LEARN · EXPLAIN · REVISE

A line has inclination 135°. Find its slope.

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.

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

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

01 / Foundation#Your turn

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^\circ-45^\circ)=-\tan45^\circ

Worked solution

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.

Teacher preparation and assessment

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?

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

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