Lesson Notes By Weeks and Term v5 - Grade 10

Lesson Video

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Performance objectives

• Calculate the distance between two points.
• Find the coordinates of the midpoint of a line segment.
• Calculate the gradient (slope) of a line.
• Interpret the meaning of a line's gradient.
• Apply these formulas to solve problems.

Lesson summary

Analytical geometry is a powerful tool that combines algebra and geometry to solve problems involving lines, shapes, and their positions in a coordinate plane. This field is crucial for various applications, from urban planning and surveying to computer graphics and navigation systems. In South Africa, analytical geometry can be applied to understand spatial relationships, develop efficient road networks, plan housing developments, and even analyse sports field layouts. Understanding analytical geometry provides a strong foundation for more advanced mathematics, physics, engineering, and computer science studies.

Lesson notes

The Cartesian Plane: The Cartesian plane is a two-dimensional coordinate system defined by two perpendicular number lines, the x-axis (horizontal) and the y-axis (vertical). The point where the axes intersect is called the origin, denoted as (0, 0). Any point in the plane can be uniquely identified by an ordered pair of coordinates (x, y), where 'x' represents the horizontal distance from the origin and 'y' represents the vertical distance.

Distance Formula: The distance 'd' between two points A(x₁, y₁) and B(x₂, y₂) in the Cartesian plane is given by the formula: ``` d = √((x₂ - x₁)² + (y₂ - y₁)² ) ``` This formula is derived from the Pythagorean theorem. Imagine a right-angled triangle with AB as the hypotenuse. The horizontal side has length |x₂ - x₁|, and the vertical side has length |y₂ - y₁|. Applying Pythagoras, we get d² = (x₂ - x₁)² + (y₂ - y₁)² , and taking the square root gives us the distance formula.

Example 1: Find the distance between points A(1, 2) and B(4, 6).

Solution: ``` d = √((4 - 1)² + (6 - 2)²) d = √((3)² + (4)²) d = √(9 + 16) d = √25 d = 5 ``` Therefore, the distance between A and B is 5 units.

Example 2: A cellular phone tower is located at point C(5, -2) on a map. Another tower is at point D(-3, 4). The map is scaled in kilometers. What is the distance between the two towers?

Solution: ``` d = √((-3 - 5)² + (4 - (-2))²) d = √((-8)² + (6)²) d = √(64 + 36) d = √100 d = 10 ``` The distance between the towers is 10 kilometers.

Midpoint Formula: The midpoint M(xₘ, yₘ) of a line segment joining points A(x₁, y₁) and B(x₂, y₂) is given by the formula: ``` xₘ = (x₁ + x₂) / 2 yₘ = (y₁ + y₂) / 2 ``` The midpoint is simply the average of the x-coordinates and the average of the y-coordinates.

Example 1: Find the midpoint of the line segment joining points A(-2, 3) and B(4, -1).

Solution: ``` xₘ = (-2 + 4) / 2 = 2 / 2 = 1 yₘ = (3 + (-1)) / 2 = 2 / 2 = 1 ``` Therefore, the midpoint is M(1, 1).

Example 2: A town, Thabazimbi, is located at coordinates (2,7) on a map and another town, Bela-Bela, is located at (8,1). A new cell phone tower is to be erected directly between the two towns for equal coverage. What is the location of the cell phone tower?

Solution: ``` xₘ = (2 + 8) / 2 = 10 / 2 = 5 yₘ = (7 + 1) / 2 = 8 / 2 = 4 ``` Therefore, the location of the cell phone tower is (5, 4).

Gradient (Slope)

Formula: The gradient (or slope) 'm' of a line passing through two points A(x₁, y₁) and B(x₂, y₂) is given by the formula: ``` m = (y₂ - y₁) / (x₂ - x₁) ``` The gradient represents the rate of change of 'y' with respect to 'x'. It tells us how much the line rises (or falls) for every unit increase in 'x'. A positive gradient indicates an upward slope, a negative gradient indicates a downward slope, a zero gradient indicates a horizontal line, and an undefined gradient indicates a vertical line.

Example 1: Find the gradient of the line passing through points A(-1, 1) and B(2, 7).

Solution: ``` m = (7 - 1) / (2 - (-1)) m = 6 / 3 m = 2 ``` Therefore, the gradient of the line is

2. This means that for every 1 unit increase in 'x', 'y' increases by 2 units.

Example 2: Calculate the gradient of a road in the Drakensberg mountains going from point P(0, 500) to point Q(1000, 2000) , where the coordinates are in meters.

Solution: ``` m = (2000 - 500) / (1000 - 0) m = 1500 / 1000 m = 1.5 ``` Therefore, the gradient of the road is 1.

5. This means that for every 1 meter increase in 'x', 'y' increases by 1.5 meters. This represents a steep uphill slope. Guided Practice (With Solutions)

Question 1: Calculate the distance between the points (3, -2) and (-1, 1).

Solution: Using the distance formula: d = √((x₂ - x₁)² + (y₂ - y₁)² ) d = √((-1 - 3)² + (1 - (-2))²) d = √((-4)² + (3)²) d = √(16 + 9) d = √25 = 5 Therefore, the distance is 5 units.

Commentary: This is a straightforward application of the distance formula. Be careful with the signs when substituting the coordinates.

Question 2: Find the midpoint of the line segment joining the points (5, 8) and (-3, 2).

Solution: Using the midpoint formula: xₘ = (x₁ + x₂) / 2 and yₘ = (y₁ + y₂) / 2 xₘ = (5 + (-3)) / 2 = 2 / 2 = 1 yₘ = (8 + 2) / 2 = 10 / 2 = 5 Therefore, the midpoint is (1, 5).

Commentary: The midpoint formula is relatively simple, but ensure you add the correct x and y values before dividing.

Question 3: Determine the gradient of the line passing through the points (2, 4) and (6, 12).

Solution: Using the gradient formula: m = (y₂ - y₁) / (x₂ - x₁) m = (12 - 4) / (6 - 2) m = 8 / 4 m = 2 Therefore, the gradient of the line is

2. Commentary: Remember that gradient is rise over run (change in y over change in x). Ensure you subtract the coordinates in the same order for both numerator and denominator.

Question 4: A straight road passes through two towns. Town A is at coordinates (1, 3) and Town B is at coordinates (5, 6) (values in km). What is the length of the road between the two towns?