Kenneth ChengMathematical modeling
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Inclination and slope of a line

Read the idea, work independently, then explain what changed.

高二選擇性必修 第一册(A版).pdf · 2.1 · PDF 56 / printed page 51

Revisit first: Exponents

TOPIC 01

Inclination and slope of a line

Connect slope, inclination and parallel/perpendicular directions, including vertical lines.

What you will be able to explain

  • Connect slope, inclination and parallel/perpendicular directions, including vertical lines.
  • Justify the method and check the conditions in a new situation.

Slope

For a nonvertical line, slope is rise divided by run and equals tan of the inclination.

m=(y2−y1)/(x2−x1)m=(y_2-y_1)/(x_2-x_1)

Exceptional directions

A vertical line has no finite slope; m1m2=−1 applies only when both slopes exist.

PREDICT → EXPLORE → EXPLAIN → TRANSFER

Make a prediction, then explore the relationship.

Lesson question: Does a steeper line always have a larger slope?

Model exploration: predict which displayed result changes with the parameters, check the values, and compare the observation with the lesson question.

y=1x+1. Angle is measured modulo 180°; distance from the origin=0.7071. This slope form excludes vertical lines.

y=1x+1. Angle is measured modulo 180°; distance from the origin=0.7071. This slope form excludes vertical lines.

Explain: Calculate two valid cases and explain the change using the defining relation.

Transfer: Compare positive and negative slopes and the vertical limiting case.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Foundation#Worked example

Find the slope of AB.

A=(1,2), B=(3,6)A=(1,2),\ B=(3,6)
  • A vertical line has no finite slope; m1m2=−1 applies only when both slopes exist.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
m=Δy/Δxm=\Delta y/\Delta x
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    m=(6−2)/(3−1)=2m=(6-2)/(3-1)=2

  3. Use consistent order in numerator and denominator.

The requested value is 2.

Checks and common pitfalls: Use consistent order in numerator and denominator.

Think first. Reveal a hint when the class is ready.

02 / Standard#Worked example

Find a so that the two nonvertical lines are parallel.

y=(2a+1)x+3,y=7x−1y=(2a+1)x+3,\quad y=7x-1
  • A vertical line has no finite slope; m1m2=−1 applies only when both slopes exist.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
2a+1=2t+12a+1=2t+1
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    2a+1=7⇒a=32a+1=7\Rightarrow a=3

  3. Different intercepts ensure distinct parallel lines once slopes match.

The requested value is 3.

Checks and common pitfalls: Different intercepts ensure distinct parallel lines once slopes match.

Think first. Reveal a hint when the class is ready.

03 / Transfer#Worked example

Lines through O meet the segment from A=(t,1) to B=(t,3). Find the slope range.

t=4t=4
  • A vertical line has no finite slope; m1m2=−1 applies only when both slopes exist.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
y=mt,1≤y≤3y=mt,\quad 1\le y\le3
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    1≤4m≤31\le 4m\le3

  3. Apply the stated relation and retain its conditions.

    1/4≤m≤3/41/4\le m\le3/4

  4. A segment gives a closed range; replacing it by the full supporting line would change the answer.

The requested relation or conclusion is shown below.

14≤m≤34\frac1{4}\le m\le\frac3{4}

Checks and common pitfalls: A segment gives a closed range; replacing it by the full supporting line would change the answer.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

Think first. Reveal a hint when the class is ready.

04 / Foundation#Your turn

Find the slope of AB.

A=(1,2), B=(3,12)A=(1,2),\ B=(3,12)
  • A vertical line has no finite slope; m1m2=−1 applies only when both slopes exist.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
m=Δy/Δxm=\Delta y/\Delta x
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    m=(12−2)/(3−1)=5m=(12-2)/(3-1)=5

  3. Use consistent order in numerator and denominator.

The requested value is 5.

Checks and common pitfalls: Use consistent order in numerator and denominator.

Think first. Reveal a hint when the class is ready.

05 / Foundation#Your turn

A line has inclination 135°. Find its slope.

m=tan⁡135∘m=\tan135^\circ
  • A vertical line has no finite slope; m1m2=−1 applies only when both slopes exist.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
tan⁡(180∘−45∘)=−tan⁡45∘\tan(180^\circ-45^\circ)=-\tan45^\circ
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    m=−1m=-1

  3. Inclination lies in [0°,180°); a second-quadrant angle has negative tangent.

The requested value is -1.

Checks and common pitfalls: Inclination lies in [0°,180°); a second-quadrant angle has negative tangent.

Think first. Reveal a hint when the class is ready.

06 / Foundation#Your turn

Describe the inclination and slope of the line.

x=7x=7
  • A vertical line has no finite slope; m1m2=−1 applies only when both slopes exist.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
Δx=0\Delta x=0
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    m=Δy/0 is not definedm=\Delta y/0\text{ is not defined}

  3. Do not replace an undefined slope by the number infinity.

The requested relation or conclusion is shown below.

α=90∘;m undefined\alpha=90^\circ;\quad m\text{ undefined}

Checks and common pitfalls: Do not replace an undefined slope by the number infinity.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

Think first. Reveal a hint when the class is ready.

07 / Foundation#Your turn

Find the slope of a line perpendicular to y=tx+1.

t=8t=8
  • A vertical line has no finite slope; m1m2=−1 applies only when both slopes exist.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
m1m2=−1m_1m_2=-1
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    m2=−1/8m_2=-1/8

  3. The given slope is nonzero, so the negative reciprocal rule applies.

The requested value is -0.125.

Checks and common pitfalls: The given slope is nonzero, so the negative reciprocal rule applies.

Think first. Reveal a hint when the class is ready.

08 / Standard#Your turn

Find the slope of a line perpendicular to y=tx+1.

t=9t=9
  • A vertical line has no finite slope; m1m2=−1 applies only when both slopes exist.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
m1m2=−1m_1m_2=-1
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    m2=−1/9m_2=-1/9

  3. The given slope is nonzero, so the negative reciprocal rule applies.

The requested value is -0.111111111111.

Checks and common pitfalls: The given slope is nonzero, so the negative reciprocal rule applies.

Think first. Reveal a hint when the class is ready.

09 / Standard#Your turn

Find a so that the two nonvertical lines are parallel.

y=(2a+1)x+3,y=21x−1y=(2a+1)x+3,\quad y=21x-1
  • A vertical line has no finite slope; m1m2=−1 applies only when both slopes exist.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
2a+1=2t+12a+1=2t+1
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    2a+1=21⇒a=102a+1=21\Rightarrow a=10

  3. Different intercepts ensure distinct parallel lines once slopes match.

The requested value is 10.

Checks and common pitfalls: Different intercepts ensure distinct parallel lines once slopes match.

Think first. Reveal a hint when the class is ready.

10 / Standard#Your turn

Find the tangent of the acute angle between slopes 0 and t.

t=11t=11
  • A vertical line has no finite slope; m1m2=−1 applies only when both slopes exist.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
tan⁡θ=∣(m2−m1)/(1+m1m2)∣\tan\theta=|(m_2-m_1)/(1+m_1m_2)|
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    tan⁡θ=∣(11−0)/(1+0)∣=11\tan\theta=|(11-0)/(1+0)|=11

  3. The denominator is nonzero here; perpendicular lines require separate handling.

The requested value is 11.

Checks and common pitfalls: The denominator is nonzero here; perpendicular lines require separate handling.

Think first. Reveal a hint when the class is ready.

11 / Standard#Your turn

Lines through O meet the segment from A=(t,1) to B=(t,3). Find the slope range.

t=12t=12
  • A vertical line has no finite slope; m1m2=−1 applies only when both slopes exist.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
y=mt,1≤y≤3y=mt,\quad 1\le y\le3
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    1≤12m≤31\le 12m\le3

  3. Apply the stated relation and retain its conditions.

    1/12≤m≤3/121/12\le m\le3/12

  4. A segment gives a closed range; replacing it by the full supporting line would change the answer.

The requested relation or conclusion is shown below.

112≤m≤312\frac1{12}\le m\le\frac3{12}

Checks and common pitfalls: A segment gives a closed range; replacing it by the full supporting line would change the answer.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

Think first. Reveal a hint when the class is ready.

12 / Transfer#Your turn

Find the tangent of the acute angle between slopes 0 and t.

t=13t=13
  • A vertical line has no finite slope; m1m2=−1 applies only when both slopes exist.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
tan⁡θ=∣(m2−m1)/(1+m1m2)∣\tan\theta=|(m_2-m_1)/(1+m_1m_2)|
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    tan⁡θ=∣(13−0)/(1+0)∣=13\tan\theta=|(13-0)/(1+0)|=13

  3. The denominator is nonzero here; perpendicular lines require separate handling.

The requested value is 13.

Checks and common pitfalls: The denominator is nonzero here; perpendicular lines require separate handling.

Think first. Reveal a hint when the class is ready.

13 / Transfer#Your turn

Lines through O meet the segment from A=(t,1) to B=(t,3). Find the slope range.

t=14t=14
  • A vertical line has no finite slope; m1m2=−1 applies only when both slopes exist.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
y=mt,1≤y≤3y=mt,\quad 1\le y\le3
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    1≤14m≤31\le 14m\le3

  3. Apply the stated relation and retain its conditions.

    1/14≤m≤3/141/14\le m\le3/14

  4. A segment gives a closed range; replacing it by the full supporting line would change the answer.

The requested relation or conclusion is shown below.

114≤m≤314\frac1{14}\le m\le\frac3{14}

Checks and common pitfalls: A segment gives a closed range; replacing it by the full supporting line would change the answer.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

Think first. Reveal a hint when the class is ready.

Focus on one question

End-of-lesson check and correction

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    Teacher preparation and assessment

    Question sequence

    • Connect slope, inclination and parallel/perpendicular directions, including vertical lines.
    • Which condition is essential in inclination and slope of a line?
    • Does a steeper line always have a larger slope?

    Board plan

    • Slope: For a nonvertical line, slope is rise divided by run and equals tan of the inclination.
      m=(y2−y1)/(x2−x1)m=(y_2-y_1)/(x_2-x_1)
    • Exceptional directions: A vertical line has no finite slope; m1m2=−1 applies only when both slopes exist.

    Anticipated thinking

    • A vertical line is perpendicular to a horizontal line even though its slope is undefined.

    Assessment checklist

    • 1 mark: choose the correct representation and conditions.
    • 1 mark: establish the intermediate relation.
    • 1 mark: complete a connected calculation or proof.
    • 1 mark: interpret and check the conclusion.

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    Curriculum and source notes ↗