Analytical Geometry (Grade 11 NSC Matric Mathematics): Revision Notes

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Inclination of a Line

What is the angle of inclination?

The angle of inclination of a straight line is the angle that the line makes with the positive x-axis. This angle is always measured from the positive x-axis in an anticlockwise direction.

Figure

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The angle of inclination is represented by the Greek letter θ (theta) and can range from 0° to 180°. Understanding this concept is crucial because it directly connects the geometric appearance of a line with its algebraic properties.

The relationship between gradient and angle of inclination

There is a fundamental relationship between the gradient of a line and its angle of inclination. From the coordinate geometry basics, we know that gradient is calculated as:

m=ΔyΔxm = \frac{\Delta y}{\Delta x}

From trigonometry, we also know that the tangent ratio is:

tan⁡θ=opposite sideadjacent side\tan \theta = \frac{\text{opposite side}}{\text{adjacent side}}

Looking at the right triangle formed by the line, we can see that:

tan⁡θ=ΔyΔx\tan \theta = \frac{\Delta y}{\Delta x}

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The Key Relationship:

m=tan⁡θ for 0°≤θ<180°m = \tan \theta \text{ for } 0° \leq \theta < 180°

This means that the gradient of any straight line equals the tangent of its angle of inclination.

Special cases you need to know

Vertical lines

For vertical lines:

  • The angle of inclination θ = 90°
  • The gradient is undefined because there is no change in x-values (Δx = 0)
  • Therefore, tan θ is also undefined (since tan 90° is undefined)

Horizontal lines

For horizontal lines:

  • The angle of inclination θ = 0°
  • The gradient equals 0 because there is no change in y-values (Δy = 0)
  • Therefore, tan θ = tan 0° = 0

Lines with negative gradients

When a straight line has a negative gradient, the angle formed between the line and the positive x-axis appears obtuse (greater than 90°).

Figure

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Process for Handling Negative Gradients:

  1. Calculate the gradient as normal: m=tan⁡−1(negative value)m = \tan^{-1}(\text{negative value})
  2. This gives a negative angle in the fourth quadrant
  3. Add 180° to convert this to the obtuse angle in the second quadrant
  4. The final angle of inclination will be between 90° and 180°

Figure

Worked examples

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Worked Example 1: Finding the angle of inclination from two points

Question: Determine the angle of inclination of the straight line passing through the points (2; 1) and (-3; -9).

Solution:

Step 1: Draw a sketch to visualise the problem.

Figure

Step 2: Assign variables to the coordinates.

  • Point 1: x1=2,y1=1x_1 = 2, y_1 = 1
  • Point 2: x2=−3,y2=−9x_2 = -3, y_2 = -9

Step 3: Calculate the gradient.

  • m=y2−y1x2−x1=−9−1−3−2=−10−5=2m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-9 - 1}{-3 - 2} = \frac{-10}{-5} = 2

Step 4: Use the gradient to find the angle of inclination.

  • tan⁡θ=m=2\tan \theta = m = 2
  • θ=tan⁡−1(2)=63.4°\theta = \tan^{-1}(2) = 63.4°

Important: Make sure your calculator is in DEG (degrees) mode.

Answer: The angle of inclination is 63.4°.

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Worked Example 2: Finding the equation from angle of inclination

Question: Determine the equation of the straight line passing through the point (3; 1) with an angle of inclination of 135°.

Solution:

Step 1: Use the angle of inclination to find the gradient.

  • m=tan⁡θ=tan⁡135°=−1m = \tan \theta = \tan 135° = -1

Step 2: Write the gradient-point form of the line equation.

  • y−y1=m(x−x1)y - y_1 = m(x - x_1)

Step 3: Substitute the known values.

  • y−1=−1(x−3)y - 1 = -1(x - 3)
  • y−1=−x+3y - 1 = -x + 3
  • y=−x+4y = -x + 4

Answer: The equation of the straight line is y = -x + 4.

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Worked Example 3: Finding the acute angle between two lines

Question: Determine the acute angle between the line passing through points M(-1; 7/4) and N(4; 3), and the straight line y = -3/2x + 4.

Solution:

Step 1: Draw a sketch to visualise the problem.

Figure

Step 2: Find the gradient of the first line (through M and N).

  • m1=3−744−(−1)=545=14m_1 = \frac{3 - \frac{7}{4}}{4 - (-1)} = \frac{\frac{5}{4}}{5} = \frac{1}{4}

Step 3: Find the angle of inclination of the first line.

  • tan⁡α=14\tan \alpha = \frac{1}{4}
  • α=tan⁡−1(14)=14.0°\alpha = \tan^{-1}\left(\frac{1}{4}\right) = 14.0°

Step 4: Find the angle of inclination of the second line.

  • From y = -3/2x + 4, the gradient is -3/2.
  • tan⁡−1(−32)=−56.3°\tan^{-1}\left(-\frac{3}{2}\right) = -56.3°

Since this is negative, we add 180°:

  • β=−56.3°+180°=123.7°\beta = -56.3° + 180° = 123.7°

Step 5: Calculate the acute angle between the lines.

  • θ=180°+α−β=180°+14.0°−123.7°=70.3°\theta = 180° + \alpha - \beta = 180° + 14.0° - 123.7° = 70.3°

Answer: The acute angle between the two straight lines is 70.3°.

Key formulas and exam tips

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Essential Formulas:

  • Gradient: m=ΔyΔxm = \frac{\Delta y}{\Delta x}
  • Angle of inclination: m=tan⁡θm = \tan \theta for 0°≤θ<180°0° \leq \theta < 180°
  • For negative gradients: Add 180° to the reference angle
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Exam Tips:

  • Always ensure your calculator is in degrees mode (DEG)
  • For negative gradients, remember to add 180° to get the obtuse angle
  • Draw a sketch when possible - it helps prevent errors
  • Double-check your answer by substituting back into m=tan⁡θm = \tan \theta
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Common Exam Traps:

  • Forgetting to convert negative angles to obtuse angles
  • Calculator set to radians instead of degrees
  • Mixing up the order of coordinates when calculating gradient
  • Not recognising when to find acute vs obtuse angles between lines
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Key Points to Remember:

  • The angle of inclination is the angle between a line and the positive x-axis, measured anticlockwise
  • The key relationship is m=tan⁡θm = \tan \theta, connecting gradient and angle of inclination
  • Vertical lines have θ = 90° and undefined gradient
  • Horizontal lines have θ = 0° and gradient = 0
  • For negative gradients, add 180° to the reference angle to get the correct obtuse angle of inclination

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