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Relate a line's gradient and angle of inclination

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Angle of inclination

The angle of inclination is measured anticlockwise from the positive x direction. The gradient equals the tangent of that angle, with the angle taken from 0° up to but not including 180°

\[m=\tan\theta\]
  1. 1Use the prescribed interval Measure the angle anticlockwise from the positive x direction, from 0° up to but not including 180° \(\displaystyle 0°\le\theta<180°\)
  2. 2Positive gradients A rising line has an acute inclination between 0° and 90°. Take the inverse tangent of the gradient directly \(\displaystyle m>0\Rightarrow 0°<\theta<90°\)
  3. 3Negative gradients A falling line has an obtuse inclination between 90° and 180°. Take the inverse tangent of the size of the gradient, then subtract from 180° \(\displaystyle \begin{gathered}m=\frac{0-6}{6-0}=-1\\\tan^{-1}(1)=45°\\\theta=180°-45°=135°\end{gathered}\)
  4. 4Use edge rules A horizontal line has an inclination of 0° and a vertical line has an inclination of 90° \(\displaystyle m=0\Rightarrow\theta=0°,\quad \text{vertical line}\Rightarrow\theta=90°\)
Year 11, VCE Units 1 & 2, Chapter 1, 1 difficulty level, Calculator allowedBuild a worksheet

Angle of inclination

The angle of inclination is measured anticlockwise from the positive horizontal direction. The gradient and inclination are related by tangent, with the angle chosen in the prescribed interval

\[m=\tan\theta\]
  1. Use the prescribed interval Choose the inclination from zero degrees up to, but not including, 180 degrees \(\displaystyle 0^\circ\le\theta<180^\circ\)
  2. Positive gradients A positive gradient gives an inclination between 0 degrees and 90 degrees \(\displaystyle m>0\Rightarrow0^\circ<\theta<90^\circ\)
  3. Negative gradients A negative gradient gives an obtuse inclination \(\displaystyle m<0\Rightarrow90^\circ<\theta<180^\circ\)
  4. Use edge rules A horizontal line has zero inclination and a vertical line has an inclination of 90 degrees \(\displaystyle m=0\Rightarrow\theta=0^\circ,\quad\text{vertical line}\Rightarrow\theta=90^\circ\)

Worked example

Line AB passes through A(1,4) and B(3,2). Find its gradient and angle of inclination

\[A(1,4),\quad B(3,2)\]
  1. \(\displaystyle m=\frac{2-4}{3-1}=\frac{-2}{2}=-1\)Use the change in y divided by the change in x to find the gradient
  2. \(\displaystyle |m|=1\)The magnitude of the negative gradient gives the acute reference calculation
  3. \(\displaystyle \alpha=\tan^{-1}(1)=45^\circ\)Use degree mode to find the acute reference angle
  4. \(\displaystyle \theta=180^\circ-45^\circ=135^\circ\)Use the obtuse supplement because the gradient is negative

Common mistakes

Using the acute angle for a negative gradient
✕\(\displaystyle m=-2\Rightarrow\theta=\tan^{-1}(2)\)✓\(\displaystyle m=-2\Rightarrow\theta=180^\circ-\tan^{-1}(2)\)
A negative gradient gives an obtuse inclination, so the acute reference angle needs its supplement
Using a signed inverse-tangent value
✕\(\displaystyle m=-1\Rightarrow\theta=-45^\circ\)✓\(\displaystyle m=-1\Rightarrow\theta=135^\circ\)
The inclination is measured anticlockwise in the interval from 0 degrees to less than 180 degrees
Using radians instead of degrees
✕\(\displaystyle m=1\Rightarrow\tan^{-1}(1)=0.785\ \text{radians}\)✓\(\displaystyle m=1\Rightarrow\tan^{-1}(1)=45^\circ\)
The required inclination is expressed in degrees, so the calculator must use degree mode

Difficulty levels

challengeFind the gradient from two points before calculating the inclination

Questions about this topic

What interval is used for an angle of inclination?+

The inclination is chosen in the interval 0 degrees to less than 180 degrees, measured anticlockwise from the positive horizontal direction

Why is the inclination obtuse for a negative gradient?+

A negative gradient means the line falls from left to right. Its anticlockwise angle from the positive horizontal direction is therefore greater than 90 degrees

What is the inclination of a vertical line?+

A vertical line has an inclination of 90 degrees. Its gradient is undefined, so the tangent formula is not used to find its gradient

Is this a separately named Unit 1 requirement?+

No. This is bounded enrichment supporting gradient and coordinate geometry. Current Methods scope does not separately name an inclination topic. The three served families calculate a gradient, select an angle description, and provide the final inclination

Do the graphs show angles faithfully?+

These diagrams use equal unit lengths on both axes. Use the given coordinates and degree-mode calculation rather than measuring a screen image

Can I use a calculator?+

A calculator is allowed unless a question is marked no calculator

Which year level and chapter is this?+

Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 1: Linear relations and coordinate geometry

In this chapter

Recognise linear equationsSubstitute values into a formulaSolve a linear equationSet up and solve a practical linear equationRearrange literal equations and construct equivalent line formsConstruct a table of values from a formulaRead coordinates and points from a straight-line graphClassify and compare gradients from graphsFind the gradient of a line through two pointsSolve and interpret a linear inequationRead the vertical-axis intercept and gradient from an equationRecognise a line or find points for its graphConstruct a line on a coordinate gridFind the equation of a line from its graph or from its intercept and slopeFind the equation of a line through two pointsConstruct a linear model from a descriptionInterpret the graph of a linear modelRead the point of intersection of two linesSolve a pair of simultaneous linear equationsSolve a pair of simultaneous linear equations by eliminationSet up a pair of simultaneous equations from a descriptionSolve a practical problem with simultaneous equationsSolve a linear system with three or four unknownsVerify coordinate incidence and determine a parameterForm an equation of a parallel or perpendicular lineRelate a line's gradient and angle of inclinationMidpoint, endpoint and distance of a coordinate segmentDetermine a perpendicular bisector