1 / 9

Coordinate Geometry II

Y-axis. 4. 2. X-axis. 0. -2. 2. 4. 6. -2. Coordinate Geometry II. y = mx + b. ax + by + c = 0. Line Equations I By Mr Porter. Assumed Knowledge. It is assumed that you have completed the basic of the power-point Coordinate Geometry I.

lei
Download Presentation

Coordinate Geometry II

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Y-axis 4 2 X-axis 0 -2 2 4 6 -2 Coordinate Geometry II y = mx + b ax + by + c = 0 Line Equations I By Mr Porter

  2. Assumed Knowledge. It is assumed that you have completed the basic of the power-point Coordinate Geometry I. You are familiar with calculating the gradient of the line joining two points. Test Yourself: Find the gradient of the line joining the points P(-2,4) and Q(3,-2).

  3. Two point formula of a line. Equation of a Line through two Points: A(x1,y1) and B(x2, y2). Select a general point on the line joining AB with coordinates P(x,y). Now, the three points are collinear (lie in the same straight line). Hence, the gradient of AB = gradient of AP = gradient of BP. Example 1: Find the equation of the line joining A(1,2) and (4,5). State the formula clearly: , rearrange to make y the subject. Substitute the values: The equation of the required line is y = x + 1. (Standard form) simplify the R.H.S. The equation can be written in standard form or general form: ax + by + c = 0

  4. Example 2: Find the equation of a line through P(-5,-2) and Q(6,4). Example 3: Find the equation of a line through H(7,-2) and K(-2,5). State the formula clearly: State the formula clearly: Substitute the values : Substitute the values : simplify the R.H.S. simplify the R.H.S. , rearrange to general form. , rearrange to general form. 11(y – 4) = 6(x– 6) -9(y + 2) = 7(x– 7) 11y – 44 = 6x– 36 -9y – 18 = 7x– 49 0 = 6x – 11y+8 0 = 7x + 9y– 31 The required line is: 6x – 11y + 8 = 0 . The required line is: 7x + 9y – 31 = 0 .

  5. Remember the formula: 2 Point Equation of a line. Exercise 1: Find the equation of the line joining the following two points: a) A(1,3) and B(5,11) b) C(-1,-3) and D(3,5) Ans: y = 2x + 1 Ans: y = 2x – 1 c) E(0,6) and F(3,-2) d) G(-5,6) and B(4,-1) Ans: 8x + 3y - 18 = 0 Ans: 7x + 9y - 19 = 0

  6. Point - Gradient formula. - This is a preferred method, used with DIFFERENTIATION. Definition: The equation of a line with gradient , m, passing through a point P(x1, y1) is given by: y – y1 = m (x – x1) Example 1: Find the equation of a line passing through A(1,3) and B(5,-3). Example 2: Find the equation of a line passing through P(-5,-3) and Q(3,4). Gradient: Gradient: Equation: y – y1 = m (x – x1) Equation: y – y1 = m (x – x1) The required line is: 3x + 2y – 9 = 0 . The required line is: 7x – 8y – 4 = 0 .

  7. Remember the formula: Point-Gradient Equation of a line. Exercise 2: Find the equation of the line joining the following two points: a) A(5,3) and B(-4,11) b) C(-4,-2) and D(8,5) Ans: 14x – 9y – 97 = 0 Ans: 7x – 12y +4 = 0 c) E(0,-2) and F(2,8) d) G(-4,-6) and B(4,3) Ans: y = 5x – 2 Ans: 9x – 8y - 12 = 0

  8. Example 2: Find the equation of a line with gradient m = passing through the point (-6,4). Do you have a gradient m? Yes , m = Substitute: y – 4 = (x – -6) More Point - Gradient Lines. Example 1: Find the equation of a line with gradient m = 3 passing through the point (1,7). Do you have a point? Yes, (1,6) Do you have a point? Yes, (-6,4) Do you have a gradient m? Yes , m = 3 Equation: y – y1 = m (x – x1) Equation: y – y1 = m (x – x1) Substitute: y – 7 = 3 (x – 1) Expand: y – 7 = 3x – 3 Expand: 2y – 8 = -3x – 18 Rearrange: y = 3x + 4 Rearrange: 3x + 2y + 10 = 0 The required line is: y = 3x + 4 . The required line is: 3x + 2y + 10 = 0 .

  9. c) C(0,-2) with gradient m = d) D(-4,-6) with gradient m = Exercise 3: Find the equation of the lines: a) A(5,3) with gradient m = -3 b) B(-4,-2) with gradient m = 4 Ans: y = -3x + 18 Ans: y = 4x + 14 Ans: 3x – 4y – 8 = 0 Ans: 5x + 3y + 38 = 0

More Related