Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

This page supports the lesson note with a companion video and a short classroom-ready summary.

For class groups and homework, share this lesson page so learners also get the summary, objectives, and full lesson context.

Performance objectives

• Find the equation of a straight line.
• Calculate the angle of inclination of a line.
• Understand and use the properties of parallel and perpendicular lines.
• Calculate the distance between two points.

Lesson summary

Analytical Geometry is a powerful tool that connects algebra and geometry. It allows us to describe geometric shapes and relationships using algebraic equations and coordinates. This means we can solve geometric problems using algebraic methods and vice versa. In South Africa, understanding spatial relationships is crucial in fields such as urban planning (designing efficient and equitable settlements), surveying (mapping land boundaries and infrastructure), and even sports analysis (understanding trajectories and angles in soccer or rugby). For example, town planners use analytical geometry to design road layouts that minimize traffic congestion and maximize accessibility for all residents.

Lesson notes

2.1 The Equation of a Straight Line A straight line can be represented by several forms of an equation.

The two most common forms are: Slope-intercept form: y = mx + c m represents the gradient (slope) of the line, indicating its steepness and direction. A positive m indicates an upward slope, while a negative m indicates a downward slope. c represents the y-intercept, the point where the line crosses the y-axis (where x = 0).

General form: ax + by + c = 0 a, b, and c are constants. This form is useful for representing lines regardless of their gradient (even vertical lines, which have an undefined gradient).

Finding the Equation of a Line: Given two points (x1, y1) and (x2, y2): Calculate the gradient: m = (y2 - y1) / (x2 - x1)

Use the point-slope form: y - y1 = m(x - x1) Rearrange to the slope-intercept form (y = mx + c) or the general form (ax + by + c = 0). Given a point (x1, y1) and the gradient m: Use the point-slope form: y - y1 = m(x - x1) Rearrange to the desired form. Given the gradient m and the y-intercept c: Directly substitute into the slope-intercept form: y = mx + c Example 1: Find the equation of the line passing through the points (2, 3) and (4, 7).

Calculate the gradient: m = (7 - 3) / (4 - 2) = 4 / 2 = 2 Use the point-slope form with the point (2, 3): y - 3 = 2(x - 2)

Rearrange to slope-intercept form: y - 3 = 2x - 4 => y = 2x - 1 Rearrange to general form: 2x - y - 1 = 0 Example 2: Find the equation of the line with a gradient of -3 that passes through the point (1, -2).

Use the point-slope form: y - (-2) = -3(x - 1)

Simplify: y + 2 = -3x + 3 Rearrange to slope-intercept form: y = -3x + 1 Rearrange to general form: 3x + y - 1 = 0 2.2 Angle of Inclination (θ) The angle of inclination (θ) of a line is the angle measured counter-clockwise from the positive x-axis to the line. The gradient (m) of a line is related to the angle of inclination by the following formula: m = tan θ Therefore, θ = arctan(m) or tan⁻¹(m). Remember that your calculator must be in degree mode when calculating the angle of inclination. Also, consider the quadrant in which the angle lies to ensure you get the correct angle between 0° and 180°.

Example 3: Find the angle of inclination of the line y = x +

2. Identify the gradient: m = 1 Calculate the angle of inclination: θ = tan⁻¹(1) = 45° Example 4: Find the angle of inclination of the line y = -√3x +

5. Identify the gradient: m = -√3 Calculate the angle of inclination: θ = tan⁻¹(-√3) = -60°. Since we need an angle between 0° and 180°, we add 180°: θ = -60° + 180° = 120° 2.3 Parallel and Perpendicular Lines Parallel lines: Parallel lines have the same gradient. If line 1 has gradient m1 and line 2 has gradient m2, then for parallel lines, m1 = m

2. Perpendicular lines: Perpendicular lines intersect at a right angle (90°). The product of their gradients is -

1. If line 1 has gradient m1 and line 2 has gradient m2, then for perpendicular lines, m1 m2 = -

1. This means m2 = -1/m

1. Example 5: Line 1 has the equation y = 3x -

1. Find the equation of a line parallel to line 1 that passes through the point (2, 5). The gradient of line 1 is m1 =

3. Since the lines are parallel, the gradient of the new line (m2) is also 3: m2 =

3. Use the point-slope form with the point (2, 5) and m2 = 3: y - 5 = 3(x - 2)

Rearrange to slope-intercept form: y - 5 = 3x - 6 => y = 3x - 1 Example 6: Line 1 has the equation y = (1/2)x +

4. Find the equation of a line perpendicular to line 1 that passes through the point (-1, 3). The gradient of line 1 is m1 = 1/

2. The gradient of the perpendicular line (m2) is the negative reciprocal of m1: m2 = -1 / (1/2) = -

2. Use the point-slope form with the point (-1, 3) and m2 = -2: y - 3 = -2(x - (-1))

Rearrange to slope-intercept form: y - 3 = -2x - 2 => y = -2x + 1 2.4 Distance Between Two Points The distance d between two points (x1, y1) and (x2, y2) is given by the distance formula, which is derived from the Pythagorean theorem: d = √[(x2 - x1)² + (y2 - y1)²] Example 7: Find the distance between the points (1, 2) and (4, 6).

Apply the distance formula: d = √[(4 - 1)² + (6 - 2)²] Simplify: d = √[(3)² + (4)²] = √(9 + 16) = √25 = 5 Guided Practice (With Solutions)

Question 1: Determine the equation of the line that passes through the points A(1, 5) and B(3, 11). Write your answer in the form y = mx + c.

Solution: Calculate the gradient: m = (11 - 5) / (3 - 1) = 6 / 2 = 3 Use the point-slope form with point A(1, 5): y - 5 = 3(x - 1)

Simplify: y - 5 = 3x - 3 Rearrange to slope-intercept form: y = 3x + 2

Commentary: This question tests the ability to find the equation of a line given two points. We first calculate the gradient and then use the point-slope form before converting to the desired slope-intercept form.

Question 2: Find the angle of inclination of the line 2y = -x +

4. Solution: Rewrite the equation in slope-intercept form: y = (-1/2)x + 2 Identify the gradient: m = -1/2 Calculate the angle of inclination: θ = tan⁻¹(-1/2) ≈ -26.57°.