Year 1 Additional Mathematics: Section 5 - Complete Study Guide on Straight Lines

Introduction to Straight Lines, Geometric Reasoning, and Measurement

Learning about straight lines is integral to understanding spatial reasoning, specifically in directing and planning routes for efficient travel and transportation. These concepts are foundational in architecture and various art forms. This section details the properties of straight lines, methods for determining orientation (parallel or perpendicular), midpoint calculations, dividing lines into ratios (internally and externally), generating equations, and measuring angles of intersection.

Learning Objectives

By the end of this study guide, the student will be able to:

• Describe properties of lines, including parallel and perpendicular relationships and midpoints.
• Calculate and generalize the midpoint of a line segment given two specific points.
• Divide line segments internally or externally using specific ratios.
• Apply the gradient formula to derive equations of straight lines in various forms.
• Use algebraic manipulations to find equations for parallel lines, perpendicular lines, and perpendicular bisectors.
• Calculate the shortest perpendicular distance from an external point to a line.
• Determine the acute angles between intersecting lines with tools like GeoGebra.

Key Concepts and Verbatim Definitions

• Straight Line: A line with no curve; the shortest distance between two points. It is one-dimensional, has no width, and can be extended infinitely in both directions.
• Parallel Lines: Straight lines that are always equidistant and never meet, no matter how far they are extended.
• Perpendicular Lines: Straight lines that meet or intersect at an angle of exactly $90^{\circ}$.
• Midpoint: A point that divides a straight line into two equal parts; it is located at an equal distance from both endpoints.
• Acute Angle: An angle measuring greater than $0^{\circ}$ but less than $90^{\circ}$.
• Right Angle: An angle measuring exactly $90^{\circ}$.
• Obtuse Angle: An angle measuring greater than $90^{\circ}$ but less than $180^{\circ}$.
• Gradient (Slope): The measure of a line's steepness and direction calculated as the ratio of vertical change (rise) to horizontal change (run).

Fundamental Properties of Lines

Straight lines possess specific traits that classify them as such. They can be horizontal, vertical, slanted, parallel, or perpendicular.

General Attributes

• Joins two points via the shortest possible distance.
• Contains no curves.
• One-dimensional with zero width.

Parallel and Perpendicular Lines

• Parallel Lines: These are always the same distance apart. Examples include the opposite ends of a goalpost, railway tracks, the edges of a ruler, and the white lines of a zebra crossing.
• Perpendicular Lines: All angles at the point of intersection are $90^{\circ}$. Examples include "T" junctions on roads and the corners of a football pitch.

Distance Constraints

1. Distances are always positive.
2. Distance is zero only if the two points coincide.
3. The distance from point $P$ to $Q$ is identical to the distance from $Q$ to $P$.

Distance Formula and Calculations

The distance ($d$) between two points $(x_1, y_1)$ and $(x_2, y_2)$ is generalized as: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Worked Examples

Example 1: Endpoints $U(-7, 23)$ and $V(6, 18)$

1. Substitute values: $|UV| = \sqrt{(6 - (-7))^2 + (18 - 23)^2}$
2. Simplify: $|UV| = \sqrt{13^2 + (-5)^2} = \sqrt{169 + 25}$
3. Final result: $|UV| = \sqrt{194} \approx 13.93$

Example 2: Points $N(12, 5)$ and $R(-3, -6)$

1. Substitute values: $|NR| = \sqrt{(-3 - 12)^2 + (-6 - 5)^2}$
2. Simplify: $|NR| = \sqrt{(-15)^2 + (-11)^2} = \sqrt{225 + 121}$
3. Final result: $|NR| = \sqrt{346} \approx 18.60$

Midpoint of a Line Segment

The midpoint ($M$) is the point of equal distance from both ends, dividing the segment into two equal parts.

Generalization:$M = \left( \frac{1}{2}(x_1 + x_2) , \frac{1}{2}(y_1 + y_2) \right)$

Worked Examples

Example 3: Line $KG$ with $K(16, 24)$ and $G(-8, 10)$

1. Substitute: $M = \left( \frac{1}{2}(16 + (-8)) , \frac{1}{2}(24 + 10) \right)$
2. Simplify: $M = \left( \frac{1}{2}(8), \frac{1}{2}(34) \right)$
3. Result: $M(4, 17)$

Example 4: Line $RP$ with $R(1, 17)$ and $P(0, 4)$

1. Substitute: $M = \left( \frac{1}{2}(1 + 0) , \frac{1}{2}(17 + 4) \right)$
2. Result: $M(0.5, 10.5)$

Example 5: Finding an endpoint ($W$ is the midpoint of $DV$) Given $D(-5, 4)$ and $W(-2, 1)$, find $V(x_v, y_v)$.

1. Equation for $x$: $-2 = \frac{1}{2}(-5 + x_v) \rightarrow -4 = -5 + x_v \rightarrow x_v = 1$
2. Equation for $y$: $1 = \frac{1}{2}(4 + y_v) \rightarrow 2 = 4 + y_v \rightarrow y_v = -2$
3. Result: $V(1, -2)$

Division of a Line Segment in a Given Ratio

Lines can be divided into specific parts using ratios either internally (between endpoints) or externally (extending outside the segment).

Internal Division

If point $P$ divides the segment joining $K(x_1, y_1)$ and $E(x_2, y_2)$ internally in the ratio $m:n$, then: $P = \left( \frac{mx_2 + nx_1}{m+n} , \frac{my_2 + ny_1}{m+n} \right)$

• A 1:1 ratio is equivalent to finding the midpoint.

External Division

If point $Q$ divides the segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ externally in the ratio $m:n$, then: $Q = \left( \frac{mx_2 - nx_1}{m-n} , \frac{my_2 - ny_1}{m-n} \right)$

• Note: In external division, you do not always negate $n$; instead, negate the smaller number in the ratio during calculation if needed for directionality.

Worked Example 6: Internal and External Division

Points $G$ and $K$ divide line $FH$ with $F(13, 9)$ and $H(6, -17)$ in ratio $11:3$.

• Internal ($G$): $G = \left( \frac{11(6) + 3(13)}{11+3} , \frac{11(-17) + 3(9)}{11+3} \right) = G(7.5, -11.43)$
• External ($K$): $K = \left( \frac{11(6) - 3(13)}{11-3} , \frac{11(-17) - 3(9)}{11-3} \right) = K(3.375, -26.75)$

Equations of Straight Lines

The equation of a line can be derived by considering an arbitrary point $P(x, y)$ on a line segment between points $A(x_1, y_1)$ and $B(x_2, y_2)$.

Gradient (Slope)

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

Standard Forms

1. Point-Slope Form: $y - y_1 = m(x - x_1)$
2. Slope-Intercept Form: $y = mx + c$ (where $c = -mx_1 + y_1$)
3. Horizontal Lines: $m = 0$, so $y = c$.
4. Vertical Lines: $m = \text{infinite}$, so $x = c$.

Relations Between Gradients

1. Parallel Lines: Have identical gradients ($m_1 = m_2$).
2. Perpendicular Lines: The product of their gradients is $-1$ ($m_1 \times m_2 = -1$, or $m_2 = -\frac{1}{m_1}$).

Worked Example 8: Comprehensive Line Equations

Points $S(4, -2)$ and $T(8, 4)$. a) Equation of line $ST$:

• $m_1 = \frac{4 - (-2)}{8 - 4} = \frac{6}{4} = 1.5$
• $y - (-2) = \frac{3}{2}(x - 4)$, expanding to $2y + 4 = 3x - 12$
• Result: $3x - 2y - 16 = 0$

b) Perpendicular line $JK$ through $W(-7, 5)$:

• $m_2 = -\frac{1}{1.5} = -\frac{2}{3}$
• $y - 5 = -\frac{2}{3}(x + 7) \rightarrow 3y - 15 = -2x - 14$
• Result: $2x + 3y - 1 = 0$

c) Parallel line $ED$ to $JK$ through $L(2, 9)$:

• $m_3 = m_2 = -\frac{2}{3}$
• $y - 9 = -\frac{2}{3}(x - 2) \rightarrow 3y - 27 = -2x + 4$
• Result: $2x + 3y - 31 = 0$

Shortest Distance Between a Point and a Line

The shortest distance ($D$) between a point $(x_1, y_1)$ and a line $ax + by + c = 0$ is the perpendicular distance.

Distance Formula ($D$):$D = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$

Worked Examples

Example 9: Point $J(1, 5)$ and line $5x + 12y + 7 = 0$

1. Substitute: $D = \frac{|5(1) + 12(5) + 7|}{\sqrt{5^2 + 12^2}}$
2. Simplify: $D = \frac{|5 + 60 + 7|}{\sqrt{25 + 144}} = \frac{72}{\sqrt{169}} = \frac{72}{13}$
3. Result: $D \approx 5.54 \text{ units}$

Example 10: Cell Tower at $C(6, -2)$ and Main Road $3x - 4y = 12$

1. Standard Form: $3x - 4y - 12 = 0$
2. Substitute: $D = \frac{|3(6) - 4(-2) - 12|}{\sqrt{3^2 + 4^2}} = \frac{|18 + 8 - 12|}{\sqrt{9 + 16}} = \frac{14}{5}$
3. Result: $D = 2.8$

Acute Angles Between Two Intersecting Lines

To find the acute angle ($\theta$) between two lines with gradients $m_1$ and $m_2$: $\tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$

Worked Example 13

Slopes: $4$ and $\frac{2}{7}$.

1. Substitute: $\tan(\theta) = \left| \frac{4 - \frac{2}{7}}{1 + 4(\frac{2}{7})} \right| = \left| \frac{\frac{26}{7}}{\frac{15}{7}} \right| = \frac{26}{15}$
2. Find $\theta$: $\theta = \tan^{-1}\left(\frac{26}{15}\right) \approx 60.02^{\circ}$

Intersection of Straight Lines

The point of intersection is found by equating the equations of two lines ($y = m_1x + c_1$ and $y = m_2x + c_2$) and solving for $x$, then substituting back for $y$.

Worked Example 12: Route Planning in Kasoa

• Route A: Through $(2, 5)$ and $(8, 17)$. Gradient $= 2$. Equation: $y = 2x + 1$.
• Route B: Through $(3, 20)$ and $(9, 2)$. Gradient $= -3$. Equation: $y = -3x + 29$.

1. Equate: $2x + 1 = -3x + 29$
2. Solve for $x$: $5x = 28 \rightarrow x = \frac{28}{5}$
3. Find $y$: $y = 2\left(\frac{28}{5}\right) + 1 = \frac{56}{5} + \frac{5}{5} = \frac{61}{5}$
4. Intersection Point: $\left( \frac{28}{5}, \frac{61}{5} \right)$

Questions and Discussion

Review Questions Summary:

1. Calculate distance between $F(3, 7)$ and $Y(-2, 8)$.
2. Find midpoint for a bridge pillar between river banks $R(3, 5)$ and $G(11, 7)$.
3. Given $A(3, 2)$, $B(8, 11)$, $C(14, 5)$: Divide $AB$ and $BC$ internally, find the resulting line equation, and analyze triangle properties.
4. Find ratio variable '$n$' for external point $M(12, -5)$ dividing $AB$ where $A(-6, 4)$ and $B(2, 0)$.
5. Find line equation through $D(25, 6)$ and $C(29, -14)$.
6. Identify constants $a$ and $b$ in $3y = ax + b$ passing through $(4, 2)$ and $(-8, -2)$.
7. Perpendicular line through $(-5, 6)$ to $5x - 3y - 18 = 0$.
8. Parallel line to $3x + 5y = 108$ through $(5, 10)$.
9. Acute angle between $y = 2x + 10$ and $y = -5x + 4$.
10. Perpendicular length from $B(-1, -7)$ to line through $E(6, -4)$ and $Y(9, -5)$.
11. Determine the equation for points $(3, 6)$ and $(1, 2)$ using point-slope form.
12. City Layout Case Study:
    • Main Street: $y = 2x + 3$.
    • Nana Botwe Street: Parallel to Main, through $(4, 1)$.
    • Bebianiha Street: Perpendicular to Main, through $(1, 5)$.
    • Park Road: Parallel to Nana Botwe Street, through $(2, 3)$.
    • Find playground coordinates where Bebianiha Street intersects Park Road.
13. Discuss geometric relationships between four roads: Road A ($y = -12x + 10$), Road B ($-x + 12y = 56$), Road C ($12x = 76 - y$), and Road D ($12y = x - 52$).

Glossary

• Straight line: Shortest distance between two points, extending infinitely.
• Perpendicular lines: Intersect at $90^{\circ}$; product of slopes is $-1$.
• Parallel lines: Never intersect; slopes are equal, y-intercepts differ.
• Gradient: Ratio of vertical change ($\Delta y$) to horizontal change ($\Delta x$).
• Acute angle: $0^{\circ} < \theta < 90^{\circ}$.
• Obtuse angle: $90^{\circ} < \theta < 180^{\circ}$.
• Right angle: Exactly $90^{\circ}$.