Presentation on theme: "Straight Line Equation. x y x y y = -2x + 3. The Video Shop. You join a video shop for a membership fee of £3 and then charges £2 for each video you hire:"— Presentation transcript:
Straight Line Equation. x y x y y = -2x + 3
The Video Shop. You join a video shop for a membership fee of £3 and then charges £2 for each video you hire: Complete the table below for the cost of hiring different numbers of videos. No of videos Cost of Videos(£) Now draw a graph of the table above.
No of Videos Cost Graph of videos hired against cost.
No of Videos Cost Now consider the structure of the graph. The graph cuts the y axis at (0,3) because it cost £3 to join the video shop before you hired any videos. For every square that you move to the right you go two squares up because the cost of each video is £2.
Finding A Formula. Look at the table of values for the video hire once again: V C Find a formula for the cost of videos (C) given the number of videos (V) : C =2V + 3 From the graph we saw that : This was the number of squares you went up for each one you went along.This is called the gradient of the line. This was the place were the graph cut the y axis. This is called the y axis intercept.
Now repeat the question again for a video shop charging £5 to join and £3 for each video hired.Start by completing the table below. No of videos Cost of Videos(£) 5 Answer the questions below: Where does the graph cut the y axis ? ( 0, 5 ) What is the gradient of the line: Gradient = 3 The full graph is shown on the next slide:
Gradient of 3. Y axis intercept (0,5) No of Videos Cost C = 3V + 5 Is the equation of the line.
More About The Gradient. The gradient (m) of a straight line is defined to be: Change in vertical height. Change in horizontal distance. We are going to use this definition to calculate the gradient of various straight lines:
3 Find the gradients of the straight lines below: (1) 4 3 m = 4 4 (2) 7 4 m = 7 4 (3) 4 4 m = 4 m = 1
(5) 8 6 m = 6 8 = 3 4 (6) 9 3 m = 3 9 = 3
Negative Gradient Consider the straight lines shown below: (a) (c) (b) (d) (e) Can you split the lines into two groups based on their gradients ? Lines (a) (c) and (d) slope upwards from left to right. Lines (b) and (e) slope downwards from left to right. Positive gradient Negative gradient
Calculate the gradients of the lines below: (1) (2) - 8 6
The Equation Of A Straight Line. To find the equation of any straight line we require to know two things: (a) The gradient of the line. (b) The y axis intercept of the line. m = gradient. c = y axis intercept. The equation of a straight line is :y = m x + c Examples. Give the gradient and the y axis intercept for each of the following lines. (1) y = 6x + 5 m = 6c = 5 (2) y = 4x + 2 m = 4c = 2 (3) y = x - 3 m = 1c = - 3
Finding The Equation. Find the equation of the straight lines below: x y (1) What is the gradient ?m = 1 What is the y axis intercept?c = 2 Now use y = m x + cy = x + 2 x y (2) c = 1
(3) x y c = -2 (4) x y m = -2 c = 3 y = -2x + 3
x y (5) c = 6 (6) x y c = 2
The Gradient Formula. Look at the diagram below: x1x1 x2x2 y1y1 y2y2 It shows a straight line passing through the points (x 1,y 1 ) and (x 2,y 2 ). We must calculate the gradient of the line using the triangle shown: Change in vertical height: y 2 -y 1 y 2 – y 1 Change in horizontal distance: x 2 -x 1 Gradient of line:
Calculate the gradient of the line through the points below: (1) A(4,6) and B( 10,12) Solution: Write down the gradient formula: Gradient of line: Substitute in your values: Calculate and simplify: (2) C(-4,8) and D(6,-10) Solution:
Straight Line From Two Points. Find the equation of the straight line passing through (4,6) and (8,12) Solution: Find the gradient of the line: Substitute gradient into y = m x + c. y = 3 x + c Now substitute one of the points into y = m x + c to find c. Sub’ (4,6) into y = 3x +c : 6 = 3 x 4 +c c + 12 = 6 c = - 6 Now write down the equation of the straight line: y = 3x - 6
Find the equation of a straight line passing through C(6,-7) and D(-12,9) Solution. Calculate the gradient: Substitute gradient into y = m x + c. Now substitute one of the points into y = m x + c to find c. Sub’ (6,-7) into equation: c = -23 Equation of the straight line: