Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• Find the equation of a straight line.
• Calculate the angle of inclination.
• Determine if lines are parallel or perpendicular.
• Find equations of parallel or perpendicular lines.

Lesson summary

Analytical geometry bridges the gap between algebra and geometry, allowing us to describe geometric shapes and solve geometric problems using algebraic equations. This is incredibly useful in various fields, from architecture and engineering to computer graphics and GPS navigation. In the South African context, analytical geometry can be applied to land surveying, urban planning, and even designing sports fields for our favourite games. Imagine trying to build a new RDP house without understanding the angles and distances involved – analytical geometry provides the tools to ensure accuracy and efficiency.

Lesson notes

2. 1. The Equation of a Straight Line The general equation of a straight line is given by: y = mx + c Where: y is the y-coordinate of any point on the line. x is the x-coordinate of any point on the line. m is the gradient (slope) of the line. This represents the steepness of the line and is defined as the change in y divided by the change in x (rise over run). c is the y-intercept of the line. This is the point where the line crosses the y-axis (i.e., the value of y when x = 0). 2.1.1 Finding the Equation Given Two Points If you are given two points, (x 1 , y 1 ) and (x 2 , y 2 ), you can find the equation of the line as follows: Calculate the gradient (m): m = (y 2 - y 1 ) / (x 2 - x 1 ) Substitute the gradient (m) and one of the points (x 1 , y 1 ) into the equation y = mx + c and solve for c: y 1 = m*x 1 + c c = y 1 - m*x 1 Write the equation of the line in the form y = mx + c, using the calculated values for m and c.

Example 1: Find the equation of the line passing through the points (1, 2) and (3, 8).

Calculate the gradient: m = (8 - 2) / (3 - 1) = 6 / 2 = 3 Substitute m = 3 and the point (1, 2) into y = mx + c: 2 = 3 * 1 + c 2 = 3 + c c = -1 The equation of the line is y = 3x - 1 2.1.2 Finding the Equation Given a Point and a Gradient If you are given a point (x 1 , y 1 ) and the gradient m, you can directly substitute these values into the equation y = mx + c and solve for c, as shown in step 2 of the previous method.

Example 2: Find the equation of the line with a gradient of -2 that passes through the point (4, 1). Substitute m = -2 and the point (4, 1) into y = mx + c: 1 = -2 * 4 + c 1 = -8 + c c = 9 The equation of the line is y = -2x + 9 2.1.3 Finding the Equation Given the Gradient and Y-intercept If you are given the gradient, m, and the y-intercept, c, you can directly substitute these values into the equation y = mx + c.

Example 3: Find the equation of the line with a gradient of 1/2 and a y-intercept of

3. The equation of the line is y = (1/2)x + 3 2.

2. Angle of Inclination The angle of inclination (θ) of a straight line is the angle that the line makes with the positive x-axis, measured in an anti-clockwise direction. It is related to the gradient (m) by the following formula: m = tan θ Therefore, to find the angle of inclination, you need to calculate the gradient (m) and then find the arctangent (inverse tangent) of m: θ = arctan(m) = tan -1 (m) Remember to set your calculator to degree mode.

Example 4: Find the angle of inclination of the line y = x +

2. Identify the gradient: m = 1 Calculate the arctangent: θ = tan -1 (1) = 45° Therefore, the angle of inclination is 45°. 2.

3. Parallel and Perpendicular Lines Parallel Lines: Two lines are parallel if they have the same gradient. If line 1 has a gradient of m 1 and line 2 has a gradient of m 2 , then the lines are parallel if m 1 = m 2 .

Perpendicular Lines: Two lines are perpendicular if the product of their gradients is -

1. If line 1 has a gradient of m 1 and line 2 has a gradient of m 2 , then the lines are perpendicular if m 1 m 2 = -

1. This can also be expressed as m 2 = -1/m 1 . This means the gradient of the perpendicular line is the negative reciprocal of the original line's gradient.

Example 5: Determine if the lines y = 2x + 3 and y = 2x - 1 are parallel. Both lines have a gradient of

2. Therefore, the lines are parallel.

Example 6: Determine if the lines y = 3x + 1 and y = (-1/3)x + 4 are perpendicular. The product of their gradients is 3 * (-1/3) = -

1. Therefore, the lines are perpendicular. 2.

4. Equation of a Line Parallel or Perpendicular to a Given Line Parallel: If you need to find the equation of a line parallel to y = mx + c and passing through a point (x 1 , y 1 ), the new line will have the same gradient (m). Substitute the gradient and the point into y = mx + c and solve for the new y-intercept.

Perpendicular: If you need to find the equation of a line perpendicular to y = mx + c and passing through a point (x 1 , y 1 ), the new gradient will be -1/m. Substitute the new gradient and the point into y = mx + c and solve for the new y-intercept.

Example 7: Find the equation of the line parallel to y = 4x - 5 and passing through the point (2, 3). The gradient of the parallel line is m =

4. Substitute m = 4 and the point (2, 3) into y = mx + c: 3 = 4 * 2 + c 3 = 8 + c c = -5 The equation of the line is y = 4x -

5. Example 8: Find the equation of the line perpendicular to y = (1/2)x + 1 and passing through the point (1, -2). The gradient of the perpendicular line is m = -1 / (1/2) = -

2. Substitute m = -2 and the point (1, -2) into y = mx + c: -2 = -2 * 1 + c -2 = -2 + c c = 0 The equation of the line is y = -2x. 2.

5. Distance Between Two Points The distance (d) between two points (x 1 , y 1 ) and (x 2 , y 2 ) is given by the distance formula: d = √[(x 2 - x 1 ) 2 + (y 2 - y 1 ) 2 ] Example 9: Find the distance between the points (1, 4) and (5, 7).