Coordinates Picture For each instruction, join up the coordinates. Do not join one instruction to the next. (1, 3)  (0, 3)  (0, 5)  (2, 7)  (6, 7)  (7, 6)  (10, 5)  (10, 3)  (8, 3) (6, 2)  (6, 4)  (8, 4)  (8, 2)  (6, 2) (1, 2)  (1, 4)  (3, 4)  (3, 2)  (1, 2) (3, 3)  (6, 3) (2, 5)  (3, 6)  (6, 6)  (7, 5)  (2, 5)

Suppose we want to plot points for: Given an equation, we can find coordinate points for that equation by constructing a table of values. Suppose we want to plot points for: y = x + 3 We can use a table as follows: Explain that when we construct a table of values, the value of y depends on the value of x. That means that we choose the values for x and substitute them into the equation to get the corresponding value for y. The minimum number of points needed to draw a straight line is two, however, it is best to plot several points to ensure that no mistakes have been made. The points given by the table can then be plotted to give the graph of the required function. x y = x +3 –3 –2 –1 1 2 3 1 2 3 4 5 6 (–3, 0) (–2, 1) (–1, 2) (0, 3) (1, 4) (2, 5) (3, 6)

1) Complete a table of values: x y = x +3 –3 –2 –1 1 2 3 (–3, 0) 4 5 6 (–2, 1) (–1, 2) (0, 3) (1, 4) (2, 5) (3, 6) 2) Plot the points on a coordinate grid. 1 2 3 4 5 6 7 8 9 10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -10 x y 3) Draw a straight line through the points. 4) Label the line. y = x + 3

1) Complete a table of values: x y = 3x + 1 –3 –2 –1 1 2 3 1) Complete a table of values: -8 -5 -2 1 4 7 10 (-3, -8) (-2, -5) (-1, -2) (0, 1) (1, 4) (2, 7) (3, 10) 2) Plot the points on a coordinate grid. 1 2 3 4 5 6 7 8 9 10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -10 x y 3) Draw a straight line through the points. y = 3x + 1 4) Label the line.

1) Complete a table of values: x –3 –2 –1 1 2 3 1) Complete a table of values: 2) Plot the points on a coordinate grid. 1 2 3 4 5 6 7 8 9 10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -10 x y 3) Draw a straight line through the points. 4) Label the line.

Finished? Sketch these in your book on your own axes! 5) y = 2x – 4 6) y = 3x + 2 7) y = ½x

Hint: Draw a table of values for each question. Pair Activity Match the equations on the dominoes to the coordinates that are on the line! Hint: Draw a table of values for each question. Extension: Can you sketch any of them?

What can you tell me about the following pairs of linear equations? y = 2x + 4 y = ¼x + 4 y = ½x – 3 y = -2x + 6 y = 3x + 2 y = x - 1 y = 4x – 2 y = 4x + 1

We are learning to calculate the equation of a linear graph from two coordinates.

What can you tell me about y = mx + c Gradient (the slope of the line) Y – intercept (where the graph cuts the y-axis)

But you also need to think about whether its +ve or –ve. Calculate the gradient of the line which passes through (2, 6) and (4, -2). The gradient is rise run But you also need to think about whether its +ve or –ve. Start by calculating the gradient. (2, 6) 8 (4, -2) 2 rise = run 8 = 2 4 The graph slopes downwards so the gradient is negative. m = -4

But you also need to think about whether its +ve or –ve. Calculate the gradient of the line which passes through (7, 12) and (15, 32). The gradient is rise run But you also need to think about whether its +ve or –ve. Start by calculating the gradient. (15, 32) 20 (7, 12) 8 rise = run 20 = 8 5 2 The graph slopes upwards so the gradient is positive. m = 5 2

Remember to sketch each pair of coordinates and this about the sign! Calculate the gradients of the lines which pass through the following pairs of points (3, 6) and (5, 12) (2, 1) and (5, 10) (3, 5) and (5, 1) (2, 7) and (6, -1) (4, 3) and (8, 5) (2, 6) and (10, 10) (-1, 3) and (3, -1) (4, 2) and (3, -4) (3, 1) and (6, -8) (5, 8) and (1, 5) (4, 9) and (1, 15) (2, 7) and (3, 4) (2, 1) and (-3, 5) (-2, 3) and (-5, -2) (6, -1) and (-3, 2) Remember to sketch each pair of coordinates and this about the sign! Extension: Does the line y = 3x + 2 go through (-2, -3)? How do you know?

Extension: No  substitute x = -2 and y = -4, not -3 Calculate the gradients of the lines which pass through the following pairs of points 3 -2 ½ ½ -1 6 -3 ¾ -2 -3 -4/5 5/3 Remember to sketch each pair of coordinates and this about the sign! Extension: No  substitute x = -2 and y = -4, not -3

Find the equation of the line which passes through (2, 6) and (4, -2). We found earlier that the gradient was –4. Substitute the gradient and one of the coordinates (x and y) into y = mx + c. We now need the y-intercept. y = -4x + c 6 = -4(2) + c 6 = -8 + c 14 = c y = -4x + 14

We found earlier that the gradient was 5. 2 Find the equation of the line which passes through (7, 12) and (15, 32). We found earlier that the gradient was 5. 2 We now need the y-intercept. Substitute the gradient and one of the coordinates (x and y) into y = mx + c. y = 5x + c 2 12 = 5(7) + c 2 12 = 35 + c 2 -9 = c 2 y = 5x – 9 2 2

Finished? Complete the extension activity!

RAG Answers 1) y = 2x + 4 2) y = 5x – 2 3) y = 4x – 14 4) y = 5x – 1 5) y = -2x + 6 6) y = -½x + 2

Spot the mistake Find the equation of the line which passes through (9, 2) and (3, -7). rise = run 6 = 9 2 3 y = 2x + c 3 2 = 2(9) + c 3 2 = 6 + c -4 = c y = 2x – 4 3