Exhaustive Guide to Straight Line Geometry

Chapter 11: Fundamentals of Gradient and Inclination

• Definition of Inclination Angle (直线 l 的倾斜角):   - The inclination angle, denoted as $\theta$, is the angle formed by a straight line $l$ and the positive direction of the x-axis.

• Definition of Slope/Gradient ($m$):   - The gradient of a line is defined as the tangent of its inclination angle: $m = \tan(\theta)$.

• Cases for Slope Based on the Angle of Inclination:   - Acute Angle (锐角): If the line forms an acute angle with the positive x-axis (0^\circ < \theta < 90^\circ), the gradient is positive (m > 0).   - Obtuse Angle (钝角): If the line forms an obtuse angle with the positive x-axis (90^\circ < \theta < 180^\circ), the gradient is negative (m < 0).   - Parallel to x-axis: If the line is parallel to the x-axis, the inclination angle $\theta = 0^\circ$. The slope $m = \tan(0^\circ) = 0$.   - Perpendicular to x-axis: If the line is perpendicular to the x-axis, the inclination angle $\theta = 90^\circ$. The slope $m = \tan(90^\circ)$ is undefined (无意义/不存在).

Gradient Formula for Two Points

• Formula: The gradient $m$ of a line passing through two distinct points $A(x_1, y_1)$ and $B(x_2, y_2)$ is given by the ratio of the change in y ($\Delta y$) to the change in x ($\Delta x$):   - $m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}$

• Example 1: Find the gradient of the line passing through $A(-2, 1)$ and $B(5, 3)$.   - Solution: $m_{AB} = \frac{3 - 1}{5 - (-2)} = \frac{2}{7}$

• Example 2: Find the gradient of the line passing through $(6, -2)$ and $(1, 3)$.   - Solution: $m = \frac{3 - (-2)}{1 - 6} = \frac{5}{-5} = -1$

• Classroom Exercise 1: Points $(-3, 6)$ and $(-1, -3)$.   - ANS: $-\frac{9}{2}$

• Classroom Exercise 2: Points $(-3, 4)$ and $(5, 4)$.   - ANS: $0$

• Classroom Exercise 3: Points $(1, -2)$ and $(3, 8)$.   - ANS: $5$

Parallel Lines and Collinearity

• Parallel Lines (两条直线的平行):   - Two distinct lines $l_1$ and $l_2$ with gradients $m_1$ and $m_2$ are parallel ($l_1 \parallel l_2$) if and only if their gradients are equal: $m_1 = m_2$.

• Three Points are Collinear (三点共线):   - Points $A$, $B$, and $C$ lie on the same straight line if the gradients between any two pairs of points are equal: $m_{AB} = m_{BC} = m_{AC}$.

• Example 3: If $A(-4, -3)$, $B(-2, 0)$, and $C(2, k)$ are collinear, find $k$.   - Solution: Setting $m_{AB} = m_{BC}$.   - $m_{AB} = \frac{0 - (-3)}{-2 - (-4)} = \frac{3}{2}$   - $m_{BC} = \frac{k - 0}{2 - (-2)} = \frac{k}{4}$   - Equating: $\frac{3}{2} = \frac{k}{4} \implies 2k = 12 \implies k = 6$

• Classroom Exercise 4: Prove $A(-1, 5)$, $B(2, 1)$, and $C(5, -3)$ are collinear.

• Example 4: Line connecting $A(-3, 4)$ and $B(-1, 1)$ is parallel to the line connecting $C(0, 5)$ and $D(2, p)$. Find $p$.   - Solution: $m_{AB} = \frac{1 - 4}{-1 - (-3)} = \frac{-3}{2}$   - $m_{CD} = \frac{p - 5}{2 - 0} = \frac{p - 5}{2}$   - Setting $m_{AB} = m_{CD} \implies -\frac{3}{2} = \frac{p - 5}{2} \implies -3 = p - 5 \implies p = 2$

• Classroom Exercise 5: Prove $A(-5, 2)$, $B(-2, 4)$, $C(6, 0)$, and $D(3, -2)$ are the vertices of a parallelogram (by showing opposite sides are parallel).

Perpendicular Lines

• Relationship of Gradients (两条直线的垂直):   - If two lines $l_1$ and $l_2$ with gradients $m_1$ and $m_2$ are perpendicular ($l_1 \perp l_2$), then the product of their gradients is $-1$: $m_1 \times m_2 = -1$.

• Geometric Proof Sketch:   - An exterior angle of a triangle equals the sum of interior angles: $\theta_2 = 90^\circ + \theta_1$.   - $\tan(\theta_2) = \tan(90^\circ + \theta_1) = -\cot(\theta_1) = -\frac{1}{\tan(\theta_1)}$.   - Therefore, $m_2 = -\frac{1}{m_1}$.

• Example 5: Line $AB \perp CD$. If gradient of $AB$ is $3$, find gradient of $CD$.   - Solution: $3 \times m_{CD} = -1 \implies m_{CD} = -\frac{1}{3}$.

• Classroom Exercise 6: Line $PQ \perp RS$. Given gradient of $RS$ is $-2$, find gradient of $PQ$.   - ANS: $\frac{1}{2}$

• Example 6: Given $A(-5, 1)$, $B(-4, 8)$, $C(1, 3)$, $D(2, k)$. If $AC \perp BD$, find $k$.   - Solution: $m_{AC} = \frac{3 - 1}{1 - (-5)} = \frac{2}{6} = \frac{1}{3}$.   - $m_{BD} = \frac{k - 8}{2 - (-4)} = \frac{k - 8}{6}$.   - Condition: $(\frac{1}{3})(\frac{k - 8}{6}) = -1 \implies \frac{k - 8}{18} = -1 \implies k - 8 = -18 \implies k = -10$.

• Example 7: Given $A(-2, 5)$, $B(k, -1)$, and $C(5, 1)$. If $AB \perp BC$, find $k$.   - Solution: $m_{AB} \times m_{BC} = -1$.   - $(\frac{-1 - 5}{k - (-2)}) \times (\frac{1 - (-1)}{5 - k}) = -1 \implies (\frac{-6}{k + 2})(\frac{2}{5 - k}) = -1$.   - $\frac{-12}{(k + 2)(5 - k)} = -1 \implies (k + 2)(5 - k) = 12$.   - $5k - k^2 + 10 - 2k = 12 \implies -k^2 + 3k + 10 = 12 \implies k^2 - 3k + 2 = 0$.   - $(k - 1)(k - 2) = 0 \implies k = 1 \text{ or } k = 2$.

• Example 8 (Rectangle Proof): Prove $A(2, 2)$, $B(4, 1)$, $C(2, -3)$, and $D(0, -2)$ are vertices of a rectangle.   - Step 1: Prove opposite sides parallel. $m_{AB} = -\frac{1}{2}$, $m_{DC} = -\frac{1}{2}$; $m_{AD} = 2$, $m_{BC} = 2$.   - Step 2: Prove adjacent sides perpendicular. $m_{AD} \times m_{DC} = 2 \times (-\frac{1}{2}) = -1$.

• Example 9 (Right-Angle Triangle Proof): Prove $A(2, 1)$, $B(3, -2)$, and $C(-4, -1)$ form a right-angle triangle.   - Gradients: $m_{AB} = -3$, $m_{BC} = -\frac{1}{7}$, $m_{CA} = \frac{1}{3}$.   - Since $m_{AB} \times m_{CA} = -1$, $AB \perp CA$, establishing the right-angle.

Various Forms of Straight Line Equations

1. Point-Slope Form (点斜式)

• Equation: Required point $(x_1, y_1)$ and gradient $m$.   - $y - y_1 = m(x - x_1)$

• Example 10: Passing through $(5, 3)$ with $m = 0$.   - Solution: $y - 3 = 0(x - 5) \implies y = 3$.

• Example 11: Passing through $(3, 5)$ with gradient not existent (vertical).   - Solution: $x - 3 = 0 \implies x = 3$.

• Example 12: Passing through $(2, -3)$ with gradient $m = \frac{1}{2}$.   - Solution: $y - (-3) = \frac{1}{2}(x - 2) \implies 2y + 6 = x - 2 \implies x - 2y - 8 = 0$.

2. Slope-Intercept Form (斜截式)

• Equation: Gradient $m$ and y-intercept $c$ (intersects y-axis at $(0, c)$).   - $y = mx + c$

• Example 13: Gradient is $\frac{2}{3}$ and y-intercept is $3$.   - Solution: $y = \frac{2}{3}x + 3 \implies 3y = 2x + 9 \implies 2x - 3y + 9 = 0$.

3. Two-Point Form (两点式)

• Equation: Line passing through $A(x_1, y_1)$ and $B(x_2, y_2)$.   - $\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$

• Example 14: Passing through $A(-1, 5)$ and $B(-3, 2)$.   - Solution: $\frac{y - 5}{x - (-1)} = \frac{2 - 5}{-3 - (-1)} \implies \frac{y - 5}{x + 1} = \frac{-3}{-2} = \frac{3}{2}$.   - $2(y - 5) = 3(x + 1) \implies 2y - 10 = 3x + 3 \implies 3x - 2y + 13 = 0$.

4. Intercept Form (截距式)

• Equation: Given x-intercept $a$ (point $(a, 0)$) and y-intercept $b$ (point $(0, b)$).   - $\frac{x}{a} + \frac{y}{b} = 1$

• Example 15: x-intercept is $-5$, y-intercept is $4$.   - Solution: $\frac{x}{-5} + \frac{y}{4} = 1 \implies \frac{-4x + 5y}{20} = 1 \implies -4x + 5y = 20 \implies 4x - 5y + 20 = 0$.

General Form of the Straight Line Equation

• General Equation: $Ax + By + C = 0$ (where $A, B, C$ are constants, and $A, B$ are not both zero).

• Properties Derived from General Form:   - Slope ($m$): $m = -\frac{A}{B}$.   - Y-intercept ($c$): $c = -\frac{C}{B}$.

• Special Cases:   - If $A = 0, B \neq 0$: $By + C = 0 \implies y = -\frac{C}{B}$ (Horizontal line parallel to the x-axis).   - If $A \neq 0, B = 0$: $Ax + C = 0 \implies x = -\frac{C}{A}$ (Vertical line parallel to the y-axis).   - If $A = 0, B \neq 0, C = 0$: $y = 0$ (The x-axis).   - If $A \neq 0, B = 0, C = 0$: $x = 0$ (The y-axis).

• Example 16: Find gradient and y-intercept for $2x - 4y + 5 = 0$.   - Solution: $4y = 2x + 5 \implies y = \frac{1}{2}x + \frac{5}{4}$. Gradient = $\frac{1}{2}$; y-intercept = $\frac{5}{4}$.

Intersection and Perpendicular Bisectors

Intersection of Two Lines (两条直线的交点)

• Method: Solve the equations of the two lines simultaneously.

• Example 21: Intersection of $l_1: 3x + 4y - 2 = 0$ and $l_2: 2x + y + 2 = 0$.   - Solving via elimination: Multiply $l_2$ by $4 \implies 8x + 4y + 8 = 0$.   - Subtracting equations: $(8x + 4y + 16) - (3x + 4y - 2) = 0 \implies 5x + 18 = 0$. [Correction from transcript data: $5x + 10 = 0 \implies x = -2$].   - Sub $x = -2$ into $l_2 \implies 2(-2) + y + 2 = 0 \implies -4 + y + 2 = 0 \implies y = 2$.   - Intersection Point: $(-2, 2)$.

• Example 23 (No Intersection): Intersection of $2x - 4y + 3 = 0$ and $x - 2y = 0$.   - Finding gradients: $m_1 = \frac{1}{2}$, $m_2 = \frac{1}{2}$. Since slopes are equal and intercepts are different, the lines are parallel and have no intersection.

Perpendicular Bisector (垂直平分线/中垂线)

• Definition: A line that passes through the midpoint of a segment and is perpendicular to it.

• Steps to Find the Equation:   1. Calculate the midpoint $M(x, y)$ of segment $AB$.   2. Calculate the gradient $m_{AB}$.   3. Calculate the gradient of the bisector ($m = -\frac{1}{m_{AB}}$).   4. Use the point-slope form with Midpoint $M$ and the calculated perpendicular gradient $m$.

• Example 21: Perpendicular bisector of $PQ$ where $P(-2, 5)$ and $Q(4, 9)$.   - Midpoint $M = (\frac{-2+4}{2}, \frac{5+9}{2}) = (1, 7)$.   - $m_{PQ} = \frac{9 - 5}{4 - (-2)} = \frac{4}{6} = \frac{2}{3}$.   - Gradient of bisector $= -\frac{3}{2}$.   - Equation: $y - 7 = -\frac{3}{2}(x - 1) \implies 2y - 14 = -3x + 3 \implies 3x + 2y - 17 = 0$.

Distance Formulas

Distance from Point to Line

• Formula: The distance $d$ from point $P(x_0, y_0)$ to the line $Ax + By + C = 0$:   - $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

• Example 25: From $P(-1, 2)$ to $2x + y - 10 = 0$.   - $d = \frac{|2(-1) + 1(2) - 10|}{\sqrt{2^2 + 1^2}} = \frac{|-10|}{\sqrt{5}} = 2\sqrt{5}$.

• Example 26: Distance from $A(\frac{5}{2}, k)$ to $6x + 8y = 7$ is $4$.   - $4 = \frac{|6(\frac{5}{2}) + 8k - 7|}{\sqrt{6^2 + 8^2}} = \frac{|15 + 8k - 7|}{10} = \frac{|8 + 8k|}{10}$.   - $|8 + 8k| = 40 \implies 8 + 8k = 40$ or $8 + 8k = -40$.   - $k = 4$ or $k = -6$.

Distance Between Two Parallel Lines

• Method: Pick an arbitrary point on one line and calculate its distance to the other line.

• Example 28: Distance between $3x - 4y = 5$ and $6x - 8y = 3$.   - Pick point on $l_1 (3x - 4y - 5 = 0)$: Set $y = 0 \implies x = \frac{5}{3}$. Point is $(\frac{5}{3}, 0)$.   - Distance to $l_2 (6x - 8y - 3 = 0)$: $d = \frac{|6(\frac{5}{3}) - 8(0) - 3|}{\sqrt{6^2 + 8^2}} = \frac{|10 - 3|}{10} = \frac{7}{10}$.

• Example 29: Distance between $3x + 4y - 3 = 0$ and $6x + my + 14 = 0$ given they are parallel.   - Parallel condition: $\frac{-3}{4} = \frac{-6}{m} \implies m = 8$.   - Pick point on $l_1$: Set $y = 0 \implies x = 1$. Point is $(1, 0)$.   - Distance to $l_2 (6x + 8y + 14 = 0)$: $d = \frac{|6(1) + 8(0) + 14|}{\sqrt{6^2 + 8^2}} = \frac{20}{10} = 2$.

Questions & Discussion

• Question on Collinearity (Ex 3): Why do we set gradients equal?   - Response: If three points lie on the same line, the steepness between any two points must be identical, as a straight line has a constant slope.

• Intersection on X-axis (Ex 24/26): If lines intersect on the x-axis, what do we know about the intersection point?   - Response: The y-coordinate of the intersection point must be zero. We let the point be $(a, 0)$, substitute it into the equations, and solve for the unknown constants.

• Standard Exam Format (UEC 统考): Problems often require finding trapizoid areas, intersection points without graphing, and proving perpendicularity using the gradient product rule ($m_1 m_2 = -1$). Examples include 2015, 2018, and 2020 examination questions.