Lesson Notes By Weeks and Term v5 - Grade 10

Lesson Video

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Performance objectives

• Calculate the gradient (slope) of a line given two points.
• Determine the equation of a line given the gradient and a point.
• Determine the equation of a line given two points.
• Use the point-slope form y - y1 = m(x - x1).
• Solve problems involving collinear points.

Lesson summary

Analytical geometry provides us with a powerful toolkit to describe and analyze geometric shapes using algebra. This is particularly important because it allows us to model and solve real-world problems involving position, distance, and spatial relationships. Consider town planning where distances between amenities need to be calculated, or land surveying where areas and boundaries are defined mathematically. Understanding analytical geometry is fundamental for various careers, including engineering, architecture, surveying, and computer graphics, all vital for South Africa's infrastructure development and technological advancement.

Lesson notes

2.1 Gradient of a Line The gradient (often denoted by 'm') of a line is a measure of its steepness. It tells us how much the y-value changes for every one unit change in the x-value. A positive gradient means the line slopes upwards from left to right, a negative gradient means it slopes downwards, a zero gradient means it is a horizontal line, and an undefined gradient means it is a vertical line.

Formula: Given two points on a line, (x1, y1) and (x2, y2), the gradient 'm' is calculated as: ``` m = (y2 - y1) / (x2 - x1) ``` Example 1: Calculating the gradient Two points in Soweto are being used as reference points for a new fibre optic cable installation. Point A is at coordinates (2, 3) and Point B is at coordinates (6, 11). Calculate the gradient of the line connecting these two points.

Solution: Let A = (x1, y1) = (2, 3) and B = (x2, y2) = (6, 11) m = (y2 - y1) / (x2 - x1) m = (11 - 3) / (6 - 2) m = 8 / 4 m = 2 Therefore, the gradient of the line connecting points A and B is

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

Example 2: Gradient with negative values Find the gradient of the line passing through the points C(-1, 4) and D(3, -2).

Solution: Let C = (x1, y1) = (-1, 4) and D = (x2, y2) = (3, -2) m = (y2 - y1) / (x2 - x1) m = (-2 - 4) / (3 - (-1)) m = -6 / 4 m = -3/2 Therefore, the gradient of the line connecting points C and D is -3/

2. This indicates a downward slope. 2.2 Equation of a Line: Slope-Intercept Form (y = mx + c) The slope-intercept form of a linear equation is y = mx + c, where: 'y' and 'x' are the coordinates of any point on the line. 'm' is the gradient of the line (as defined above). 'c' is the y-intercept (the point where the line crosses the y-axis; the y-value when x = 0). Finding the equation given the gradient and a point: If we know the gradient 'm' and a point (x1, y1) on the line, we can substitute these values into y = mx + c to solve for 'c', and thus find the equation of the line.

Example 3: Finding the equation given gradient and a point A straight road has a gradient of 1.

5. It passes through the point (4, 7), which is the location of a traffic light. Determine the equation of the road.

Solution: We know m = 1.5 and (x1, y1) = (4, 7). Substitute these values into y = mx + c: 7 = 1.5 * 4 + c 7 = 6 + c c = 7 - 6 c = 1 Therefore, the equation of the line (the road) is y = 1.5x +

1. Finding the equation given two points: If we are given two points, (x1, y1) and (x2, y2), we first calculate the gradient 'm' using the formula above. Then, we substitute 'm' and one of the points into the equation y = mx + c and solve for 'c'.

Example 4: Finding the equation given two points A water pipe runs in a straight line between two points: (1, 2) and (3, 8). Determine the equation of the line representing the water pipe.

Solution: First, find the gradient: m = (8 - 2) / (3 - 1) m = 6 / 2 m = 3 Now, substitute m = 3 and one of the points (let's use (1, 2)) into y = mx + c: 2 = 3 * 1 + c 2 = 3 + c c = 2 - 3 c = -1 Therefore, the equation of the line representing the water pipe is y = 3x - 1. 2.3 Equation of a Line: Point-Slope Form (y - y1 = m(x - x1)) The point-slope form of a linear equation is: ``` y - y1 = m(x - x1) ``` where: (x1, y1) is a known point on the line. 'm' is the gradient of the line. 'x' and 'y' are the coordinates of any other point on the line. This form is particularly useful when you have a gradient and a point and you need to quickly determine the equation of the line.

Example 5: Using the point-slope form A cable car travels up a mountain with a gradient of 0.

8. It passes through a station at coordinates (5, 6). Use the point-slope form to find the equation of the line representing the cable car's path.

Solution: We know m = 0.8 and (x1, y1) = (5, 6). Substitute these values into y - y1 = m(x - x1): y - 6 = 0.8(x - 5) y - 6 = 0.8x - 4 y = 0.8x - 4 + 6 y = 0.8x + 2 Therefore, the equation of the line representing the cable car's path is y = 0.8x + 2. 2.4 Collinear Points Collinear points are points that lie on the same straight line. To determine if three or more points are collinear, we can check if the gradient between any two pairs of points is the same. If the gradients are equal, the points are collinear.

Example 6: Determining if points are collinear Are the points P(1, 1), Q(3, 5), and R(4, 7) collinear?

Solution: Calculate the gradient between points P and Q: mPQ = (5 - 1) / (3 - 1) = 4 / 2 = 2 Calculate the gradient between points Q and R: mQR = (7 - 5) / (4 - 3) = 2 / 1 = 2 Since mPQ = mQR = 2, the points P, Q, and R are collinear. Guided Practice (With Solutions)

Question 1: Calculate the gradient of the line passing through the points (2, -1) and (5, 5).

Solution: Let (x1, y1) = (2, -1) and (x2, y2) = (5, 5) m = (y2 - y1) / (x2 - x1) m = (5 - (-1)) / (5 - 2) m = 6 / 3 m = 2 Therefore, the gradient of the line is

2. Question 2: A line has a gradient of -1 and passes through the point (3, 2).