_2 (θ∈(0∘,90∘]\theta \in (0^\circ, 90^\circ]):

cos⁡θ=∣n⃗1⋅n⃗2∣∣n⃗1∣∣n⃗2∣=∣A1A2+B1B2∣A12+B12A22+B22\cos \theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|} = \frac{|A_1 A_2 + B_1 B_2|}{\sqrt{A_1^2 + B_1^2} \sqrt{A_2^2 + B_2^2}}

If slopes k1,k2k_1, k_2 exist and k1k2≠−1k_1 k_2 \neq -1:

tan⁡θ=∣k2−k11+k1k2∣\tan \theta = \left| \frac{k_2 - k_1}{1 + k_1 k_2} \right|

4. Distance Formulas and Distance Geometry

Point-to-Line Distance Formula

The perpendicular distance dd from a point P(x0,y0)P(x_0, y_0) to a line l:Ax+By+C=0l: Ax + By + C = 0 is:

d=∣Ax0+By0+C∣A2+B2d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}

Rigorous Proof: Let Q(x1,y1)Q(x_1, y_1) be the projection of P(x0,y0)P(x_0, y_0) onto ll. The vector QP⃗\vec{QP} is parallel to the normal vector n⃗=(A,B)\vec{n} = (A, B). Since QQ lies on ll, Ax1+By1+C=0  ⟹  Ax1+By1=−CA x_1 + B y_1 + C = 0 \implies A x_1 + B y_1 = -C. The scalar projection of P0P⃗\vec{P_0 P} (where P0P_0 is any point on ll) onto n⃗\vec{n} gives:

d=∣n⃗⋅P0P⃗∣∣n⃗∣=∣A(x0−x1)+B(y0−y1)∣A2+B2=∣Ax0+By0−(Ax1+By1)∣A2+B2=∣Ax0+By0+C∣A2+B2d = \frac{|\vec{n} \cdot \vec{P_0 P}|}{|\vec{n}|} = \frac{|A(x_0 - x_1) + B(y_0 - y_1)|}{\sqrt{A^2 + B^2}} = \frac{|Ax_0 + By_0 - (Ax_1 + By_1)|}{\sqrt{A^2 + B^2}} = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}

Distance Between Parallel Lines

For two parallel lines l1:Ax+By+C1=0l_1: Ax + By + C_1 = 0 and l2:Ax+By+C2=0l_2: Ax + By + C_2 = 0 (C1≠C2C_1 \neq C_2):

d=∣C1−C2∣A2+B2d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

Requirement: Both line equations must be normalized to share identical coefficients for xx and yy (AA and BB) before applying this formula.

Family of Lines Passing Through an Intersection (Line Pencil)

Given two intersecting lines l1:A1x+B1y+C1=0l_1: A_1 x + B_1 y + C_1 = 0 and l2:A2x+B2y+C2=0l_2: A_2 x + B_2 y + C_2 = 0, the equation:

A1x+B1y+C1+λ(A2x+B2y+C2)=0(λ∈R)A_1 x + B_1 y + C_1 + \lambda (A_2 x + B_2 y + C_2) = 0 \quad (\lambda \in \mathbb{R})

represents the family of all straight lines passing through the intersection point of l1l_1 and l2l_2, excluding line l2l_2 itself.


5. Reflection and Symmetry Problems in Cartesian Coordinates

Symmetry is a frequently tested topic in Gaokao analytic geometry. Two core scenarios exist:

Point Reflection Across a Line

To find the reflection point P′(x′,y′)P'(x', y') of P(x0,y0)P(x_0, y_0) across mirror line l:Ax+By+C=0l: Ax + By + C = 0:

  1. Perpendicularity Condition: Slope of PP′PP' is perpendicular to line ll, so PP′⃗∥n⃗=(A,B)\vec{PP'} \parallel \vec{n} = (A, B), giving y′−y0x′−x0=BA\frac{y' - y_0}{x' - x_0} = \frac{B}{A} (or (x′−x0)B−(y′−y0)A=0(x' - x_0) B - (y' - y_0) A = 0).
  2. Midpoint Condition: The midpoint M(x0+x′2,y0+y′2)M\left(\frac{x_0 + x'}{2}, \frac{y_0 + y'}{2}\right) of segment PP′PP' lies on ll: A(x0+x′2)+B(y0+y′2)+C=0A\left(\frac{x_0 + x'}{2}\right) + B\left(\frac{y_0 + y'}{2}\right) + C = 0

Solving this 2×22 \times 2 system uniquely yields coordinates (x′,y′)(x', y').

Line Reflection Across a Line

To find the reflection of line l1l_1 across mirror line ll:

  • Find the intersection point I=l1∩lI = l_1 \cap l (if l1∦ll_1 \nparallel l).
  • Pick a non-intersection point P∈l1P \in l_1, calculate its point reflection P′P' across ll.
  • The reflected line is the unique line passing through II and P′P'.
Loading diagram...
Classification and Geometric Metrics of Straight Lines
Test Your Knowledge

Given two lines l1:ax+2y+1=0l_1: a x + 2 y + 1 = 0 and l2:x+(a+1)y+1=0l_2: x + (a+1) y + 1 = 0. What is the necessary and sufficient condition for l1∥l2l_1 \parallel l_2?

A

a=−2a = -2

B

a=1a = 1

C

a=1a = 1 or a=−2a = -2

D

a=−1a = -1

Test Your Knowledge

What is the distance between the two parallel straight lines l1:3x−4y+5=0l_1: 3x - 4y + 5 = 0 and l2:6x−8y−10=0l_2: 6x - 8y - 10 = 0?

A

155\frac{15}{5}

B

12\frac{1}{2}

C

22

D

44

Test Your Knowledge

Find the coordinates of the reflection point P′P' when the point P(3,1)P(3, 1) is reflected across the line l:x−y+2=0l: x - y + 2 = 0.

A

(−1,5)(-1, 5)

B

(−1,4)(-1, 4)

C

(1,5)(1, 5)

D

(0,4)(0, 4)

Sections you finish are checked off in the contents.