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Coordinate Geometry II Line Equations I By Mr Porter y = mx + b ax + by + c = 0 0 246-2 X-axis Y-axis -2 2 4.

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Presentation on theme: "Coordinate Geometry II Line Equations I By Mr Porter y = mx + b ax + by + c = 0 0 246-2 X-axis Y-axis -2 2 4."— Presentation transcript:

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

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 Equation of a Line through two Points: A(x 1,y 1 ) and B(x 2, y 2 ). 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: Substitute the values: simplify the R.H.S., rearrange to make y the subject. The equation of the required line is y = x + 1. (Standard form) The equation can be written in standard form or general form: ax + by + c = 0 Two point formula of a line.

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: Substitute the values : simplify the R.H.S., rearrange to general form. 11(y – 4) = 6(x – 6) 11y – 44 = 6x – 36 0 = 6x – 11y +8 The required line is: 6x – 11y + 8 = 0. State the formula clearly: Substitute the values : simplify the R.H.S., rearrange to general form. -9(y + 2) = 7(x – 7) -9y – 18 = 7x – 49 0 = 7x + 9y – 31 The required line is: 7x + 9y – 31 = 0.

5 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) c) E(0,6) and F(3,-2)d) G(-5,6) and B(4,-1) Ans: y = 2x + 1Ans: y = 2x – 1 Ans: 8x + 3y - 18 = 0Ans: 7x + 9y - 19 = 0 Remember the formula: 2 Point Equation of a line.

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(x 1, y 1 ) is given by: y – y 1 = m (x – x 1 ) Example 1: Find the equation of a line passing through A(1,3) and B(5,-3). Gradient: Equation: y – y 1 = m (x – x 1 ) The required line is: 3x + 2y – 9 = 0. Example 2: Find the equation of a line passing through P(-5,-3) and Q(3,4). Gradient: Equation: y – y 1 = m (x – x 1 ) The required line is: 7x – 8y – 4 = 0.

7 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) c) E(0,-2) and F(2,8)d) G(-4,-6) and B(4,3) Ans: 14x – 9y – 97 = 0Ans: 7x – 12y +4 = 0 Ans: y = 5x – 2Ans: 9x – 8y - 12 = 0 Remember the formula: Point-Gradient Equation of a line.

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

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


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