Lesson Notes By Weeks and Term v5 - Grade 10

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

• Calculate the distance between two points.
• Determine the midpoint of a line segment.
• Calculate the gradient of a line.
• Determine the equation of a straight line.
• Apply these concepts to solve problems.

Lesson summary

This week's revision focuses on consolidating our understanding of Analytical Geometry, specifically relating to the distance formula, midpoint formula, gradient of a line, and equation of a straight line. These concepts are fundamental building blocks for more advanced mathematical studies and have practical applications in various fields, from architecture and engineering to navigation and computer graphics.

Lesson notes

2. 1. Distance Formula The distance d between two points A(x₁; y₁) and B(x₂; y₂) in a Cartesian plane is given by: d = √((x₂ - x₁)² + (y₂ - y₁)²)

Explanation: This formula is derived directly from the Pythagorean theorem. Imagine a right-angled triangle where the line segment AB is the hypotenuse. The horizontal side has length |x₂ - x₁| and the vertical side has length |y₂ - y₁|. Applying the Pythagorean theorem (a² + b² = c²), we get the distance formula.

Example 1: Calculate the distance between the points A(2; 3) and B(5; 7).

Solution: d = √((5 - 2)² + (7 - 3)²) d = √((3)² + (4)²) d = √(9 + 16) d = √25 d = 5 units Example 2: A soccer field has corner flags at points C(-3, 4) and D(1, -2) relative to a reference point. Calculate the length of the side CD of the soccer field. (Assume units are in meters)

Solution: d = √((1 - (-3))² + (-2 - 4)²) d = √((4)² + (-6)²) d = √(16 + 36) d = √52 d ≈ 7.21 meters 2.

2. Midpoint Formula The midpoint M(x; y) of a line segment joining the points A(x₁; y₁) and B(x₂; y₂) is given by: x = (x₁ + x₂)/2 y = (y₁ + y₂)/2 Explanation: The midpoint formula finds the average of the x-coordinates and the average of the y-coordinates. It essentially finds the "middle" point between the two given points.

Example 1: Find the midpoint of the line segment joining the points A(-1; 5) and B(3; -1).

Solution: x = (-1 + 3)/2 = 2/2 = 1 y = (5 + (-1))/2 = 4/2 = 2 Midpoint M(1; 2)

Example 2: A farmer plants two trees at coordinates P(2,5) and Q(8, 1). He wants to place a scarecrow exactly in the middle of the two trees. What are the coordinates where the farmer should place the scarecrow? (Assume units are in meters)

Solution: x = (2 + 8)/2 = 10/2 = 5 y = (5 + 1)/2 = 6/2 = 3 Scarecrow location (5; 3) 2.

3. Gradient of a Line The gradient m of a line passing through two points A(x₁; y₁) and B(x₂; y₂) is given by: m = (y₂ - y₁)/(x₂ - x₁)

Explanation: The gradient represents the steepness of the line. It's the change in the y-coordinate divided by the change in the x-coordinate ("rise over run"). A positive gradient indicates an increasing line, a negative gradient indicates a decreasing line, a gradient of 0 indicates a horizontal line, and an undefined gradient indicates a vertical line.

Example 1: Calculate the gradient of the line passing through the points C(0; 2) and D(4; 6).

Solution: m = (6 - 2)/(4 - 0) = 4/4 = 1 Example 2: A road climbs from a height of 100m to a height of 150m over a horizontal distance of 500m. What is the gradient (slope) of the road?

Solution: m = (150 - 100)/(500 - 0) = 50/500 = 1/10 = 0.1 2.

4. Equation of a Straight Line The equation of a straight line can be represented in various forms. The most common form is the slope-intercept form: y = mx + c where m is the gradient and c is the y-intercept (the point where the line crosses the y-axis).

Finding the equation given two points: Calculate the gradient m using the two points. Substitute one of the points and the gradient m into the equation y = mx + c and solve for c. Write the equation in the form y = mx + c, substituting the values for m and c. Finding the equation given the gradient and one point: Substitute the gradient m and the point into the equation y = mx + c and solve for c. Write the equation in the form y = mx + c, substituting the values for m and c.

Example 1: Find the equation of the line passing through the points (1; 3) and (3; 7).

Solution: m = (7 - 3)/(3 - 1) = 4/2 = 2 Substitute (1; 3) and m = 2 into y = mx + c: 3 = 2(1) + c => c = 1 Equation: y = 2x + 1 Example 2: Find the equation of the line with a gradient of -1 and passing through the point (2; 4).

Solution: Substitute m = -1 and (2; 4) into y = mx + c: 4 = -1(2) + c => c = 6 Equation: y = -x + 6 2.

5. Collinearity Points are collinear if they lie on the same straight line. To determine if three points are collinear: Calculate the gradient between the first two points. Calculate the gradient between the second and third points. If the gradients are equal, the points are collinear. Guided Practice (With Solutions)

Question 1: Calculate the distance between the points P(-2; 1) and Q(3; -3).

Solution: d = √((3 - (-2))² + (-3 - 1)²) d = √((5)² + (-4)²) d = √(25 + 16) d = √41 ≈ 6.40 units

Commentary: Direct application of the distance formula. Pay attention to signs when substituting values.

Question 2: Find the midpoint of the line segment joining the points R(4; -2) and S(-6; 8).

Solution: x = (4 + (-6))/2 = -2/2 = -1 y = (-2 + 8)/2 = 6/2 = 3 Midpoint M(-1; 3)

Commentary: Use the midpoint formula correctly by averaging the respective x and y coordinates.

Question 3: Determine the gradient of the line passing through the points T(1; -4) and U(5; 2).

Solution: m = (2 - (-4))/(5 - 1) = 6/4 = 3/2

Commentary: Gradient is rise over run. Ensure you subtract the y-coordinates and x-coordinates in the same order.

Question 4: Find the equation of the line passing through the point (0; -2) and having a gradient of 3.