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**Created by Mr.Lafferty Maths Dept**

The Gradient Difference in y -coordinates S4 Credit The gradient is the measure of steepness of a line Change in vertical height Change in horizontal distance Difference in x -coordinates The steeper a line the bigger the gradient 28-Mar-17 Created by Mr.Lafferty Maths Dept

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**Created by Mr.Lafferty Maths Dept**

The Gradient S4 Credit 3 4 3 2 3 5 2 6 28-Mar-17 Created by Mr.Lafferty Maths Dept

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**The gradient using coordinates**

Mr. Lafferty The gradient using coordinates S4 Credit m = gradient y-axis y2 We start by find the equation of a circle centre the origin. First draw set axises x,y and then label the origin O. Next we plot a point P say, which as coordinates x,y. Next draw a line from the origin O to the point P and label length of this line r. If we now rotate the point P through 360 degrees keep the Origin fixed we trace out a circle with radius r and centre O. Remembering Pythagoras’s Theorem from Standard grade a square plus b squared equal c squares we can now write down the equal of any circle with centre the origin. y1 O x-axis x1 x2 28-Mar-17

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**The gradient using coordinates**

Mr. Lafferty The gradient using coordinates S4 Credit Find the gradient of the line. m = gradient y-axis We start by find the equation of a circle centre the origin. First draw set axises x,y and then label the origin O. Next we plot a point P say, which as coordinates x,y. Next draw a line from the origin O to the point P and label length of this line r. If we now rotate the point P through 360 degrees keep the Origin fixed we trace out a circle with radius r and centre O. Remembering Pythagoras’s Theorem from Standard grade a square plus b squared equal c squares we can now write down the equal of any circle with centre the origin. O x-axis 28-Mar-17

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**The gradient using coordinates**

Mr. Lafferty The gradient using coordinates S4 Credit Find the gradient of the two lines. y-axis We start by find the equation of a circle centre the origin. First draw set axises x,y and then label the origin O. Next we plot a point P say, which as coordinates x,y. Next draw a line from the origin O to the point P and label length of this line r. If we now rotate the point P through 360 degrees keep the Origin fixed we trace out a circle with radius r and centre O. Remembering Pythagoras’s Theorem from Standard grade a square plus b squared equal c squares we can now write down the equal of any circle with centre the origin. O x-axis 28-Mar-17

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**The gradient using coordinates**

S4 Credit The gradient formula is : It is a measure of how steep a line is A line sloping up from left to right is a positive gradient A line sloping down from left to right is a negative gradient 28-Mar-17 Created by Mr.Lafferty Maths Dept

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**Created by Mr.Lafferty Maths Dept**

Straight Line Graphs S4 Credit Key Points Make a table Calculate and plot 3 coordinates 3. Draw a line through points 28-Mar-17 Created by Mr.Lafferty Maths Dept

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**Drawing Straight Line Graphs y = ax + b y = x**

1 2 3 4 5 6 7 8 9 10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -10 x y x y 3 -4 y = 3x+1 3 -4 x y -2 2 -5 1 7 y = x - 3 y = 2x x y 4 8 x y 3 -4 -3 1 5 6 -8 28-Mar-17 Created by Mr. Lafferty Maths Dept

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**Created by Mr. Lafferty Maths Dept**

Straight line equation and the gradient connection 1 2 3 4 5 6 7 8 9 10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -10 x y y = 2x - 5 y = 2x + 1 x y 1 3 x y 1 3 -5 -3 1 1 3 7 m = 2 m = 2 28-Mar-17 Created by Mr. Lafferty Maths Dept

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**Straight Line Equation**

S4 Credit y 10 lines are parallel if they have the same gradient All straight lines have the equation of the form 9 8 y = mx + c 7 6 5 4 3 Where line meets y-axis 2 Gradient 1 x 1 2 3 4 5 6 7 8 9 10 Find the equations of the following lines y = x y = x+4 y = 4x+2 y = -0.5x+2 28-Mar-17 Created by Mr.Lafferty Maths Dept

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**x = -6 Mark the points on your grid below and then join them together**

Equations for Vertical and Horizontal lines 1 2 3 4 5 6 7 8 9 10 -9 -8 -7 -6 -5 -4 -3 -2 -1 -10 x y x = -6 Mark the points on your grid below and then join them together Vertical lines have equations of the form x = 8 Horizontal lines have equations of the form y = 2 y = -9 28-Mar-17 Created by Mr. Lafferty Maths Dept

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**Straight Line Equation**

lines are parallel if same gradient Straight Line Equation S4 Credit All straight lines have the equation of the form Slope left to right upwards positive gradient y = mx + c y - intercept Gradient y intercept is were line cuts y axis Slope left to right downwards negative gradient 28-Mar-17 Created by Mr.Lafferty Maths Dept

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**Equation of a Straight Line**

y = mx+c S4 Credit To find the equation of a straight line we need to know Two points that lie on the line ( x1, y1) and ( x2, y2) Or The gradient and a point on the line m and (a,b) 28-Mar-17

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**Equation of a Straight Line**

Mr. Lafferty Equation of a Straight Line y = mx+c S4 Credit Find the equation of the straight line passing through the points (4, 4) and (8,24). Solution Using the point (4,4) and the gradient m = 5 sub into straight line equation We are now in a position to find the equation of any circle with centre A,B. All we have to do is repeat the process in shown in slide 2, but this time the centre is chosen to be (a,b). First plot a point C and label it’s coordinates (a,b), next we plot another point P and label it’s coordinates (x,y). Next draw a line from C to P and call this length (r). (r) will be the radius of our circle with centre (a,b). Again we rotate the point P through 360 degrees keeping the point C fixed. Using Pythagoras Theorem a squared plus b squared equal c squared we can write down the equation of any circle with centre (a,b) and radius (r). The equation is (x - a) all squared plus (y-b) all squared equals (r) squared. Finally to write down the equation of a circle we need to know the co-ordinates of the centre and the length of the radius or co-ordinates of the centre and the co-ordinates of a point on the circumference of the circle. y = mx + c 4 = 5 x 4 + c c = = -16 Equation : y = 5x - 16

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**Equation of a Straight Line**

Mr. Lafferty Equation of a Straight Line y = mx+c S4 Credit Find the equation of the straight line passing through the points (3, -5) and (6,4). Solution Using the point (6,4) and the gradient m = 3 sub into straight line equation We are now in a position to find the equation of any circle with centre A,B. All we have to do is repeat the process in shown in slide 2, but this time the centre is chosen to be (a,b). First plot a point C and label it’s coordinates (a,b), next we plot another point P and label it’s coordinates (x,y). Next draw a line from C to P and call this length (r). (r) will be the radius of our circle with centre (a,b). Again we rotate the point P through 360 degrees keeping the point C fixed. Using Pythagoras Theorem a squared plus b squared equal c squared we can write down the equation of any circle with centre (a,b) and radius (r). The equation is (x - a) all squared plus (y-b) all squared equals (r) squared. Finally to write down the equation of a circle we need to know the co-ordinates of the centre and the length of the radius or co-ordinates of the centre and the co-ordinates of a point on the circumference of the circle. y = mx + c 4 = 3 x 6 + c c = = -14 Equation : y = 3x - 14

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**Created by Mr.Lafferty Maths Dept**

Starter Questions S4 Credit Q1. The points ( 1, 4) and (3, 11) lie on a line. Find the gradient of the line. Q2. Complete the table given : y = 3x + 1 x -3 3 y Q3. Are the two lines parallel. Explain your answer y = x and y = 2x + 2 28-Mar-17 Created by Mr.Lafferty Maths Dept

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**The cost for hiring a plumber per hour is shown below**

Pick any 2 points Modelling Real – life The cost for hiring a plumber per hour is shown below 10 1 2 3 4 5 6 7 8 9 10 T C 20 30 40 50 60 70 80 90 100 Cost £ Time (hours) (7,100) (a) Calculate the gradient. What is the value of C when T = 0 30 30 Write down an equation connecting C and T. (0,30) C = T + (d) Find the cost for a plumber for 10 hours? T = 10 C = 10x = £130

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