Lesson Notes By Weeks and Term v5 - Grade 10

Lesson Video

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Performance objectives

• Plot and identify points on the Cartesian plane.
• Calculate the distance between two points.
• Find the midpoint of a line segment.
• Calculate the gradient (slope) of a line.
• Interpret the meaning of a line's gradient.

Lesson summary

Analytical geometry is the study of geometric properties and relationships using algebraic equations. It bridges the gap between algebra and geometry, allowing us to describe shapes and positions with numbers. This is incredibly useful in various fields, from mapping and construction to computer graphics and even understanding complex data patterns. Imagine, for example, using it to plan the most efficient route for a delivery service in Johannesburg or designing a safe and stable bridge across the Orange River. Analytical geometry provides the tools to solve practical problems involving distance, midpoint, gradient, and equations of lines, all concepts vital in real-world applications.

Lesson notes

2.1 The Cartesian Plane The Cartesian plane (also known as the coordinate plane) is formed by two perpendicular number lines: the horizontal x-axis and the vertical y-axis. Their point of intersection is called the origin, represented by the coordinates (0, 0). Each point on the Cartesian plane can be uniquely identified by an ordered pair of numbers, (x, y), called its coordinates. The first number, x, is the x-coordinate or abscissa, which represents the point's horizontal distance from the y-axis. The second number, y, is the y-coordinate or ordinate, which represents the point's vertical distance from the x-axis. The Cartesian plane is divided into four quadrants: Quadrant I: x > 0, y > 0 (Top Right)

Quadrant II: x 0 (Top Left)

Quadrant III: x 0, y 1 , y 1 ) and B(x 2 , y 2 ) on the Cartesian plane is given by the distance formula: d = √((x 2 - x 1 ) 2 + (y 2 - y 1 ) 2 )

Explanation: The distance formula is derived from the Pythagorean theorem. Consider the right-angled triangle formed by the points A, B, and a point C(x 2 , y 1 ). The length of AC is |x 2 - x 1 | and the length of BC is |y 2 - y 1 |. The distance AB (the hypotenuse) is then found using Pythagoras' theorem: AB 2 = AC 2 + BC 2 . Taking the square root of both sides gives the distance formula.

Example 1: Find the distance between the points P(1, 2) and Q(4, 6). d = √((4 - 1) 2 + (6 - 2) 2 ) d = √((3) 2 + (4) 2 ) d = √(9 + 16) d = √25 d = 5 Therefore, the distance between P and Q is 5 units.

Example 2: A farm owner in KwaZulu-Natal wants to fence a triangular piece of land. The vertices of the triangle are at points R(-2, -1), S(1, 3), and T(4, -1). Calculate the total length of fencing needed. First, we need to find the lengths of the sides RS, ST, and TR using the distance formula: RS = √((1 - (-2)) 2 + (3 - (-1)) 2 ) = √((3) 2 + (4) 2 ) = √25 = 5 ST = √((4 - 1) 2 + (-1 - 3) 2 ) = √((3) 2 + (-4) 2 ) = √25 = 5 TR = √((-2 - 4) 2 + (-1 - (-1)) 2 ) = √((-6) 2 + (0) 2 ) = √36 = 6 Total length of fencing = RS + ST + TR = 5 + 5 + 6 = 16 units. 2.3 The Midpoint Formula The midpoint, M(x m , y m ), of a line segment connecting two points A(x 1 , y 1 ) and B(x 2 , y 2 ) on the Cartesian plane is given by the midpoint formula: x m = (x 1 + x 2 ) / 2 y m = (y 1 + y 2 ) / 2 Therefore, M((x 1 + x 2 ) / 2, (y 1 + y 2 ) / 2)

Explanation: The midpoint formula finds the average of the x-coordinates and the average of the y-coordinates of the two endpoints. This effectively finds the point exactly halfway between the two given points.

Example 1: Find the midpoint of the line segment connecting the points A(-2, 5) and B(4, 1). x m = (-2 + 4) / 2 = 2 / 2 = 1 y m = (5 + 1) / 2 = 6 / 2 = 3 Therefore, the midpoint is M(1, 3).

Example 2: A surveyor in Limpopo needs to find the exact centre point between two landmarks represented by coordinates C(0, -3) and D(6, 2). Determine the coordinates of this centre point. x m = (0 + 6) / 2 = 6 / 2 = 3 y m = (-3 + 2) / 2 = -1 / 2 = -0.5 The centre point is M(3, -0.5). 2.4 The Gradient of a Line The gradient (or slope), m, of a line passing through two points A(x 1 , y 1 ) and B(x 2 , y 2 ) on the Cartesian plane is given by the gradient formula: m = (y 2 - y 1 ) / (x 2 - x 1 )

Explanation: The gradient represents the "steepness" of the line. It is the ratio of the vertical change (rise) to the horizontal change (run) between any two points on the line. A positive gradient indicates that the line slopes upwards from left to right. A negative gradient indicates that the line slopes downwards from left to right. A gradient of 0 indicates a horizontal line. An undefined gradient (division by zero) indicates a vertical line.

Example 1: Find the gradient of the line passing through the points E(2, 1) and F(5, 7). m = (7 - 1) / (5 - 2) = 6 / 3 = 2 The gradient of the line is

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

Example 2: Calculate the gradient of the line segment joining G(-3, 4) and H(1, -2). m = (-2 - 4) / (1 - (-3)) = -6 / 4 = -3/2 The gradient is -3/2, indicating a downward slope. Guided Practice (With Solutions)

Question 1: Plot the points A(1, 5) and B(4, 1) on a Cartesian plane. Calculate the distance between A and

B. Solution: First plot the points A and B. d = √((4 - 1) 2 + (1 - 5) 2 ) d = √((3) 2 + (-4) 2 ) d = √(9 + 16) d = √25 d = 5 The distance between A and B is 5 units. We applied the distance formula directly after correctly identifying the coordinates.

Question 2: Find the midpoint of the line segment connecting the points C(-5, 2) and D(1, -4).

Solution: x m = (-5 + 1) / 2 = -4 / 2 = -2 y m = (2 + (-4)) / 2 = -2 / 2 = -1 The midpoint is M(-2, -1). This is a straightforward application of the midpoint formula.

Question 3: Calculate the gradient of the line passing through the points E(-2, -3) and F(0, 1).

Solution: m = (1 - (-3)) / (0 - (-2)) = 4 / 2 = 2 The gradient of the line is

2. We substituted the coordinates into the gradient formula.