Week 2 Notes

Mathematics for
Data Science 1

Straight lines: the coordinate plane, slope, the equations of a line, distances, angles, and fitting a line to data.

8 chapters45 worked examples~45 min read

CoordinatesSlopeEquations of a lineStraight-line fit

Chapters

01Rectangular Coordinate System and DistanceSections 2.1, 2.2
02Section Formula and Area of a TriangleSections 2.3, 2.4
03Slope, Parallel and Perpendicular LinesSection 2.5.1
04Equations and Forms of a Straight LineSections 2.5.2, 2.6
05Angle Between Two LinesSection 2.5.3
06General Equation: Conditions and DistancesSections 2.7 to 2.9
07Straight-Line Fit and Sum Squared ErrorSection 2.10
08Week 2 Summary SheetAll sections

Coordinates and points

Rectangular Coordinate System and Distance

Straight lines live in the coordinate plane. This part sets up that plane and gives the first tool for working in it: the distance between two points.

1.1The Cartesian coordinate system

Definition: Cartesian coordinate system

A Cartesian coordinate system fixes every point of a plane by numbers. The coordinates are the signed distances from the point to two perpendicular reference lines, measured in the same unit. Each reference line is an axis, and the point where they meet is the origin, $(0, 0)$ .

The horizontal axis is the X-axis; the first coordinate of a point is its abscissa.
The vertical axis is the Y-axis; the second coordinate is its ordinate.
The axes split the plane into four quadrants, numbered anticlockwise from the top right.

Figure 1. The plane, its four quadrants and one example point in each

1.2Which quadrant?

Look only at the signs of the two coordinates.

Point	Signs of (x, y)	Quadrant
$(3, 2)$	+, +	I
$(-4, 2)$	−, +	II
$(-4, -2)$	−, −	III
$(3, -2)$	+, −	IV
$(0, 3)$	x is 0	on the Y-axis, in no quadrant

Extra. A point with a zero coordinate lies on an axis, and axes belong to no quadrant.

1.3The distance between two points

Distance formula

The distance between $(x_1, y_1)$ and $(x_2, y_2)$ is

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

It is the hypotenuse of a right triangle whose legs are the horizontal gap $x_2 - x_1$ and the vertical gap $y_2 - y_1$ .

Figure 2. The distance between (2, 4) and (−4, 12) is the hypotenuse of a 6, 8, 10 triangle

Why the formula works

Draw the horizontal leg from the first point and the vertical leg from the second; they meet at a right angle.
The legs have lengths $|x_2 - x_1|$ and $|y_2 - y_1|$ .
Pythagoras: $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$ . Take the square root.

Remember

The gaps are squared, so it does not matter which point is called 1 and which 2, and negative gaps are harmless.

1.4Going deeper

Distance is symmetric and non-negative. It is 0 only when the two points are the same point.
All points at distance $r$ from a point form a circle of radius $r$ (as in Example 2).
Classifying triangles (extra). Find the three side lengths. For $(0, 0), (6, 0), (0, 8)$ they are 6, 8 and 10, and $6^2 + 8^2 = 10^2$ , so it is a right triangle with perimeter 24.
Unknown coordinate means two answers. Squaring a distance equation gives $\pm$ , so always check both roots.

1.5Exam tips and common mistakes

Exam tips

Write the formula first, then substitute with brackets around negative numbers.
For an unknown coordinate set up the squared equation and solve for both signs.
For quadrants look at the signs only.

Common mistakes

Forgetting the square root, and giving 100 instead of 10.
Losing the second answer when solving $(4 - y)^2 = 9$ .
Writing $(x_2 - x_1) + (y_2 - y_1)$ inside the root instead of the sum of squares.

Practice questions (graded-assignment style)

Q1Distance

Multiple Choice Question (MCQ): choose the one correct option.

What is the distance between $(-1, 3)$ and $(5, -5)$ ?

Solution

Idea

Find the two gaps, square and add.

Gaps: $5 - (-1) = 6$ and $-5 - 3 = -8$ .
$d = \sqrt{36 + 64} = \sqrt{100} = 10$ .
Answer: option 3.

Q2Unknown coordinate

Numerical Answer Type (NAT).

The distance between $(k, 2)$ and $(-1, 6)$ is 5. Find the sum of all possible values of $k$ .

Solution

Idea

Square the distance equation and keep both roots.

$(k + 1)^2 + (2 - 6)^2 = 25$ , so $(k + 1)^2 = 9$ and $k + 1 = \pm 3$ .
$k = 2$ or $k = -4$ . Sum: $2 + (-4) = -2$ .
Answer: -2

Q3Quadrants

Multiple Select Question (MSQ): select all that apply.

Choose the correct statements.

(-3, 4)

lies in quadrant II.

(5, -2)

lies in quadrant IV.

(-1, -1)

lies in quadrant II.

(0, 3)

lies in quadrant I.

Solution

Idea

Read the signs: I (+,+), II (−,+), III (−,−), IV (+,−).

Option 1: signs (−, +), quadrant II. ✓
Option 2: signs (+, −), quadrant IV. ✓
Option 3: signs (−, −) is quadrant III, not II.
Option 4: $x = 0$ , so the point is on the Y-axis, in no quadrant.
Answer: options 1 and 2.

Q4Perimeter of a triangle

Numerical Answer Type (NAT).

Find the perimeter of the triangle with vertices $(0, 0)$ , $(6, 0)$ and $(0, 8)$ .

Solution

Idea

Find all three side lengths.

$\sqrt{(6 - 0)^2 + 0^2} = 6$ and $\sqrt{0^2 + (8 - 0)^2} = 8$ .
$\sqrt{(0 - 6)^2 + (8 - 0)^2} = \sqrt{36 + 64} = 10$ .
Perimeter: $6 + 8 + 10 = 24$ .
Answer: 24

Revision overview

CoordinatesTwo perpendicular axes
Origin is (0, 0)
Signed distances

QuadrantsI: +, + II: −, +
III: −, − IV: +, −
Abscissa x, ordinate y

Typical problemsDistance of two points
Unknown coordinate: ± cases
Check each answer

CoordinatesWeek 2 · Sections 2.1, 2.2

Distance formula√((x₂−x₁)² + (y₂−y₁)²)
Hypotenuse of a right triangle

Order is freeEither point can be first
Negative gaps are harmless

CirclePoints at distance r
from a centre form a circle

Revision: the coordinate plane and the distance formula on one page

Next · Coordinates and pointsSection Formula and Area of a Triangle

Coordinates and points

Section Formula and Area of a Triangle

Two more tools for points: the section formula finds a point that divides a segment in a given ratio, and the area formula measures the triangle formed by three points.

2.1The section formula

Let $A(x_1, y_1)$ and $B(x_2, y_2)$ be the ends of a segment and let $P(x, y)$ divide it in the ratio $m : n$ , meaning $PA : PB = m : n$ .

Internal section

$P$ lies between $A$ and $B$ .

x = \frac{mx_2 + nx_1}{m + n}, \quad y = \frac{my_2 + ny_1}{m + n}

External section

$P$ lies outside the segment, on its extension (needs $m \neq n$ ).

x = \frac{mx_2 - nx_1}{m - n}, \quad y = \frac{my_2 - ny_1}{m - n}

Figure 3. Internal and external division of the segment AB in the ratio m : n

Midpoint

The midpoint divides the segment in the ratio $1 : 1$ . Putting $m = n = 1$ in the internal formula:

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

2.2Reading the formulas

Weights. In $m : n$ the number $m$ goes with the far end $B$ and $n$ with $A$ : $x = \frac{m x_2 + n x_1}{m + n}$ . A larger $m$ pulls $P$ towards $B$ .
External needs $m \neq n$ , because $m - n$ is in the denominator. Check: $A(0, 0)$ , $B(2, 0)$ , ratio $3 : 1$ gives $x = \frac{3(2) - 1(0)}{3 - 1} = 3$ , and indeed $PA = 3$ , $PB = 1$ .
Centroid (extra). The average of the three vertices, $\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$ , is the centre of mass of a triangle; the midpoint is the two-point version of the same averaging.

2.3Area of a triangle

Area formula

The area $\Delta$ of the triangle with vertices $(x_1, y_1)$ , $(x_2, y_2)$ , $(x_3, y_3)$ is

\Delta = \frac{1}{2}\left|\,x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\,\right|

The modulus is used because an area is never negative.

Collinear points

If the area of the triangle formed by three points is zero, the three points are collinear (lie on one line).

2.4Worked examples: area and collinearity

Figure 4. The midpoint triangle of PQR has area 1.5

2.5Exam tips and common mistakes

Exam tips

Say which point is $A$ and which is $B$ before using the ratio $m : n$ ; the ratio is read from $A$ .
Test a section answer by computing $PA$ and $PB$ with the distance formula.
For collinear questions set the area to 0 and solve for the unknown.

Common mistakes

Swapping $m$ and $n$ : the answer then divides in $n : m$ .
Using the internal formula with $m - n$ (or the external with $m + n$ ).
Forgetting the $\tfrac{1}{2}$ or the modulus in the area formula.

Practice questions (graded-assignment style)

Q1A dividing point

Numerical Answer Type (NAT).

The point $P(x, y)$ divides the segment from $(1, 2)$ to $(7, 11)$ internally in the ratio $1 : 2$ . Find $x + y$ .

Solution

Idea

Use the internal formula with $m = 1$ , $n = 2$ .

$x = \dfrac{1(7) + 2(1)}{3} = 3$ .
$y = \dfrac{1(11) + 2(2)}{3} = 5$ .
$x + y = 3 + 5 = 8$ .
Answer: 8

Q2Midpoint

Numerical Answer Type (NAT).

Find the sum of the coordinates of the midpoint of $(-4, 6)$ and $(10, -2)$ .

Solution

Idea

Average each coordinate.

Midpoint: $\left(\dfrac{-4 + 10}{2}, \dfrac{6 - 2}{2}\right) = (3, 2)$ .
$3 + 2 = 5$ .
Answer: 5

Q3Area of a triangle

Numerical Answer Type (NAT).

Find the area of the triangle with vertices $(1, 1)$ , $(5, 1)$ and $(3, 5)$ .

Solution

Idea

Use the area formula; check with base × height.

$\Delta = \tfrac{1}{2}\left|1(1 - 5) + 5(5 - 1) + 3(1 - 1)\right| = \tfrac{1}{2}\left|-4 + 20 + 0\right| = 8$ .
Check: the base from $(1, 1)$ to $(5, 1)$ is 4 and the height is 4, so $\tfrac{1}{2} \cdot 4 \cdot 4 = 8$ .
Answer: 8

Q4Collinear points

Numerical Answer Type (NAT).

The points $(1, 2)$ , $(2, 4)$ and $(3, k)$ lie on one line. Find $k$ .

Solution

Idea

Collinear means the area is zero.

$\tfrac{1}{2}\left|1(4 - k) + 2(k - 2) + 3(2 - 4)\right| = \tfrac{1}{2}\left|k - 6\right|$ .
Set it to 0: $k = 6$ . (The points have slope 2 each step, so the next point is $(3, 6)$ .)
Answer: 6

Q5An external point

Numerical Answer Type (NAT).

The point $P$ divides the segment from $A(1, 1)$ to $B(3, 5)$ externally in the ratio $3 : 2$ . Find the sum of the coordinates of $P$ .

Solution

Idea

Use the external formula with $m = 3$ , $n = 2$ .

$x = \dfrac{3(3) - 2(1)}{3 - 2} = 7$ and $y = \dfrac{3(5) - 2(1)}{3 - 2} = 13$ .
Check: $PA = \sqrt{180}$ , $PB = \sqrt{80}$ , and $\sqrt{180}/\sqrt{80} = 1.5 = 3 : 2$ .
$x + y = 7 + 13 = 20$ .
Answer: 20

Revision overview

Internal section(mx₂ + nx₁) / (m + n)
(my₂ + ny₁) / (m + n)
P between A and B

External section(mx₂ − nx₁) / (m − n)
(my₂ − ny₁) / (m − n)
P outside, m ≠ n

MidpointRatio 1 : 1
Average of the coordinates

PointsWeek 2 · Sections 2.3, 2.4

Area of a triangle½ |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|

ModulusArea is always positive

CollinearArea 0 means the three points
lie on one line

Revision: the section formula, midpoint and triangle area on one page

Next · Straight linesSlope, Parallel and Perpendicular Lines

Straight lines

Slope, Parallel and Perpendicular Lines

The slope of a line is the single number that tells you how it leans. Parallel and perpendicular lines are recognised from their slopes alone.

3.1Slope of a straight line

Definition: slope

The slope $m$ of a line describes both its direction and its steepness:

m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1} = \tan\theta

where $(x_1, y_1)$ and $(x_2, y_2)$ are two points on the line and $\theta$ is its inclination: the angle from the positive X-axis to the line.

Figure 5. Slope is the vertical change divided by the horizontal change between two points

3.2What the slope tells you

Slope	What the formula shows	The line
$m > 0$	$\Delta y$ and $\Delta x$ have the same sign	rises from left to right
$m < 0$	they have opposite signs	falls from left to right
$m = 0$	$\Delta y = 0$ , so $\tan\theta = 0$	horizontal
no slope	$\Delta x = 0$ : the fraction is undefined	vertical

The slope rules (parallel, perpendicular) are for non-vertical lines. A vertical line has no slope.

Example 7Slope through the origin

What is the slope of the line through the origin and the point $(-3, 5)$ ?

Solution

Take $(x_1, y_1) = (0, 0)$ and $(x_2, y_2) = (-3, 5)$ : $m = \frac{5 - 0}{-3 - 0} = -\frac{5}{3}$ The slope is negative, so the line falls from left to right.

Figure 6. The line through the origin and (−3, 5)

3.3Parallel and perpendicular lines

Let $l_1$ and $l_2$ be non-vertical lines with slopes $m_1$ , $m_2$ .

Parallel

$l_1 \parallel l_2$ if and only if the slopes are equal: $m_1 = m_2$ .

Perpendicular

$l_1 \perp l_2$ if and only if $m_1 m_2 = -1$ .

Figure 7. Parallel lines lean the same way; perpendicular lines have slopes that multiply to −1

Remember

If $m_1 m_2 = -1$ then $m_2 = -\dfrac{1}{m_1}$ : the perpendicular slope is the negative reciprocal. A line of slope 2 is perpendicular to a line of slope $-\dfrac{1}{2}$ .

3.4Going deeper

Slope is a rate of change. A slope of 3 means y goes up by 3 for every 1 step in x. This is exactly the idea behind a fitted line in Part 7.
Use the same point first. In $\dfrac{y_2 - y_1}{x_2 - x_1}$ both the top and the bottom must start from the same point. Swapping both gives the same answer, swapping only one flips the sign.
Equal slopes: parallel or the same line. To separate the two cases check whether one point of the first line lies on the second.
Collinearity by slope. Three points are collinear exactly when the slope from the first to the second equals the slope from the second to the third.

3.5Exam tips and common mistakes

Exam tips

For perpendicular, flip the fraction and change the sign: slope $\tfrac{3}{4}$ becomes $-\tfrac{4}{3}$ .
Check any slope answer by plugging both points back into $\dfrac{\Delta y}{\Delta x}$ .

Common mistakes

Mixing up the order: $\dfrac{y_2 - y_1}{x_1 - x_2}$ .
Using the slope rules for a vertical line, which has no slope.
Thinking perpendicular means $m_1 = -m_2$ . It means $m_1 m_2 = -1$ .

Practice questions (graded-assignment style)

Q1A slope

Numerical Answer Type (NAT).

Find the slope of the line through $(2, -3)$ and $(6, 5)$ .

Solution

Idea

Rise over run.

$m = \dfrac{5 - (-3)}{6 - 2} = \dfrac{8}{4} = 2$ .
Answer: 2

Q2Parallel condition

Numerical Answer Type (NAT).

The line through $(1, 1)$ and $(3, k)$ is parallel to the line through $(0, 0)$ and $(2, 6)$ . Find $k$ .

Solution

Idea

Parallel lines have equal slopes.

Slope of the second line: $\dfrac{6 - 0}{2 - 0} = 3$ .
$\dfrac{k - 1}{3 - 1} = 3$ gives $k - 1 = 6$ , so $k = 7$ .
Answer: 7

Q3Perpendicular condition

Numerical Answer Type (NAT).

The line through $(0, 0)$ and $(a, 6)$ is perpendicular to a line of slope $\dfrac{3}{2}$ . Find $a$ .

Solution

Idea

The product of slopes is −1.

Required slope: $-\dfrac{2}{3}$ .
$\dfrac{6}{a} = -\dfrac{2}{3}$ gives $a = -9$ .
Check: $\dfrac{6}{-9} = -\dfrac{2}{3}$ and $-\dfrac{2}{3} \times \dfrac{3}{2} = -1$ .
Answer: -9

Q4Slope facts

Multiple Select Question (MSQ): select all that apply.

Choose the correct statements.

The line through

(1, 2)

and

(3, 6)

rises from left to right.

Lines of slope 2 and

-\dfrac{1}{2}

are perpendicular.

Lines of slope 2 and 3 are parallel.

A vertical line has slope 0.

Solution

Idea

Use the sign of the slope and the two rules.

Option 1: slope $2 > 0$ . ✓
Option 2: $2 \times (-\tfrac{1}{2}) = -1$ . ✓
Option 3: parallel needs equal slopes, and $2 \neq 3$ .
Option 4: a vertical line has no slope (the fraction is undefined); a horizontal line has slope 0.
Answer: options 1 and 2.

Revision overview

Definitionm = Δy / Δx
= (y₂ − y₁) / (x₂ − x₁)
= tan θ

MeaningDirection and steepness
θ from the positive X-axis

Signsm > 0 rises, m < 0 falls
m = 0 horizontal
vertical: no slope

SlopeWeek 2 · Section 2.5.1

Parallelm₁ = m₂
If and only if
(or the same line)

Perpendicularm₁ × m₂ = −1
Negative reciprocal
If and only if

Non-verticalThe rules are for lines
that have a slope

Revision: slope and its two key consequences

Next · Straight linesEquations and Forms of a Straight Line

Straight lines

Equations and Forms of a Straight Line

A line is described by an equation: a relation between x and y that is true for exactly the points on the line. There are several standard forms, and each suits different given information.

4.1The three basic forms

A line is described by an equation: a relation between $x$ and $y$ that is true for exactly the points on the line. Each standard form suits different given information.

Slope-intercept

y = mx + c

$m$ is the slope and $c$ is the y-intercept.

Point-slope

y - y_1 = m(x - x_1)

A line of slope $m$ through $(x_1, y_1)$ .

Two-point

y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)

The line joining two points. It is the point-slope form with the slope $m = \frac{y_2 - y_1}{x_2 - x_1}$ worked out from the two points.

4.2All the forms of a straight line

Form	Representation	Use it when you know
General	$ax + by + c = 0$	a standard shape for any line
Slope-point	$(y - y_0) = m(x - x_0)$	slope and one point
Slope-intercept (y)	$y = mx + c$	slope and the y-intercept
Slope-intercept (x)	$y = m(x - d)$	slope and the x-intercept $d$
Intercept	$\dfrac{x}{a} + \dfrac{y}{b} = 1$	both intercepts $a$ and $b$
Two-point	$(y - y_1) = \dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)$	two points on the line

In the x-intercept form $y = m(x - d)$ , putting $y = 0$ gives $x = d$ , so $d$ is the x-intercept. In the intercept form $a$ and $b$ are the x- and y-intercepts.

Figure 8. One line, 2x + 3y − 6 = 0, written in every form

4.3Choosing the form

What you know	Form to use
Slope and the y-intercept	$y = mx + c$
Slope and one point	$(y - y_1) = m(x - x_1)$
Two points	$(y - y_1) = \dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)$
Slope and the x-intercept	$y = m(x - d)$
Both intercepts	$\dfrac{x}{a} + \dfrac{y}{b} = 1$
Any line, in a standard shape	$ax + by + c = 0$

4.4Going deeper

Read the general form directly (extra). For $ax + by + c = 0$ with $b \neq 0$ : slope $= -\dfrac{a}{b}$ , y-intercept $= -\dfrac{c}{b}$ , and (if $a \neq 0$ ) x-intercept $= -\dfrac{c}{a}$ . For $2x + 3y - 6 = 0$ : $-\tfrac{2}{3}$ , $2$ and $3$ .
Horizontal and vertical lines (extra). A horizontal line is $y = k$ (slope 0). A vertical line is $x = h$ and has no slope, so it cannot be written as $y = mx + c$ .
A point lies on a line exactly when its coordinates satisfy the equation. Substitute and see whether both sides agree.
The six forms describe the same line. Conversion is only algebra: expand, move terms, divide.

4.5Exam tips and common mistakes

Exam tips

Pick the form from what is given (Section 4.3), then convert only if the question asks for another form.
Always verify by substituting a given point into your final equation.

Common mistakes

Mixing $x_1$ and $y_1$ in point-slope form: it is $y - y_1$ with $m(x - x_1)$ .
Sign errors when moving terms to the other side.
Using the intercept form when an intercept is 0 (the line passes through the origin).

Practice questions (graded-assignment style)

Q1Slope and a point

Multiple Choice Question (MCQ): choose the one correct option.

Which is the line of slope $-2$ through the point $(3, 1)$ ?

y = -2x + 7

y = -2x + 5

y = 2x - 5

y = -2x - 7

Solution

Idea

Use the point-slope form and expand.

$y - 1 = -2(x - 3) = -2x + 6$ , so $y = -2x + 7$ .
Check: at $x = 3$ , $-2(3) + 7 = 1$ . ✓
Answer: option 1.

Q2The y-intercept

Numerical Answer Type (NAT).

Find the y-intercept of the line through $(2, 3)$ and $(4, 7)$ .

Solution

Idea

Slope first, then the equation.

$m = \dfrac{7 - 3}{4 - 2} = 2$ .
$y - 3 = 2(x - 2)$ , so $y = 2x - 1$ .
The y-intercept is $c = -1$ .
Answer: -1

Q3The x-intercept

Numerical Answer Type (NAT).

Find the x-intercept of the line $3x - 2y + 12 = 0$ .

Solution

Idea

Put $y = 0$ .

$y = 0$ gives $3x + 12 = 0$ , so $x = -4$ .
Answer: -4

Q4Intercept form

Multiple Choice Question (MCQ): choose the one correct option.

A line has x-intercept 4 and y-intercept $-2$ . Which is its intercept form?

\dfrac{x}{4} + \dfrac{y}{-2} = 1

\dfrac{x}{4} + \dfrac{y}{2} = 1

\dfrac{x}{2} + \dfrac{y}{4} = 1

\dfrac{x}{-4} + \dfrac{y}{2} = 1

Solution

Idea

The form is $\frac{x}{a} + \frac{y}{b} = 1$ with $a = 4$ , $b = -2$ .

Option 1 puts $a = 4$ and $b = -2$ in the right places.
Check: $(4, 0)$ gives $1 + 0 = 1$ ✓ and $(0, -2)$ gives $0 + 1 = 1$ ✓.
Answer: option 1.

Q5Points on a line

Multiple Select Question (MSQ): select all that apply.

Which points lie on the line $y = 2x - 1$ ?

(1, 1)

(2, 3)

(0, 1)

(-1, -3)

Solution

Idea

Substitute the coordinates.

$(1, 1)$ : $2(1) - 1 = 1$ . ✓
$(2, 3)$ : $2(2) - 1 = 3$ . ✓
$(0, 1)$ : $2(0) - 1 = -1 \neq 1$ . ✗
$(-1, -3)$ : $2(-1) - 1 = -3$ . ✓
Answer: options 1, 2 and 4.

Revision overview

Slope-intercepty = mx + c
m slope, c y-intercept
y = m(x − d), d x-intercept

Point-slope(y − y₁) = m(x − x₁)
Slope plus one point

Two-pointSlope from the two points
Then use point-slope

Equations of a lineWeek 2 · Sections 2.5.2, 2.6

Intercept formx/a + y/b = 1
a and b are the intercepts

General formax + by + c = 0
Every form converts to it

CheckingSubstitute a known point
into the final equation

Revision: the six forms and when to use each

Next · Straight linesAngle Between Two Lines

Straight lines

Angle Between Two Lines

Two lines that meet form angles at the point of intersection. Their slopes alone are enough to find those angles.

5.1The angle between two lines

Let $l_1$ and $l_2$ be lines with slopes $m_1$ , $m_2$ and inclinations $\alpha_1$ , $\alpha_2$ , and assume $\alpha_2 > \alpha_1$ . If they intersect and form adjacent angles $\theta$ and $\varphi$ , then:

Angle formulas

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

Figure 9. The lines l₁ and l₂, their inclinations and the angles θ and φ they form

5.2Reading the two formulas

The numerators are opposite, so $\tan\varphi = -\tan\theta$ . That fits two adjacent angles on a straight line, which add up to $180^\circ$ .
$\theta = \alpha_2 - \alpha_1$ , so $\tan\theta = \tan(\alpha_2 - \alpha_1)$ : $\theta$ is the gap between the two inclinations.
The perpendicular case. If $m_1 m_2 = -1$ , the denominator $1 + m_1 m_2$ is zero, so the tangent is undefined: the angle is exactly $90^\circ$ . This agrees with the perpendicularity rule of Part 3.

5.3Going deeper

The acute angle (extra). Taking the modulus, $\tan\theta = \left|\dfrac{m_2 - m_1}{1 + m_1 m_2}\right|$ always gives the acute angle between the lines, so you do not have to worry which slope is larger.
Equal slopes. $m_1 = m_2$ gives $\tan\theta = 0$ , angle $0^\circ$ : the lines are parallel.
Check with inclinations. For slopes $\tfrac{1}{2}$ and 3: $\alpha_1 \approx 26.57^\circ$ , $\alpha_2 \approx 71.57^\circ$ , and $71.57 - 26.57 = 45$ .

5.4Exam tips and common mistakes

Exam tips

Compute $1 + m_1 m_2$ first: if it is 0 the answer is $90^\circ$ and you are done.
Give both $\theta$ and $\varphi$ when the question asks for "the angles", and check that they add to $180^\circ$ .

Common mistakes

Writing $1 - m_1 m_2$ in the denominator.
Putting $m_1 - m_2$ in the numerator of $\tan\theta$ (it is $m_2 - m_1$ when $\alpha_2 > \alpha_1$ ).
Forgetting the negative sign of $\tan\varphi$ and reporting $45^\circ$ twice.

Practice questions (graded-assignment style)

Q1Slopes 0 and 1

Numerical Answer Type (NAT).

Find the angle $\theta$ (in degrees) between lines of slopes 0 and 1.

Solution

Idea

Use $m_1 = 0$ , $m_2 = 1$ .

$\tan\theta = \dfrac{1 - 0}{1 + 1 \cdot 0} = 1$ , so $\theta = 45^\circ$ .
Answer: 45

Q2Reciprocal of the tangent

Numerical Answer Type (NAT).

For lines of slopes 2 and 3 find $\dfrac{1}{\tan\theta}$ .

Solution

Idea

Compute $\tan\theta$ first.

$\tan\theta = \dfrac{3 - 2}{1 + 3 \cdot 2} = \dfrac{1}{7}$ .
So $\dfrac{1}{\tan\theta} = 7$ .
Answer: 7

Q3Which angle?

Multiple Choice Question (MCQ): choose the one correct option.

What is the angle between lines of slopes 3 and $-\dfrac{1}{3}$ ?

0^\circ

45^\circ

90^\circ

It cannot be found

Solution

Idea

Check $1 + m_1 m_2$ .

$1 + 3 \cdot \left(-\dfrac{1}{3}\right) = 1 - 1 = 0$ : the tangent is undefined.
So the lines are perpendicular and the angle is $90^\circ$ .
Answer: option 3.

Q4The adjacent angle

Numerical Answer Type (NAT).

For slopes $\dfrac{1}{2}$ and $3$ we found $\theta = 45^\circ$ . Find $\varphi$ in degrees.

Solution

Idea

Adjacent angles on a line add to $180^\circ$ .

$\varphi = 180^\circ - 45^\circ = 135^\circ$ .
Also $\tan\varphi = -\tan\theta = -1$ , which gives $135^\circ$ .
Answer: 135

Q5A non-standard angle

Numerical Answer Type (NAT).

Find the angle $\theta$ between lines of slopes 1 and 3, rounded to the nearest degree.

Solution

Idea

$\tan\theta$ first, then $\tan^{-1}$ .

$\tan\theta = \dfrac{3 - 1}{1 + 3} = \dfrac{1}{2}$ .
$\theta = \tan^{-1}\tfrac{1}{2} \approx 26.57^\circ$ , which rounds to $27^\circ$ .
Answer: 27

Revision overview

SettingSlopes m₁, m₂
Inclinations α₁ < α₂
Adjacent angles θ and φ

Formula for θtan θ = (m₂ − m₁) / (1 + m₂m₁)
θ = α₂ − α₁

Formula for φtan φ = (m₁ − m₂) / (1 + m₁m₂)
tan φ = − tan θ

Angle between linesWeek 2 · Section 2.5.3

Sumθ + φ = 180°

Special case1 + m₁m₂ = 0
Lines are perpendicular
Angle is 90°

Parallelm₁ = m₂ gives tan θ = 0
Angle 0°

Revision: the angle between two lines from their slopes

Next · General equation and fittingGeneral Equation: Conditions and Distances

General equation and fitting

General Equation: Conditions and Distances

When a line is written in the general form ax + by + c = 0, its coefficients answer questions directly: are two lines parallel or perpendicular, how far is a point from the line, and how far apart are two parallel lines.

6.1Parallel and perpendicular lines in general form

Given two lines $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$ , with $b_1, b_2 \neq 0$ :

Parallel

$a_1 b_2 = a_2 b_1$

Same slope: $-\dfrac{a_1}{b_1} = -\dfrac{a_2}{b_2}$ , cross-multiplied.

Perpendicular

$a_1 a_2 = -b_1 b_2$

Slopes multiply to $-1$ : $\dfrac{a_1 a_2}{b_1 b_2} = -1$ , cross-multiplied.

Where the conditions come from. Solving $ax + by + c = 0$ for $y$ gives $y = -\dfrac{a}{b}x - \dfrac{c}{b}$ , so the slope is $-\dfrac{a}{b}$ (this needs $b \neq 0$ ). Equal slopes give $a_1 b_2 = a_2 b_1$ ; slopes with product $-1$ give $a_1 a_2 = -b_1 b_2$ .

6.2Distance of a point from a line

Point-to-line distance

The distance $d$ of the line $ax + by + c = 0$ from the point $(x_1, y_1)$ is

d = \frac{|a x_1 + b y_1 + c|}{\sqrt{a^2 + b^2}}

Put the point into the left side of the equation, take the modulus, and divide by $\sqrt{a^2 + b^2}$ .

The result is the shortest (perpendicular) distance, so it is never negative.
If the numerator is $0$ , the point lies on the line.

6.3Distance between two parallel lines

Parallel-line distance

The distance $D$ between the parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is

D = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}

Remember

Both equations must have the same $a$ and $b$ . If not, multiply or divide one equation so that they match (as in Example 21).

Figure 10. The point-to-line distance and the distance between two parallel lines

6.4Going deeper

The vector $(a, b)$ is perpendicular to the line (extra). That is why the perpendicular distance has $\sqrt{a^2 + b^2}$ underneath: it is the length of $(a, b)$ . It also explains the perpendicularity test: two lines are perpendicular when their $(a, b)$ vectors are, $a_1 a_2 + b_1 b_2 = 0$ .
Special cases. The distance of $(x_1, y_1)$ from the X-axis ( $y = 0$ ) is $|y_1|$ , and from the Y-axis ( $x = 0$ ) it is $|x_1|$ .
Parallel distance is a point distance. Pick any point on one line and use the point formula on the other (Example 21 does this as a check).
Equal $c$ after matching $a, b$ means the two lines are the same line, with distance 0.

6.5Perpendicular bisector, foot and reflection, and lines through an intersection point

A. Perpendicular bisector of a segment

The perpendicular bisector of $AB$ passes through the midpoint of $AB$ and is perpendicular to $AB$ . Two ways to get it:

Midpoint and slope. For $A(1, 2)$ , $B(5, 4)$ : midpoint $(3, 3)$ ; slope of $AB$ is $\dfrac{4 - 2}{5 - 1} = \dfrac12$ ; the perpendicular slope is $-2$ . So $y - 3 = -2(x - 3)$ , i.e. $2x + y = 9$ .
Equal distances. Every point $P(x, y)$ on it satisfies $PA = PB$ . Squaring: $(x - 1)^2 + (y - 2)^2 = (x - 5)^2 + (y - 4)^2$ , which simplifies to $8x + 4y - 36 = 0$ , the same line $2x + y = 9$ ✓. This also works when the segment is vertical or horizontal, where slopes cause trouble.

B. Foot of the perpendicular and reflection

Let $P(x_1, y_1)$ and the line $ax + by + c = 0$ . Put $t = \dfrac{ax_1 + by_1 + c}{a^2 + b^2}$ . Then

\text{foot} = (x_1 - at,\ y_1 - bt), \qquad \text{reflection} = (x_1 - 2at,\ y_1 - 2bt), \qquad \text{distance} = |t|\sqrt{a^2 + b^2}

Why. The direction $(a, b)$ is perpendicular to the line, so the foot is $P$ moved along $(a, b)$ by some multiple $-t$ . Putting $(x_1 - at,\ y_1 - bt)$ into the line equation and solving gives that $t$ . The reflection goes twice as far. The distance formula $\dfrac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$ is the same as $|t|\sqrt{a^2 + b^2}$ .

Example. $P(1, 2)$ and $x + y - 6 = 0$ . Here $t = \dfrac{1 + 2 - 6}{2} = -\dfrac32$ . The foot is $\left(1 + \tfrac32,\ 2 + \tfrac32\right) = \left(\tfrac52, \tfrac72\right)$ ; check: $\tfrac52 + \tfrac72 = 6$ ✓. The reflection is $(1 + 3,\ 2 + 3) = (4, 5)$ , and the midpoint of $P$ and $(4, 5)$ is $\left(\tfrac52, \tfrac72\right)$ , the foot ✓. The distance is $\dfrac32\sqrt2 = \dfrac{3}{\sqrt2} \approx 2.12$ .

C. Intersection of lines, concurrency and the family of lines

Two lines that are not parallel meet at one point: solve the two equations together.
Three lines are concurrent (pass through one point) when the $3 \times 3$ determinant of their coefficients $\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$ is $0$ , provided the first two are not parallel. Example: $x + y - 3 = 0$ and $2x - y = 0$ meet at $(1, 2)$ . The line $x - 2y + 3 = 0$ passes through $(1, 2)$ since $1 - 4 + 3 = 0$ , and the determinant $\begin{vmatrix} 1 & 1 & -3 \\ 2 & -1 & 0 \\ 1 & -2 & 3 \end{vmatrix} = 0$ ✓.
Family of lines. Every line through the intersection of $L_1 = 0$ and $L_2 = 0$ can be written $L_1 + \lambda L_2 = 0$ (for some number $\lambda$ , or $L_2 = 0$ itself). Use one more condition to find $\lambda$ . Example: the line through the intersection of $2x + 3y - 5 = 0$ and $x - y + 1 = 0$ and the origin: $(2x + 3y - 5) + \lambda(x - y + 1) = 0$ at $(0, 0)$ gives $-5 + \lambda = 0$ , so $\lambda = 5$ , and the line is $7x - 2y = 0$ . Check: the lines meet at $\left(\tfrac25, \tfrac75\right)$ and $7 \cdot \tfrac25 - 2 \cdot \tfrac75 = 0$ ✓.

6.6Exam tips and common mistakes

Exam tips

Write each line as $ax + by + c = 0$ first (everything on one side) and list $a, b, c$ .
For the parallel distance make $a, b$ equal before reading the $c$ values.

Common mistakes

Forgetting the modulus, which can give a negative distance.
Using $|c_1 - c_2|$ when $a, b$ differ between the two equations.
Forgetting to move the constant to the left so that the equation has $= 0$ .

Practice questions (graded-assignment style)

Q1Parallel or perpendicular?

Multiple Select Question (MSQ): select all that apply.

Pair A: $2x + y - 3 = 0$ and $4x + 2y + 5 = 0$ . Pair B: $3x - 2y + 1 = 0$ and $2x + 3y - 4 = 0$ . Choose the correct statements.

Pair A is parallel.

Pair B is perpendicular.

Pair A is perpendicular.

Pair B is parallel.

Solution

Idea

Use $a_1 b_2 = a_2 b_1$ for parallel and $a_1 a_2 = -b_1 b_2$ for perpendicular.

Pair A: $a_1 b_2 = 2 \cdot 2 = 4$ and $a_2 b_1 = 4 \cdot 1 = 4$ . Equal, so parallel.
Pair B: $a_1 a_2 = 3 \cdot 2 = 6$ and $-b_1 b_2 = -(-2)(3) = 6$ . Equal, so perpendicular.
So options 3 and 4 are false.
Answer: options 1 and 2.

Q2A parallel condition

Numerical Answer Type (NAT).

For which $k$ is $kx + 2y - 1 = 0$ parallel to $3x + 4y + 2 = 0$ ?

Solution

Idea

Use $a_1 b_2 = a_2 b_1$ .

$k \cdot 4 = 3 \cdot 2$ , so $k = \dfrac{6}{4} = 1.5$ .
Check: slopes $-\dfrac{1.5}{2} = -0.75$ and $-\dfrac{3}{4} = -0.75$ .
Answer: 1.5

Q3Point to line

Numerical Answer Type (NAT).

Find the distance of the point $(1, 2)$ from the line $3x + 4y - 5 = 0$ .

Solution

Idea

Substitute the point and divide by $\sqrt{a^2 + b^2}$ .

$|3(1) + 4(2) - 5| = |3 + 8 - 5| = 6$ .
$\sqrt{3^2 + 4^2} = 5$ , so $d = \dfrac{6}{5} = 1.2$ .
Answer: 1.2

Q4Between parallel lines

Numerical Answer Type (NAT).

Find the distance between $x + y - 2 = 0$ and $x + y + 4 = 0$ , correct to two decimals.

Solution

Idea

Both lines already have the same $a = b = 1$ .

$D = \dfrac{|-2 - 4|}{\sqrt{1^2 + 1^2}} = \dfrac{6}{\sqrt{2}} = 3\sqrt{2} \approx 4.24$ .
Answer: 4.24

Q5Another parallel condition

Numerical Answer Type (NAT).

For which $k$ is $2x + ky - 3 = 0$ parallel to $6x + 9y + 1 = 0$ ?

Solution

Idea

$a_1 b_2 = a_2 b_1$ .

$2 \cdot 9 = 6 \cdot k$ , so $k = 3$ .
Check: slopes $-\dfrac{2}{3}$ and $-\dfrac{6}{9} = -\dfrac{2}{3}$ .
Answer: 3

P1A perpendicular bisector

Numerical Answer Type (NAT).

The perpendicular bisector of the segment joining $A(1, 2)$ and $B(5, 4)$ is $2x + y = k$ . Find $k$ .

Solution

Idea

The bisector passes through the midpoint, and every point on it is equidistant from $A$ and $B$ .

Midpoint $(3, 3)$ lies on $2x + y = k$ , so $k = 2(3) + 3 = 9$ .
Check with a second point on the line, $(4, 1)$ : $PA^2 = 9 + 1 = 10$ and $PB^2 = 1 + 9 = 10$ ✓ (and $2(4) + 1 = 9$ ).
Answer: 9

P2Foot of a perpendicular

Numerical Answer Type (NAT).

Find the $x$ -coordinate of the foot of the perpendicular from $P(1, 2)$ to the line $x + y - 6 = 0$ .

Solution

Idea

The foot is $(x_1 - at,\ y_1 - bt)$ with $t = \dfrac{ax_1 + by_1 + c}{a^2 + b^2}$ .

$t = \dfrac{1 + 2 - 6}{1 + 1} = -\dfrac32$ .
Foot $= \left(1 + \tfrac32,\ 2 + \tfrac32\right) = (2.5, 3.5)$ . It lies on the line ( $2.5 + 3.5 = 6$ ) and $P$ to the foot has direction $(1.5, 1.5)$ , perpendicular to the line (slope $-1$ ) ✓.
Answer: 2.5

P3Reflection of a point

Multiple Choice Question (MCQ): choose the one correct option.

The reflection of $P(1, 2)$ in the line $x + y = 6$ is

(5, 4)

(4, 5)

(2.5, 3.5)

(3, 3)

Solution

Idea

The foot of the perpendicular is the midpoint of $P$ and its reflection.

The foot is $(2.5, 3.5)$ (see P2). If the image is $(u, v)$ : $\tfrac{1 + u}{2} = 2.5$ and $\tfrac{2 + v}{2} = 3.5$ , so $(u, v) = (4, 5)$ .
Check: $4 + 5 = 9$ is as far above $6$ as $1 + 2 = 3$ is below it. Option 3 is the foot itself, not the image.
Answer: option 2.

P4Concurrent lines

Numerical Answer Type (NAT).

The lines $x + y - 3 = 0$ , $2x - y = 0$ and $x - 2y + k = 0$ are concurrent. Find $k$ .

Solution

Idea

Find where the first two meet; the third must pass through that point.

Adding the first two equations: $3x - 3 = 0$ , so $x = 1$ and $y = 2$ . The lines meet at $(1, 2)$ .
$1 - 2(2) + k = 0 \Rightarrow k = 3$ .
Determinant check: $\begin{vmatrix} 1 & 1 & -3 \\ 2 & -1 & 0 \\ 1 & -2 & k \end{vmatrix} = 1(-k) - 1(2k) + (-3)(-4 + 1) = -3k + 9$ , which is $0$ exactly when $k = 3$ ✓.
Answer: 3

P5A line through an intersection point and the origin

Numerical Answer Type (NAT).

Find the slope of the line through the origin and the point where $2x + 3y - 5 = 0$ meets $x - y + 1 = 0$ .

Solution

Idea

Either solve for the intersection and use two points, or use the family $L_1 + \lambda L_2 = 0$ .

From $x = y - 1$ : $2y - 2 + 3y - 5 = 0$ , so $y = \tfrac75$ and $x = \tfrac25$ . The slope from the origin is $\dfrac{7/5}{2/5} = \dfrac72 = 3.5$ .
Family method: $(2x + 3y - 5) + \lambda(x - y + 1) = 0$ through $(0, 0)$ gives $\lambda = 5$ , line $7x - 2y = 0$ , slope $\tfrac72$ ✓.
Answer: 3.5

Revision overview

General formax + by + c = 0
Slope is −a / b
b ≠ 0 for the tests

Parallela₁ b₂ = a₂ b₁
Same slope

Perpendiculara₁ a₂ = −b₁ b₂
Slopes multiply to −1

General equationWeek 2 · Sections 2.7 to 2.9

Point to line|ax₁ + by₁ + c| / √(a² + b²)
Insert the point, take modulus

Between parallels|c₁ − c₂| / √(a² + b²)
Same a and b first

AlwaysDistances are never negative
Numerator 0: point is on the line

Revision: conditions and distances from the general equation

Next · General equation and fittingStraight-Line Fit and Sum Squared Error

General equation and fitting

Straight-Line Fit and Sum Squared Error

Real data rarely lies exactly on a line. A line of fit is a straight line chosen to follow the data, and the Sum Squared Error measures how far the data is from it.

7.1Sum squared error

Real data rarely lies exactly on a line. A line of fit is a straight line chosen to follow the data, and the Sum Squared Error measures how far the data is from it.

Definition: SSE

SSE is the sum of the squares of the deviations of the predicted linear model from the actual data. For $n$ points $(x_i, y_i)$ and a line of fit $y = mx + c$ :

\text{SSE} = \sum_{i=1}^{n}\left(y_i - m x_i - c\right)^2

For each point, $m x_i + c$ is the value the line predicts and $y_i$ is the actual value. Their difference is the deviation; each deviation is squared, and the squares are added.

Why square?

Deviations above the line are positive and below are negative, so adding them directly lets them cancel. In the example below the plain deviations add up to $1 + 0 - 1 - 2 = -2$ , but the SSE is 6. Squares are never negative, so every miss counts.

7.2Computing the SSE

Example 22Computing the SSE

A line of fit $y = 2x + 2$ is given for the data $(1, 5), (2, 6), (4, 9), (9, 18)$ . Compute the sum squared error.

Solution

There are 4 points and $m = 2$ , $c = 2$ , so $\text{SSE} = \sum (y_i - 2x_i - 2)^2$ .
Write the four terms: $(5 - 2 - 2)^2 + (6 - 4 - 2)^2 + (9 - 8 - 2)^2 + (18 - 18 - 2)^2 = (1)^2 + (0)^2 + (-1)^2 + (-2)^2$ .
$= 1 + 0 + 1 + 4 = 6$ . The sum squared error is 6.

x	y	2x + 2	y − (2x + 2)	square
1	5	4	1	1
2	6	6	0	0
4	9	10	−1	1
9	18	20	−2	4

Figure 11. The deviations of four data points from the line y = 2x + 2

7.3Comparing lines of fit

The SSE turns "how well does this line follow the data?" into a number. For the same data, a line with a smaller SSE lies closer to the points.

Example 23Which line fits better?

For the same data find the SSE of the lines $y = 2x + 3$ , $y = 2x + 2$ and $y = 2x + 1.5$ , and say which fits best.

Solution

$y = 2x + 3$ : deviations $5 - 5, 6 - 7, 9 - 11, 18 - 21 = 0, -1, -2, -3$ ; squares $0 + 1 + 4 + 9$ ; SSE = 14.
$y = 2x + 2$ : deviations $1, 0, -1, -2$ ; SSE = 6.
$y = 2x + 1.5$ : deviations $1.5, 0.5, -0.5, -1.5$ ; squares $2.25 + 0.25 + 0.25 + 2.25$ ; SSE = 5.
The line $y = 2x + 1.5$ has the smallest SSE, so of the three it follows the data best.

Figure 12. Three candidate lines for the same data, compared by SSE

Beyond the source. The source defines the SSE and computes it for a given line. How to find the line with the smallest SSE, the least-squares line, is added in 7.4 below as an extra.

7.4The least-squares line: the best line of fit

The SSE tells you how good a given line is. The least-squares line is the line $y = mx + c$ whose SSE is the smallest possible. This is the line a computer would fit to the data, so it is worth knowing how it is found.

A. The formulas

For $n$ data points $(x_i, y_i)$ , write $\bar x = \tfrac1n\sum x_i$ and $\bar y = \tfrac1n\sum y_i$ . Then

m = \frac{n\sum x_iy_i - \left(\sum x_i\right)\left(\sum y_i\right)}{n\sum x_i^2 - \left(\sum x_i\right)^2}, \qquad c = \bar y - m\bar x

(The slope can also be written $m = \dfrac{\sum (x_i - \bar x)(y_i - \bar y)}{\sum (x_i - \bar x)^2}$ .) These come from asking that the SSE, a quadratic in $m$ and $c$ , stops decreasing in both directions. That gives two equations, the normal equations:

\sum y_i = m\sum x_i + nc, \qquad \sum x_iy_i = m\sum x_i^2 + c\sum x_i

B. Worked example

Data $(1, 2), (2, 3), (3, 5)$ . Here $n = 3$ , $\sum x = 6$ , $\sum y = 10$ , $\sum xy = 2 + 6 + 15 = 23$ , $\sum x^2 = 1 + 4 + 9 = 14$ .

$m = \dfrac{3 \cdot 23 - 6 \cdot 10}{3 \cdot 14 - 36} = \dfrac{9}{6} = \dfrac32$ .
$\bar x = 2$ , $\bar y = \tfrac{10}{3}$ , so $c = \dfrac{10}{3} - \dfrac32 \cdot 2 = \dfrac13$ . The line is $y = 1.5x + \tfrac13$ .
Predictions: $\tfrac{11}{6}, \tfrac{10}{3}, \tfrac{29}{6}$ ; residuals (actual $-$ predicted): $\tfrac16, -\tfrac13, \tfrac16$ . SSE $= \tfrac1{36} + \tfrac19 + \tfrac1{36} = \tfrac16$ .
Compare $y = x + 1$ (the simple line from the first practice question): SSE $= 1$ . The least-squares value $\tfrac16$ is smaller, as it must be.

C. Facts to remember

The least-squares line always passes through $(\bar x, \bar y)$ (that is what $c = \bar y - m\bar x$ says).
Its residuals add up to zero: $\sum (y_i - mx_i - c) = \sum y_i - m\sum x_i - nc = 0$ by the first normal equation. (For other lines the sum need not be 0, and this is why the SSE squares the deviations.)
No other line has a smaller SSE. The SSE is $0$ only if all points lie exactly on one line (for example $(0, 1), (1, 3), (2, 5)$ gives $y = 2x + 1$ with SSE $0$ ).
Mean squared error $\text{MSE} = \dfrac{\text{SSE}}{n}$ , and the root mean squared error is $\sqrt{\text{MSE}}$ . Dividing by $n$ makes errors comparable across data sets of different sizes. The line minimising SSE also minimises MSE.
The formula needs $n\sum x^2 - (\sum x)^2 \neq 0$ : that fails only when all $x_i$ are equal (a vertical line of points), where there is no sensible slope.

7.5Exam tips and common mistakes

Exam tips

Make a table: x, y, prediction, deviation, square. Add the last column.
Compare lines only on the same data; a smaller SSE then means a closer fit.

Common mistakes

Adding the plain deviations (they cancel).
Computing $m x_i + c - y_i$ for some points and $y_i - m x_i - c$ for others. Squaring hides the sign, but be consistent.
Skipping a data point in the sum.

Practice questions (graded-assignment style)

Q1SSE of a line

Numerical Answer Type (NAT).

For the data $(1, 2), (2, 3), (3, 5)$ and the line $y = x + 1$ , find the SSE.

Solution

Idea

Deviation = actual − predicted, then square.

Predictions: $2, 3, 4$ . Deviations: $0, 0, 1$ .
SSE $= 0 + 0 + 1 = 1$ .
Answer: 1

Q2Deviations that cancel

Numerical Answer Type (NAT).

For the data $(1, 3), (2, 3), (3, 7)$ and the line $y = 2x$ , find the SSE.

Solution

Idea

Squares do not cancel.

Predictions: $2, 4, 6$ . Deviations: $1, -1, 1$ .
SSE $= 1 + 1 + 1 = 3$ . (The plain deviations add to 1, which hides the misses.)
Answer: 3

Q3Which fits best?

Multiple Choice Question (MCQ): choose the one correct option.

For the data $(0, 1), (1, 3), (2, 5)$ , which line has the smallest SSE?

y = 2x + 1

y = 2x

y = x + 1

All are equal

Solution

Idea

Compute the SSE of each line.

A: predictions $1, 3, 5$ , deviations $0, 0, 0$ , SSE $= 0$ .
B: predictions $0, 2, 4$ , deviations $1, 1, 1$ , SSE $= 3$ .
C: predictions $1, 2, 3$ , deviations $0, 1, 2$ , SSE $= 5$ .
Line A passes through every point, so its SSE is 0.
Answer: option 1.

Q4SSE facts

Multiple Select Question (MSQ): select all that apply.

Choose the correct statements about the SSE.

The SSE is never negative.

Squaring stops positive and negative deviations from cancelling.

For the same data, a smaller SSE means a closer fit.

An SSE of 0 means the line passes through no data point.

Solution

Idea

Think about what a sum of squares can and cannot be.

Options 1, 2 and 3 are true.
Option 4 is false: an SSE of 0 means every deviation is 0, so the line passes through every data point.
Answer: options 1, 2 and 3.

L1The slope of the least-squares line

Numerical Answer Type (NAT).

Find the slope of the least-squares line for the data $(1, 2), (2, 3), (3, 5)$ .

Solution

Idea

Use $m = \dfrac{n\sum xy - \sum x\sum y}{n\sum x^2 - (\sum x)^2}$ .

$n = 3$ , $\sum x = 6$ , $\sum y = 10$ , $\sum xy = 23$ , $\sum x^2 = 14$ .
$m = \dfrac{69 - 60}{42 - 36} = \dfrac96 = 1.5$ .
Cross-check with deviations from the means: $\bar x = 2$ , $\bar y = \tfrac{10}{3}$ ; $\sum (x - \bar x)(y - \bar y) = (-1)(-\tfrac43) + 0 + (1)(\tfrac53) = 3$ and $\sum (x - \bar x)^2 = 2$ , giving $\tfrac32$ ✓.
Answer: 1.5

L2The minimum SSE

Numerical Answer Type (NAT).

For the data $(1, 2), (2, 3), (3, 5)$ , what is the SSE of the least-squares line? (Give a decimal rounded to 3 places.)

Solution

Idea

Get $m$ and $c$ , then compute the SSE as usual.

$m = 1.5$ (L1) and $c = \bar y - m\bar x = \tfrac{10}{3} - 3 = \tfrac13$ .
Residuals: $2 - \tfrac{11}{6} = \tfrac16$ , $3 - \tfrac{10}{3} = -\tfrac13$ , $5 - \tfrac{29}{6} = \tfrac16$ . Squares: $\tfrac1{36} + \tfrac4{36} + \tfrac1{36} = \tfrac6{36} = \tfrac16$ .
Answer: 0.167

L3Properties of the least-squares line

Multiple Select Question (MSQ): select all that apply.

Choose the correct statements about the least-squares line $y = mx + c$ for a data set.

It passes through the point

(\bar x, \bar y)

The residuals

y_i - mx_i - c

add up to

0

Its SSE is always

0

No other line has a smaller SSE on the same data.

\text{MSE} = n \times \text{SSE}

Solution

Idea

Use $c = \bar y - m\bar x$ and the normal equations.

1 ✓ $m\bar x + c = \bar y$ .
2 ✓ $\sum (y_i - mx_i - c) = n\bar y - mn\bar x - nc = n(\bar y - m\bar x - c) = 0$ .
3 ✗ The SSE is $0$ only when the points are exactly collinear (the example $(1,2),(2,3),(3,5)$ has SSE $\tfrac16$ ).
4 ✓ That is what "least squares" means.
5 ✗ $\text{MSE} = \text{SSE}/n$ .
Answer: options 1, 2 and 4.

L4The intercept from the sums

Numerical Answer Type (NAT).

For the data $(1, 5), (2, 6), (4, 9), (9, 18)$ the sums are $\sum x = 16$ , $\sum y = 38$ , $\sum xy = 215$ , $\sum x^2 = 102$ . Find the intercept $c$ of the least-squares line, rounded to 2 decimals.

Solution

Idea

Find $m$ first, then $c = \bar y - m\bar x$ .

$m = \dfrac{4 \cdot 215 - 16 \cdot 38}{4 \cdot 102 - 16^2} = \dfrac{860 - 608}{408 - 256} = \dfrac{252}{152} = \dfrac{63}{38} \approx 1.658$ .
$\bar x = 4$ , $\bar y = 9.5$ , so $c = 9.5 - \tfrac{63}{38}\cdot 4 = \tfrac{361 - 252}{38} = \tfrac{109}{38} \approx 2.87$ .
Check with the first normal equation: $38 = 16m + 4c = 16 \cdot \tfrac{63}{38} + 4 \cdot \tfrac{109}{38} = \tfrac{1008 + 436}{38} = \tfrac{1444}{38} = 38$ ✓.
The SSE of this line is $\tfrac{21}{38} \approx 0.55$ , smaller than the $5$ of the best of the three lines tried in Example 23.
Answer: 2.87

L5Mean squared error

Numerical Answer Type (NAT).

For the data $(1, 5), (2, 6), (4, 9), (9, 18)$ and the line $y = 2x + 2$ , find the mean squared error.

Solution

Idea

MSE = SSE divided by the number of points.

From Example 22, $\text{SSE} = 6$ , and there are $n = 4$ points.
$\text{MSE} = 6 / 4 = 1.5$ .
Answer: 1.5

Revision overview

Line of fity = mx + c
Predicted value m xᵢ + c
Actual value yᵢ

Deviationyᵢ − m xᵢ − c
Can be positive or negative

Square itSo none cancel
Plain sum −2, SSE 6

Straight-line fitWeek 2 · Section 2.10

SSEΣ (yᵢ − m xᵢ − c)²
Sum over all n points
Example: 1 + 0 + 1 + 4 = 6

UseSmaller SSE: closer fit
Compare lines on the same data

Exampley = 2x + 3: 14
y = 2x + 2: 6
y = 2x + 1.5: 5

Revision: measuring how well a line fits data

Next · All sectionsWeek 2 Summary Sheet

All sections

Week 2 Summary Sheet

One page to revise all of Week 2: points in the plane, lines and their slopes, the forms of a line, distances and angles, and the line of fit.

The big picture

Topic 1 · Coordinates and distance

Four quadrants by the signs of $(x, y)$ .
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ .

Topic 2 · Section formula and area

Divide a segment in $m : n$ ; midpoint at $1 : 1$ .
Triangle area; zero area means collinear.

Topic 3 · Slope

$m = \Delta y / \Delta x = \tan\theta$ .
Parallel: $m_1 = m_2$ . Perpendicular: $m_1 m_2 = -1$ .

Topic 4 · Equations of a line

Slope-intercept, point-slope, two-point, intercept and general forms: all the same line.

Topic 5 · Angle between lines

$\tan\theta = \dfrac{m_2 - m_1}{1 + m_2 m_1}$ .
If the denominator is 0 the lines are perpendicular.

Topics 6 and 7 · Distances and SSE

Point-to-line and parallel-line distances from $ax + by + c = 0$ .
SSE measures how well a line of fit follows the data.

Formulas to remember

Topic	Formula
Distance between two points	$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Internal section $m : n$	$\left(\dfrac{mx_2 + nx_1}{m + n}, \dfrac{my_2 + ny_1}{m + n}\right)$
External section $m : n$	$\left(\dfrac{mx_2 - nx_1}{m - n}, \dfrac{my_2 - ny_1}{m - n}\right)$
Midpoint	$\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$
Area of a triangle	$\tfrac{1}{2}\left\|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right\|$
Slope	$m = \dfrac{y_2 - y_1}{x_2 - x_1} = \tan\theta$
Parallel / perpendicular (slopes)	$m_1 = m_2$ and $m_1 m_2 = -1$
Angle between lines	$\tan\theta = \dfrac{m_2 - m_1}{1 + m_2 m_1}$ , $\tan\varphi = \dfrac{m_1 - m_2}{1 + m_1 m_2}$
Parallel / perpendicular (general form)	$a_1 b_2 = a_2 b_1$ and $a_1 a_2 = -b_1 b_2$
Point to line	$d = \dfrac{\|a x_1 + b y_1 + c\|}{\sqrt{a^2 + b^2}}$
Between parallel lines	$D = \dfrac{\|c_1 - c_2\|}{\sqrt{a^2 + b^2}}$
Sum squared error	$\text{SSE} = \sum (y_i - m x_i - c)^2$

Forms of the equation of a line

Form	Representation	Use when you know
General	$ax + by + c = 0$	a standard shape for any line
Slope-point	$(y - y_0) = m(x - x_0)$	slope and one point
Slope-intercept (y)	$y = mx + c$	slope and y-intercept
Slope-intercept (x)	$y = m(x - d)$	slope and x-intercept
Intercept	$\dfrac{x}{a} + \dfrac{y}{b} = 1$	both intercepts
Two-point	$(y - y_1) = \dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)$	two points on the line

Which tool when?

Question	Tool
How far apart are two points?	distance formula
A point dividing a segment in $m : n$	section formula, internal or external
Are three points on one line?	triangle area is zero
How steep is a line?	$m = \Delta y / \Delta x = \tan\theta$
Parallel or perpendicular?	$m_1 = m_2$ or $m_1 m_2 = -1$
Write the line from given facts	pick the form (point-slope, two-point, intercept, ...)
Angle between two lines	$\tan\theta = \dfrac{m_2 - m_1}{1 + m_2 m_1}$
Distance from a point or between parallels	$\dfrac{\|a x_1 + b y_1 + c\|}{\sqrt{a^2 + b^2}}$ , $\dfrac{\|c_1 - c_2\|}{\sqrt{a^2 + b^2}}$
How well does a line fit data?	$\text{SSE} = \sum (y_i - m x_i - c)^2$

Common mistakes

Mixing up the order in the slope

$\dfrac{y_2 - y_1}{x_2 - x_1}$ needs the same point first in the numerator and the denominator.

Losing the second answer

$(4 - y)^2 = 9$ gives $4 - y = \pm 3$ , so two values of $y$ .

Forgetting the modulus

The area and distance formulas use the modulus so the result is never negative.

Vertical lines with slope rules

A vertical line has no slope. The parallel and perpendicular slope rules are for non-vertical lines.

Different a and b for parallel distance

Make the two equations share the same $a$ and $b$ before using $|c_1 - c_2|$ .

Skipping the square in SSE

Plain deviations cancel. Square each one, then add.

Mathematics forData Science 1

Chapters

Rectangular Coordinate System and Distance

1.1The Cartesian coordinate system

1.2Which quadrant?

1.3The distance between two points

Why the formula works

1.4Going deeper

1.5Exam tips and common mistakes

Practice questions (graded-assignment style)

Section Formula and Area of a Triangle

2.1The section formula

2.2Reading the formulas

2.3Area of a triangle

2.4Worked examples: area and collinearity

2.5Exam tips and common mistakes

Practice questions (graded-assignment style)

Slope, Parallel and Perpendicular Lines

3.1Slope of a straight line

3.2What the slope tells you

3.3Parallel and perpendicular lines

3.4Going deeper

3.5Exam tips and common mistakes

Practice questions (graded-assignment style)

Equations and Forms of a Straight Line

4.1The three basic forms

4.2All the forms of a straight line

4.3Choosing the form

4.4Going deeper

4.5Exam tips and common mistakes

Practice questions (graded-assignment style)

Angle Between Two Lines

5.1The angle between two lines

5.2Reading the two formulas

5.3Going deeper

5.4Exam tips and common mistakes

Practice questions (graded-assignment style)

General Equation: Conditions and Distances

6.1Parallel and perpendicular lines in general form

6.2Distance of a point from a line

6.3Distance between two parallel lines

6.4Going deeper

6.5Perpendicular bisector, foot and reflection, and lines through an intersection point

A. Perpendicular bisector of a segment

B. Foot of the perpendicular and reflection

C. Intersection of lines, concurrency and the family of lines

6.6Exam tips and common mistakes

Practice questions (graded-assignment style)

Straight-Line Fit and Sum Squared Error

7.1Sum squared error

7.2Computing the SSE

7.3Comparing lines of fit

7.4The least-squares line: the best line of fit

A. The formulas

B. Worked example

C. Facts to remember

7.5Exam tips and common mistakes

Practice questions (graded-assignment style)

Week 2 Summary Sheet

The big picture

Formulas to remember

Forms of the equation of a line

Which tool when?

Common mistakes

Mathematics for
Data Science 1