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2. Refresher Straight Lines: Gradients and Points Remember: From previous work we know that the equation of a straight line is of the form y = mx + c, where m is the gradient of the line and c is the y-intercept. Consider the lines y = 2x + 1 and y = x - 4 2 1 Gradient = 2/1 = 2 y-intercept = + 1 y = 2x + 1 y = mx + c 1 1 Gradient = 1/1 = 1 y-intercept = - 4 y = x - 4

3. Straight Lines: Gradients and Points y = mx + c The coordinates of any point lying on the line satisfy the equation of the line. Consider the line y = 2x - 3. 0 1234567 8 910 -9-8 -7 -6 -5 -4-3-2 -10 x y 1 2 3 4 5 6 7 8 9 10 -2 -3 -4 -5 -6 -7 -8 -9 -10 y = 2x - 3 (4,5) (1,-1) Substituting x = 4 into y = 2x – 3  y = 2 x 4 - 3 = 5 Thus showing that the point (4,5) lies on the line. Similarly substituting y = -1 into y = 2x – 3 Thus showing that the point (1,-1) lies on the line.  -1 = 2x -3  -2x = 2  x = 1

4. Y-intercept Straight Lines: Gradients and Points y = mx + c Finding the y-intercept, given the gradient and the co-ordinates of a point P, on the line. 0 1234567 8 910 -9-8 -7 -6 -5 -4-3-2 -10 x y 1 2 3 4 5 6 7 8 9 10 -2 -3 -4 -5 -6 -7 -8 -9 -10 y = 2x + c (4,5) We can use the previous example to demonstrate how this works but ignoring the fact that we know the value of c. Substituting known values into: y = mx +c gives: 5 = 2 x 4 + c 5 = 8 + c -3 = c  y = 2x - 3

5. Straight Lines: Gradients and Points y = mx + c Finding the y-intercept, given the gradient and the co-ordinates of a point P, on the line. 0 1234567 8 910 -9-8 -7 -6 -5 -4-3-2 -10 x y 1 2 3 4 5 6 7 8 9 10 -2 -3 -4 -5 -6 -7 -8 -9 -10 y = 2x + c (4,5) Question (1): Find the equation of the line in each of the cases listed below. P is a point on the line. y = 3x + 6 y = 2x - 5 (a) y = 3x + c (P = 4,18) (b) y = 2x + c (P = 1,-3) (c) y = -4x + c (P = 3,-11) y = -4x + 1

6. Gradient Straight Lines: Gradients and Points y = mx + c Finding the equation of a straight line, given the co-ordinates of two points on the line. 0 1234567 8 910 -9-8 -7 -6 -5 -4-3-2 -10 x y 1 2 3 4 5 6 7 8 9 10 -2 -3 -4 -5 -6 -7 -8 -9 -10 Again we can use the previous example to demonstrate how this works. Remember from slide 1 that the gradient of a line is simply the difference in y values/difference in x values. y = mx + c (4,5) (1,-1) Step 1. Find the gradient 5 – (-1) = 6 4 – 1 = 3 So the gradient is 6/3 = 2 and so the equation of the line is y = 2x + c Step 2. Use a point on the line and proceed as previously. Substituting the point (4,5) into y = 2x + c 5 = 2 x 4 + c 5 = 8 + c -3 = c  y = 2x - 3

7. Straight Lines: Gradients and Points y = mx + c Finding the equation of a straight line, given the co-ordinates of two points on the line. Step 1. Find the gradient using both points. (y 2 – y 1 ) Step 2. Substitute values from one of the points. y = mx + c (x 1,y 1 ) (x 2,y 2 ) (x 2 – x 1 ) More formally and in general, the gradient of a line, is determined as follows: (y 2 – y 1 ) (x 2 – x 1 ) m =

8. Straight Lines: Gradients and Points y = mx + c Step 1. Find the gradient using both points. (y 2 – y 1 ) Step 2. Substitute values from one of the points. y = mx + c (x 1,y 1 ) (x 2,y 2 ) (x 2 – x 1 ) (y 2 – y 1 ) (x 2 – x 1 ) m = Example Question: Find the equation of the line that passes through the points (-2,-4) and (1,5). m = (5 – (-4)/(1- (-2)) = 9/3 = 3  y = 3x + c (Substituting (1,5))  5 = 3 x 1 + c  5 = 3 + c  2 = c  y = 3x + 2

9. Straight Lines: Gradients and Points y = mx + c Step 1. Find the gradient using both points. (y 2 – y 1 ) Step 2. Substitute values from one of the points. y = mx + c (x 1,y 1 ) (x 2,y 2 ) (x 2 – x 1 ) (y 2 – y 1 ) (x 2 – x 1 ) m = Question (2): Find the equation of the lines that pass through the given points. y = -2x + 3 y = 4x - 3 (a) (-1,5) and (2,-1) (b) (-1,-7) and (2,5) (c) (-5,0) and (3,8) y = x + 5

10. Worksheet Questions (1): Find the equation of the line in each of the cases listed below. P is a point on the line. (a) y = 3x + c (P = 4, 18) (b) y = 2x + c (P = 1, -3) (c) y = -4x + c (P = 3, -11) Questions (2): Find the equation of the lines that pass through the given points. (a) (-1,5) and (2,-1) (b) (-1,-7) and (2,5) (c) (-5,0) and (3,8) Questions (1): Find the equation of the line in each of the cases listed below. P is a point on the line. (a) y = 3x + c (P = 4, 18) (b) y = 2x + c (P = 1, -3) (c) y = -4x + c (P = 3, -11) Questions (2): Find the equation of the lines that pass through the given points. (a) (-1,5) and (2,-1) (b) (-1,-7) and (2,5) (c) (-5,0) and (3,8) Questions (1): Find the equation of the line in each of the cases listed below. P is a point on the line. (a) y = 3x + c (P = 4, 18) (b) y = 2x + c (P = 1, -3) (c) y = -4x + c (P = 3, -11) Questions (2): Find the equation of the lines that pass through the given points. (a) (-1,5) and (2,-1) (b) (-1,-7) and (2,5) (c) (-5,0) and (3,8) Questions (1): Find the equation of the line in each of the cases listed below. P is a point on the line. (a) y = 3x + c (P = 4, 18) (b) y = 2x + c (P = 1, -3) (c) y = -4x + c (P = 3, -11) Questions (2): Find the equation of the lines that pass through the given points. (a) (-1,5) and (2,-1) (b) (-1,-7) and (2,5) (c) (-5,0) and (3,8) Questions (1): Find the equation of the line in each of the cases listed below. P is a point on the line. (a) y = 3x + c (P = 4, 18) (b) y = 2x + c (P = 1, -3) (c) y = -4x + c (P = 3, -11) Questions (2): Find the equation of the lines that pass through the given points. (a) (-1,5) and (2,-1) (b) (-1,-7) and (2,5) (c) (-5,0) and (3,8) Questions (1): Find the equation of the line in each of the cases listed below. P is a point on the line. (a) y = 3x + c (P = 4, 18) (b) y = 2x + c (P = 1, -3) (c) y = -4x + c (P = 3, -11) Questions (2): Find the equation of the lines that pass through the given points. (a) (-1,5) and (2,-1) (b) (-1,-7) and (2,5) (c) (-5,0) and (3,8) Worksheet