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Straight Lines

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Straight Lines

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Straight Lines

🎯 Learning Objectives

By the end of this topic, you will be able to:

Compute the slope of a line given two points or an equation
Write the equation of a line in point-slope, slope-intercept, two-point, intercept, and general forms
Determine whether two lines are parallel or perpendicular using slope conditions
Compute the angle between two intersecting lines
Calculate the distance from a point to a line, and between parallel lines
Understand the line of best fit using the sum of squared errors (SSE)

📋 Prerequisites

Coordinate Geometry — points, distance formula
Basic algebra — solving linear equations, working with fractions

📖 Core Content

2.1 The Slope of a Line

2.1.1 Intuition: Steepness

If you're hiking up a mountain trail, some sections are steep and some are gentle. The slope of a line captures exactly this idea of steepness — how much the line rises (or falls) for each unit you move horizontally. A flat road has slope 0. A staircase might have slope 0.5 (rise 1 unit for every 2 units forward). A cliff is nearly vertical (slope approaches infinity).

💡 Why this matters: Slope is the foundation of rate of change — the core idea in calculus (derivatives). In data science, the slope of a regression line tells you how much the target variable changes per unit change in a feature.

2.1.2 Definition

The slope

m

of a line passing through points

P_1(x_1, y_1)

and

P_2(x_2, y_2)

is:

m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}}

As inclination: If a line makes angle

\theta

with the positive

x

-axis:

m = \tan\theta

Key facts:

Horizontal line ( $\theta = 0^\circ$ ): $m = \tan 0^\circ = 0$
Vertical line ( $\theta = 90^\circ$ ): slope is undefined ( $\tan 90^\circ$ is undefined)
Positive slope: line goes up as $x$ increases (rising)
Negative slope: line goes down as $x$ increases (falling)

2.1.3 Worked Examples

Example 1.1 (Easy): Find the slope of the line through

(1, 2)

and

(4, 8)

. Step 1:

m = \frac{8 - 2}{4 - 1} = \frac{6}{3} = 2

\boxed{m = 2}

Example 1.2 (Medium): Find the slope of the line through

(-3, 5)

and

(2, -7)

. Step 1:

m = \frac{-7 - 5}{2 - (-3)} = \frac{-12}{5} = -2.4

\boxed{m = -\frac{12}{5}}

Example 1.3 (Edge case): Find the slope of the line through

(3, 4)

and

(3, 9)

. Step 1:

m = \frac{9 - 4}{3 - 3} = \frac{5}{0}

, which is undefined.

\boxed{\text{Slope is undefined (vertical line)}}

2.2 Equations of a Straight Line

2.2.1 Intuition: Describing a Line

Just as a point is described by its coordinates, a line is described by an equation — a condition that every point on the line satisfies. If

(x, y)

lies on the line

y = 2x + 1

, then plugging in gives

y = 2x + 1

. There are several forms, each useful in different situations: (Diagram)

2.2.2 Forms Summary

Form	Equation	When to Use
Horizontal	$y = a$	Line parallel to x-axis
Vertical	$x = a$	Line parallel to y-axis
Point-slope	$y - y_1 = m(x - x_1)$	Given a point $(x_1,y_1)$ and slope $m$
Slope-intercept	$y = mx + c$	Given slope $m$ and y-intercept $c$
Two-point	$y - y_1 = \frac{y_2-y_1}{x_2-x_1}(x - x_1)$	Given two points
Intercept	$\frac{x}{a} + \frac{y}{b} = 1$	Given x-intercept $a$ and y-intercept $b$
General	$Ax + By + C = 0$	Standard form, all lines
X-intercept form	$y = m(x - d)$	Given x-intercept $d$ and slope $m$

2.2.3 Converting Between Forms

From general form

Ax + By + C = 0

Slope: $m = -\frac{A}{B}$
y-intercept: $-\frac{C}{B}$
x-intercept: $-\frac{C}{A}$ Recipe for writing a line equation:

Identify what information you have (point + slope? two points? intercepts?)
Choose the corresponding form
Substitute the known values
Simplify to the required form

2.2.4 Worked Examples

Example 2.1 (Easy — Point-Slope): Find the equation of the line through

(2, 3)

with slope

4

. Step 1: Use point-slope:

y - y_1 = m(x - x_1)

Step 2:

y - 3 = 4(x - 2)

Step 3: Simplify:

y - 3 = 4x - 8 \implies y = 4x - 5

\boxed{y = 4x - 5}

Example 2.2 (Medium — Two-Point): Find the equation of the line through

(1, 2)

and

(4, 8)

. Step 1: Find slope:

m = \frac{8 - 2}{4 - 1} = 2

Step 2: Use point-slope with

(1, 2)

y - 2 = 2(x - 1)

Step 3: Simplify:

y - 2 = 2x - 2 \implies y = 2x

\boxed{y = 2x}

Example 2.3 (Medium — Intercept): Find the equation of the line with x-intercept

3

and y-intercept

4

. Step 1:

a = 3

b = 4

\frac{x}{3} + \frac{y}{4} = 1

Step 2: Multiply by

12

4x + 3y = 12

\boxed{4x + 3y = 12}

Example 2.4 (Hard — General Form): Convert

3x - 2y + 6 = 0

to slope-intercept form and find its slope and intercepts. Step 1 — Slope-intercept:

3x - 2y + 6 = 0 \implies -2y = -3x - 6 \implies y = \frac{3}{2}x + 3

Step 2 — Slope:

m = \frac{3}{2}

Step 3 — y-intercept:

c = 3

(from

y = \frac{3}{2}x + 3

) Step 4 — x-intercept: Set

y = 0

3x + 6 = 0 \implies x = -2

\boxed{y = \frac{3}{2}x + 3,\ m = \frac{3}{2},\ \text{x-int} = -2,\ \text{y-int} = 3}

2.3 Parallel and Perpendicular Lines

2.3.1 Intuition

Parallel lines run side by side, never meeting. They have the same steepness (equal slopes).
Perpendicular lines meet at a right angle ( $90^\circ$ ). Their slopes are negative reciprocals.

2.3.2 Conditions

For two non-vertical lines with slopes

m_1

and

m_2

Condition	Formula	Example
Parallel	$m_1 = m_2$	$y = 2x + 1$ , $y = 2x - 5$
Perpendicular	$m_1 \cdot m_2 = -1$	$y = 2x + 1$ , $y = -\frac{1}{2}x + 3$

For lines in general form

A_1x + B_1y + C_1 = 0

and

A_2x + B_2y + C_2 = 0

Parallel: $A_1 B_2 = A_2 B_1$
Perpendicular: $A_1 A_2 + B_1 B_2 = 0$

2.3.3 Worked Examples

Example 3.1 (Easy): Are

y = 3x + 2

and

y = 3x - 7

parallel? Step 1:

m_1 = 3

m_2 = 3

Step 2: Since

m_1 = m_2

, the lines are parallel.

\boxed{\text{Yes}}

Example 3.2 (Medium): Find the equation of the line perpendicular to

y = -\frac{1}{2}x + 3

passing through

(4, 1)

. Step 1: Slope of given line:

m_1 = -\frac{1}{2}

Step 2: Slope of perpendicular:

m_2 = -\frac{1}{m_1} = -\frac{1}{-1/2} = 2

Step 3: Point-slope:

y - 1 = 2(x - 4)

Step 4: Simplify:

y - 1 = 2x - 8 \implies y = 2x - 7

\boxed{y = 2x - 7}

Example 3.3 (Hard): Are the lines

2x + 3y - 6 = 0

and

4x + 6y + 12 = 0

parallel? Step 1 — Method 1 (Slopes): Line 1:

y = -\frac{2}{3}x + 2 \implies m_1 = -\frac{2}{3}

Line 2:

y = -\frac{4}{6}x - 2 = -\frac{2}{3}x - 2 \implies m_2 = -\frac{2}{3}

m_1 = m_2

→ parallel. Step 2 — Method 2 (General form):

A_1=2, B_1=3

A_2=4, B_2=6

A_1B_2 = 2 \times 6 = 12

A_2B_1 = 4 \times 3 = 12

A_1B_2 = A_2B_1

→ parallel.

\boxed{\text{Yes, parallel}}

2.4 Angle Between Two Lines

2.4.1 Formula

The acute angle

\theta

between two lines with slopes

m_1

and

m_2

(

m_1 m_2 \neq -1

) is:

\tan\theta = \frac{m_2 - m_1}{1 + m_1 m_2}

The obtuse angle

\phi = 180^\circ - \theta

satisfies:

\tan\phi = \frac{m_1 - m_2}{1 + m_1 m_2} = -\tan\theta

2.4.2 Worked Example

Find the acute angle between

y = 2x + 3

and

y = \frac{1}{3}x - 1

. Step 1:

m_1 = 2

m_2 = \frac{1}{3}

Step 2:

\tan\theta = \frac{\frac{1}{3} - 2}{1 + 2 \cdot \frac{1}{3}} = \frac{-\frac{5}{3}}{1 + \frac{2}{3}} = \frac{-\frac{5}{3}}{\frac{5}{3}} = -1

Step 3:

\theta = \arctan(-1) = 45^\circ

(acute angle, take absolute value)

\boxed{\theta = 45^\circ}

2.5 Distance Formulas

2.5.1 Distance from a Point to a Line

The perpendicular distance from point

(x_1, y_1)

to line

Ax + By + C = 0

d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}

2.5.2 Distance Between Two Parallel Lines

For parallel lines

Ax + By + C_1 = 0

and

Ax + By + C_2 = 0

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

2.5.3 Worked Examples

Example 5.1 (Easy): Distance from

(1, 2)

to the line

3x + 4y - 5 = 0

. Step 1:

d = \frac{|3(1) + 4(2) - 5|}{\sqrt{3^2 + 4^2}} = \frac{|3 + 8 - 5|}{5} = \frac{|6|}{5} = \frac{6}{5}

\boxed{d = 1.2 \text{ units}}

Example 5.2 (Medium): Distance between parallel lines

2x - y + 3 = 0

and

2x - y - 4 = 0

. Step 1:

A=2

B=-1

C_1=3

C_2=-4

Step 2:

d = \frac{|3 - (-4)|}{\sqrt{2^2 + (-1)^2}} = \frac{|7|}{\sqrt{5}} = \frac{7}{\sqrt{5}}

\boxed{d = \frac{7}{\sqrt{5}} \text{ units}}

2.6 Line of Best Fit (Straight-Line Fit)

2.6.1 Intuition

In the real world, data points rarely lie exactly on a straight line. But often they roughly follow a linear trend. Linear regression finds the line that best approximates the data by minimizing the sum of squared errors (SSE).

2.6.2 Formula for Best-Fit Line $y = mx + c$

For

n

data points

(x_i, y_i)

m = \frac{\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})}{\sum_{i=1}^n (x_i - \bar{x})^2}

c = \bar{y} - m\bar{x}

where

\bar{x}

and

\bar{y}

are the means of the

x

-values and

y

-values. The sum of squared errors is:

SSE = \sum_{i=1}^n (y_i - (mx_i + c))^2

2.6.3 Worked Example

Find the line of best fit for the points

(1, 2)

(2, 3)

(3, 5)

(4, 4)

$x_i$	$y_i$	$x_i - \bar{x}$	$y_i - \bar{y}$	$(x_i-\bar{x})(y_i-\bar{y})$	$(x_i-\bar{x})^2$
1	2	-1.5	-1.5	2.25	2.25
2	3	-0.5	-0.5	0.25	0.25
3	5	0.5	1.5	0.75	0.25
4	4	1.5	0.5	0.75	2.25

\bar{x} = 2.5

\bar{y} = 3.5

\sum(x_i-\bar{x})(y_i-\bar{y}) = 2.25 + 0.25 + 0.75 + 0.75 = 4

\sum(x_i-\bar{x})^2 = 2.25 + 0.25 + 0.25 + 2.25 = 5

m = \frac{4}{5} = 0.8

c = 3.5 - 0.8(2.5) = 3.5 - 2 = 1.5

\boxed{y = 0.8x + 1.5}

📐 Key Formulas — Summary Table

Concept	Formula	Notes
Slope	$m = \frac{y_2 - y_1}{x_2 - x_1} = \tan\theta$	Undefined for vertical lines
Point-slope	$y - y_1 = m(x - x_1)$	Given slope + point
Slope-intercept	$y = mx + c$	$c$ = y-intercept
Two-point	$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$	Given two points
Intercept	$\frac{x}{a} + \frac{y}{b} = 1$	$a$ = x-int, $b$ = y-int
General	$Ax + By + C = 0$	All lines
Parallel condition	$m_1 = m_2$ or $A_1B_2 = A_2B_1$
Perpendicular	$m_1 \cdot m_2 = -1$ or $A_1A_2 + B_1B_2 = 0$
Angle between lines	$\tan\theta = \frac{m_2 - m_1}{1 + m_1 m_2}$	Acute angle
Point-to-line distance	$d = \frac{	Ax_1 + By_1 + C
Parallel lines distance	$d = \frac{	C_1 - C_2
Best-fit slope	$m = \frac{\sum(x_i-\bar{x})(y_i-\bar{y})}{\sum(x_i-\bar{x})^2}$	Linear regression
Best-fit intercept	$c = \bar{y} - m\bar{x}$
Sum of squared errors	$SSE = \sum(y_i - mx_i - c)^2$	Minimized by best-fit

⚠️ Common Pitfalls

Pitfall 1: Confusing $x$ -intercept with $y$ -intercept

Mistake: Saying

y = mx + c

gives the

x

-intercept as

c

. Correct:

c

is the

y

-intercept (the point

(0, c)

). The

x

-intercept is found by setting

y = 0

0 = mx + c \implies x = -c/m

Pitfall 2: Thinking Slope is the Same as the Angle

Mistake: If slope is

2

, the angle is

2

radians or degrees. Correct: Slope

m = \tan\theta

. So

\theta = \arctan(m)

. For

m = 2

\theta \approx 63.4^\circ

, not

2^\circ

Pitfall 3: Forgetting the Vertical Line Has No Slope

Mistake: Saying a vertical line has slope

0

(confusing with horizontal). Correct: Horizontal (

y = a

): slope

0

. Vertical (

x = a

): slope undefined.

Pitfall 4: Using Wrong Ratio for Perpendicular Lines

Mistake: Using

m_2 = -m_1

for perpendicular lines (that's for reflection across x-axis). Correct:

m_1 \cdot m_2 = -1

, so

m_2 = -1/m_1

. Example: if

m_1 = 2

, then

m_2 = -1/2

📝 Practice Questions

Q1: Find the slope of the line through $(3, -2)$ and $(7, 6)$ .

Strategy Hint: Use $m = (y_2 - y_1)/(x_2 - x_1)$ .

$m = \frac{6 - (-2)}{7 - 3} = \frac{8}{4} = 2$

$\boxed{m = 2}$ Q2: Write the equation of the line with slope $-3$ passing through $(2, 5)$ .

Strategy Hint: Use point-slope form.

$y - 5 = -3(x - 2) \implies y - 5 = -3x + 6 \implies y = -3x + 11$

$\boxed{y = -3x + 11}$ Q3: Are $y = -\frac{2}{3}x + 4$ and $3x - 2y + 5 = 0$ perpendicular?

Strategy Hint: Find slopes and check $m_1 \cdot m_2 = -1$ .

Line 1: $m_1 = -\frac{2}{3}$ Line 2: $y = \frac{3}{2}x + \frac{5}{2} \implies m_2 = \frac{3}{2}$ $m_1 \cdot m_2 = -\frac{2}{3} \cdot \frac{3}{2} = -1$ ✓

$\boxed{\text{Yes, perpendicular}}$ Q4: Find the distance from $(4, -1)$ to $5x - 12y + 3 = 0$ .

Strategy Hint: Use the point-to-line distance formula.

$d = \frac{|5(4) - 12(-1) + 3|}{\sqrt{5^2 + (-12)^2}} = \frac{|20 + 12 + 3|}{13} = \frac{35}{13}$

$\boxed{\frac{35}{13} \text{ units}}$ Q5: Find the equation of the line parallel to $3x + 4y = 12$ passing through $(0, 0)$ .

Strategy Hint: Parallel lines have the same $A$ and $B$ (different $C$ ).

Since lines are parallel, $3x + 4y + C = 0$ . Passes through $(0,0)$ : $3(0) + 4(0) + C = 0 \implies C = 0$ .

$\boxed{3x + 4y = 0}$ Q6: Find the acute angle between $y = x + 1$ and $y = \sqrt{3}x - 2$ .

Strategy Hint: Use $\tan\theta = \frac{m_2 - m_1}{1 + m_1 m_2}$ .

$m_1 = 1$ , $m_2 = \sqrt{3}$ $\tan\theta = \frac{\sqrt{3} - 1}{1 + 1 \cdot \sqrt{3}} = \frac{\sqrt{3} - 1}{1 + \sqrt{3}} = \frac{(\sqrt{3} - 1)^2}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$

$\theta = \arctan(2 - \sqrt{3}) = 15^\circ$

$\boxed{15^\circ}$ Q7: Write $2x - 5y + 10 = 0$ in intercept form.

Strategy Hint: Get $\frac{x}{a} + \frac{y}{b} = 1$ .

$2x - 5y + 10 = 0 \implies 2x - 5y = -10 \implies \frac{x}{-5} + \frac{y}{2} = 1$

$\boxed{\frac{x}{-5} + \frac{y}{2} = 1}$ Q8: Find the line through $(3, 2)$ perpendicular to $x = 5$ .

Strategy Hint: $x = 5$ is vertical. Perpendicular to vertical is horizontal.

$x = 5$ is vertical. A perpendicular line is horizontal: $y = k$ . Through $(3, 2)$ : $y = 2$ .

$\boxed{y = 2}$ Q9: Find the distance between $2x - 3y + 7 = 0$ and $4x - 6y - 5 = 0$ .

Strategy Hint: First check parallel (they are), then make A, B same.

Line 2: $4x - 6y - 5 = 0 \implies 2x - 3y - \frac{5}{2} = 0$ $d = \frac{|7 - (-5/2)|}{\sqrt{2^2 + (-3)^2}} = \frac{|7 + 2.5|}{\sqrt{13}} = \frac{9.5}{\sqrt{13}} = \frac{19}{2\sqrt{13}}$

$\boxed{\frac{19}{2\sqrt{13}} \text{ units}}$ Q10: Find the line with x-intercept $2$ and y-intercept $-3$ .

Strategy Hint: Intercept form directly.

$\frac{x}{2} + \frac{y}{-3} = 1 \implies \frac{x}{2} - \frac{y}{3} = 1 \implies 3x - 2y = 6$

$\boxed{3x - 2y = 6}$ Q11: For points $(0,1)$ , $(2,3)$ , $(4,5)$ , find the best-fit line.

Strategy Hint: Use the formulas for $m$ and $c$ .

$\bar{x} = 2$ , $\bar{y} = 3$ $\sum(x_i-\bar{x})(y_i-\bar{y}) = (-2)(-2) + 0(0) + 2(2) = 4 + 0 + 4 = 8$ $\sum(x_i-\bar{x})^2 = 4 + 0 + 4 = 8$ $m = \frac{8}{8} = 1$ , $c = 3 - 1(2) = 1$

$\boxed{y = x + 1}$ Q12: If a line passes through $(2,5)$ and $(-1,2)$ , write its equation and find its intercepts.

Strategy Hint: Find slope, then use point-slope.

$m = \frac{2-5}{-1-2} = \frac{-3}{-3} = 1$ $y - 5 = 1(x - 2) \implies y = x + 3$ y-intercept: $3$ (when $x=0$ ) x-intercept: $-3$ (when $y=0$ )

$\boxed{y = x + 3,\ \text{y-int}=3,\ \text{x-int}=-3}$

🔗 Cross-References

Previous topic: Coordinate Geometry
Next topic: Quadratic Functions — non-linear curves
Related: Derivatives — slope of a curve = derivative
Across courses: BSMA1002 Stats 1 (linear regression, correlation); BSMA1003 Maths 2 (linear algebra, vector spaces) Join Discord Previous2.1 Coordinate GeometryNext3.1 Quadratic Functions

Straight Lines

Straight Lines

🎯 Learning Objectives

📋 Prerequisites

📖 Core Content

2.1 The Slope of a Line

2.1.1 Intuition: Steepness

2.1.2 Definition

2.1.3 Worked Examples

2.2 Equations of a Straight Line

2.2.1 Intuition: Describing a Line

2.2.2 Forms Summary

2.2.3 Converting Between Forms

2.2.4 Worked Examples

2.3 Parallel and Perpendicular Lines

2.3.1 Intuition

2.3.2 Conditions

2.3.3 Worked Examples

2.4 Angle Between Two Lines

2.4.1 Formula

2.4.2 Worked Example

2.5 Distance Formulas

2.5.1 Distance from a Point to a Line

2.5.2 Distance Between Two Parallel Lines

2.5.3 Worked Examples

2.6 Line of Best Fit (Straight-Line Fit)

2.6.1 Intuition

2.6.2 Formula for Best-Fit Line y=mx+cy = mx + cy=mx+c

2.6.3 Worked Example

📐 Key Formulas — Summary Table

⚠️ Common Pitfalls

Pitfall 1: Confusing xxx-intercept with yyy-intercept

Pitfall 2: Thinking Slope is the Same as the Angle

Pitfall 3: Forgetting the Vertical Line Has No Slope

Pitfall 4: Using Wrong Ratio for Perpendicular Lines

📝 Practice Questions

🔗 Cross-References

Intelligence Hub

2.6.2 Formula for Best-Fit Line $y = mx + c$

Pitfall 1: Confusing $x$ -intercept with $y$ -intercept