Lesson Notes By Weeks and Term v5 - Grade 10
Analytical geometry – Week 2 focus
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Analytical geometry – Week 2 focus
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Subject: Mathematics
Class: Grade 10
Term: 3rd Term
Week: 2
Theme: General lesson support
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Analytical geometry is a branch of mathematics that uses algebraic methods to study geometric objects. In simpler terms, it's about using coordinates to describe and analyze lines, shapes, and figures. This week, we will focus on applying the distance formula, gradient formula, and midpoint formula to solve problems related to lines and line segments in the Cartesian plane. Understanding these concepts is crucial for further studies in Mathematics, Physics, and Engineering. Analytical geometry has practical applications in fields like surveying, navigation, computer graphics, and architecture, all relevant to South Africa's development and infrastructure projects.
2. 1. The Distance Formula The distance between two points, A(x₁, y₁) and B(x₂, y₂), in the Cartesian plane is given by the distance formula: d = √((x₂ - x₁)² + (y₂ - y₁)² ) Why does this work? The distance formula is derived from the Pythagorean theorem. Imagine a right-angled triangle where the hypotenuse is the line segment AB. The lengths of the other two sides are (x₂ - x₁) and (y₂ - y₁). Applying Pythagoras' theorem (a² + b² = c²), we get (x₂ - x₁)² + (y₂ - y₁)² = d², and taking the square root gives us the distance formula.
Example 1: Calculate the distance between the points A(2, 3) and B(5, 7). x₁ = 2, y₁ = 3, x₂ = 5, y₂ = 7 d = √((5 - 2)² + (7 - 3)²) d = √((3)² + (4)²) d = √(9 + 16) d = √25 d = 5 units Example 2: A rural South African farmer wants to fence a rectangular plot of land. Two opposite corners of the plot are at coordinates (1,1) and (4,5) on a map. What is the length of one side of this rectangle if these corners are on one of its diagonals? x₁ = 1, y₁ = 1, x₂ = 4, y₂ = 5 d = √((4 - 1)² + (5 - 1)²) d = √((3)² + (4)²) d = √(9 + 16) d = √25 d = 5 units 2.
2. The Gradient Formula The gradient (or slope) of a line segment joining two points, A(x₁, y₁) and B(x₂, y₂), is given by: m = (y₂ - y₁) / (x₂ - x₁) where x₁ ≠ x₂ Why does this work? The gradient represents the "rise over run," i.e., the change in the y-coordinate (vertical change) divided by the change in the x-coordinate (horizontal change). A positive gradient indicates an increasing line (going uphill from left to right), while a negative gradient indicates a decreasing line (going downhill from left to right). A gradient of 0 represents a horizontal line, and an undefined gradient (division by zero) represents a vertical line.
Example 1: Find the gradient of the line segment joining the points C(1, 4) and D(3, 8). x₁ = 1, y₁ = 4, x₂ = 3, y₂ = 8 m = (8 - 4) / (3 - 1) m = 4 / 2 m = 2 Example 2: A road in the Drakensberg mountains rises from an altitude of 1200m to 1500m over a horizontal distance of 1.5km. What is the gradient of the road? Let (0, 1200) be the starting point and (1500, 1500) be the ending point. m = (1500 - 1200) / (1500 - 0) m = 300 / 1500 m = 0.2 2.
3. The Midpoint Formula The midpoint of a line segment joining two points, A(x₁, y₁) and B(x₂, y₂), is given by: M = ((x₁ + x₂) / 2 , (y₁ + y₂) / 2) Why does this work? The midpoint formula simply calculates the average of the x-coordinates and the average of the y-coordinates. This gives you the point that lies exactly halfway between the two endpoints.
Example 1: Find the midpoint of the line segment joining the points E(-2, 1) and F(4, 5). x₁ = -2, y₁ = 1, x₂ = 4, y₂ = 5 M = ((-2 + 4) / 2 , (1 + 5) / 2) M = (2 / 2 , 6 / 2) M = (1, 3)
Example 2: A cellphone tower is to be built equidistant from two towns in the Eastern Cape, located at map coordinates (2, 6) and (8, 2). What are the coordinates of the ideal location for the tower, assuming the map uses a Cartesian coordinate system? x₁ = 2, y₁ = 6, x₂ = 8, y₂ = 2 M = ((2 + 8) / 2 , (6 + 2) / 2) M = (10 / 2 , 8 / 2) M = (5, 4) 2.4 Parallel and Perpendicular Lines Parallel Lines: Two lines are parallel if and only if their gradients are equal. If line 1 has gradient m₁ and line 2 has gradient m₂, then m₁ = m₂.
Perpendicular Lines: Two lines are perpendicular if and only if the product of their gradients is -
1. If line 1 has gradient m₁ and line 2 has gradient m₂, then m₁ m₂ = -
1. This also means that the gradient of the second line is the negative reciprocal of the first line: m₂ = -1/m₁. 2.5 Collinearity Three points A, B, and C are collinear if they lie on the same straight line. To determine if three points are collinear: Calculate the gradient of the line segment AB. Calculate the gradient of the line segment BC. If the gradients are equal, then the points are collinear. Alternatively, you can check if the gradient of AC is equal to the gradient of AB (or BC). Guided Practice (With Solutions)
Question 1: Calculate the distance between the points P(-3, 2) and Q(1, -1).
Solution: x₁ = -3, y₁ = 2, x₂ = 1, y₂ = -1 d = √((1 - (-3))² + (-1 - 2)²) d = √((4)² + (-3)²) d = √(16 + 9) d = √25 d = 5 units
Commentary: This is a direct application of the distance formula. Pay attention to the signs.
Question 2: Find the gradient of the line passing through the points R(0, -2) and S(4, 6).
Solution: x₁ = 0, y₁ = -2, x₂ = 4, y₂ = 6 m = (6 - (-2)) / (4 - 0) m = 8 / 4 m = 2
Commentary: This is a direct application of the gradient formula. Remember that a positive gradient means the line is increasing.
Question 3: Determine the midpoint of the line segment joining the points T(5, -4) and U(-1, 2).
Solution: x₁ = 5, y₁ = -4, x₂ = -1, y₂ = 2 M = ((5 + (-1)) / 2 , (-4 + 2) / 2) M = (4 / 2 , -2 / 2) M = (2, -1)
Commentary: This is a direct application of the midpoint formula. Ensure you add the coordinates correctly before dividing by
2. Question 4: Line L1 passes through points (1, 2) and (3, 6).