1. SKETCHING LINEAR GRAPHS (3A)

2. THE METHODS  There are many methods for sketching linear graphs:  plotting  x and y intercept method  gradient intercept method  using the form y=mx  using the form y=a or x=b  In tests, sometimes a question will specify which method to use (e.g. Using the x and y intercept method, sketch...) – but if this isn’t specified, then use whichever method is appropriate or easiest

3. PLOTTING  What it involves:  You will be given a whole set of x values, and you need to find their corresponding y values. Then you plot these pairs of coordinates (x,y) along a line, then connect these dots together to form a straight line  Advantages  gives you several points along a line which makes the line easy to draw (connecting the dots)  Disadvantages  more time consuming  When to use it  When the question gives you a set of x values that you need to plot (e.g. Using the x values -3, -1, 0, 2, 4, plot the graph of y = 2 x + 5)

4. STEPS FOR PLOTTING  Draw up a table with the x values along the top row  Substitute these values in to the linear equation to find their corresponding y values and put these in the row underneath  Plot these pairs of points (x and y coordinates) on the Cartesian plane  Connect the dots with a straight line Question: Plot the graph y = x - 3 with the x values of -1, 0, 1, 2, 3 Our coordinates are: (-1,-4) (0,-3) (1,-2) (2,-1) (3,0) So we find these points on the graph, draw them on, and connect the dots with a straight line. x0123 y-4-3-20

5. X AND Y INTERCEPT METHOD  What it involves:  Finding the points where the line crosses the x axis and y axis, then drawing the line to connect these two points. The line will cross the x axis when y=0, and the line will cross the y axis when x=0  Advantages  You only need 2 points to draw the line  It involves only two substitutions of putting x=0 then y=0 into the equation  Disadvantages  Need to rearrange the equation to find the x intercept  When to use it  When the question tells you to use this specific method to sketch the graph (e.g. Find the x and y intercepts, then sketch the graph of y = 2 x + 5)

6. STEPS FOR USING THE X AND Y INTERCEPT METHOD  Find the y-intercept first. This means substituting x=0 into the equation to find the point where the line crosses the y axis. This gives you the first point (0, y-intercept)  Then find the x-intercept. Go back to the original equation and make y=0, then solve for x. This involves undoing/reversing the steps to get the x value. This gives you the second point on the graph (x-intercept, 0)  Draw the line that connects these two points Question: Plot the graph y = x - 3 by finding the intercepts. y-int: x=0 y = x – 3 y = 0 – 3 y = -3 First point: (0,-3) x-int: y=0 y = x – 3 0 = x – 3 3 = x Second point: (3,0) Plot these two points along the x and y axes, then draw the line that connects them.

7. GRADIENT INTERCEPT METHOD  What it involves:  Finding the y-intercept (the point along the y axis when x=0) and plotting this point. Then, using the gradient (m in the standard form of equation y=mx+c), find the second point along the graph and connect these two points with a straight line.  Advantages  You only need 2 points to draw the line  Disadvantages  Finding the rise and run can be tricky  Need to rearrange the equation into standard form y=mx+c  When to use it  When the question tells you to use this specific method to sketch the graph (e.g. Use the gradient intercept method to sketch the graph of y = 2 x + 5). Otherwise, use a more efficient method.

8. STEPS FOR USING THE GRADIENT INTERCEPT METHOD  Check if the equation is in the standard form (y = mx+c). If it isn’t rearrange it until it looks like this.  Find the y-intercept. This means substituting x=0 into the equation to find the point where the line crosses the y axis. This gives you the first point (0, y- intercept)  Then use the gradient (m, or the number in front of the x) to do the rise and run. The numerator of the gradient is the rise (how far up to go on the y axis), and the denominator is the run (how far across to go on the x axis). Plot this new point.  Draw the line that connects these two points Question: Plot the graph 2y - 6x + 4 = 0 by using the gradient intercept method. Rearrange to look like y = mx+c 2y = 6x – 4 (I took the -6x + 4 to the other side of the = sign so they became their opposites) (y has been multiplied by 2 to get to 2y, so we need to divide the other side of the equation to get y on its own) y = 3x – 2 (now it is in standard form) Find the y-intercept (put x=0 into the equation, or simply use the ‘c’ part of the equation y=mx+c. In this case, it is -2 so our y-intercept is (0,-2). We plot this point. Our gradient (‘m’) is the number in front of the x, which in this case is 3. Remembering that 3 as a fraction is 3/1, our RISE is 3 and our RUN is 1. So, from the y-intercept, move UP 3 points, ACROSS 1 point, and plot this new point. This is also the same as adding 1 to our x value and adding 3 to our y value from the y-intercept of (0,-2), so we get the point (0+1, - 2+3) = (1,1). This is our second point. Then we connect these points with a straight line.

9. USING THE FORM Y=MX  What it involves:  When there is no ‘+c’ on the equation, we know straight away that the first point on the graph is (0,0). Then we only need to find one more point to draw the line.  Advantages  You only need 2 points to draw the line  It involves only one substitution of putting x=1 into the equation to get the second point  Disadvantages  Not all equations will be in this form  When to use it  You can only use this if the equation doesn’t have a ‘c’ value on the end (like y=mx+c)

10. STEPS FOR USING THE FORM Y=MX  Plot the first point of (0,0) at the origin.  Then choose any x value (x=1 is easiest though!) and substitute this into the equation to find a second point to plot.  Draw the line that connects these two points Question: Plot the graph y = 6x Our first point is (0,0), so we plot this straight away. Easy! Then substitute x=1 into the equation to get our second point. y = 6x y = 6 times 1 y = 6 Second point: (1,6) Plot this second point on the graph, then draw the line that connects them.

11. USING THE FORM Y=A OR X=B  What it involves:  Sometimes the equation will only have an x OR a y in it, and not both. If this is the case, our graph will either be completely vertical (up and down) or horizontal (going across) and will not be on a diagonal.  Advantages  You only need 1 point on either the x or y axis to draw the line  Disadvantages  Not all equations will be in this form  When to use it  You can only use this if the equation is y=? or x=? (rather than having an x and a y in the equation)

12. STEPS FOR USING THE FORM Y=A OR X=B  If the equation is y=?, then find the number that the y equals along the y-axis. Draw a horizontal line at this point.  If the equation is x=?, then find the number that the x equals along the x-axis. Draw a vertical line at this point. Question: Plot the graphs (a) y = -3 and (b) x=2.5 (a) Plot the point along the y-axis where y is -3. Draw a horizontal line across this point. (b) Plot the point along the x-axis where x is 2.5. Draw a vertical line through this point.

13. DETERMINING LINEAR EQUATIONS (3B)

14. SO FAR...  We have been given equations in order to sketch the graphs  Now, we are given graphs and we need to find the equation of the straight line  In order to find the equation, we need two points that the line goes through. If we know two points, we can find the gradient (the slope of the line that joins these points)  Or we might be given the gradient and a point, which we can also use to find the equation y=mx+c)

15. TO WRITE THE EQUATION Y=MX+C, WE NEED TO FIND THE ‘M’ AND THE ‘C’ 1. The y-intercept (we can sometimes read this from the graph. This is the point where x=0 and the line crosses the y axis)  This will give us the ‘c’ part of the equation y=mx+c 2. The gradient (this is the ‘rise’ over ‘run’)  This will give us the ‘m’ part of the equation y=mx+c

16. STEPS TO FOLLOW 1. Determine two points on the graph  These might be given to you in the question (e.g. Find the equation of the line that passes through (1,2) and (3,5)), or you might need to read them off the graph  Preferably, one of these points will be the y-intercept so that we already know the ‘c’ part of the equation  If not, you will need to find the y-intercept at a later stage 2. Use these two points (x 1, y 1 ) and (x 2, y 2 ) and this formula to find the gradient: 3. Substitute the ‘m’ and ‘c’ values into the equation y=mx+c

17. SOMETIMES...  A question will say: Find the equation of the line with gradient 3 and y-intercept of 4.  This is easy! Remember, the gradient is the ‘m’ in the equation y=mx+c, and the ‘c’ is the y-intercept, so you just need to put these values in!  For this question, the answer would be y=3x+4  A question will give you the gradient and a point [e.g. Find the equation of the line with gradient 3 and point (5, -1)]  In this case, you will need to put the gradient straight in to the equation as the ‘m’ value  y=3x+c  To find ‘c’ substitute the point (5, -1) where y is -1 and x is 5, and solve for ‘c’  y=3x + c -1 = 3(5) + c -1 = 15 + c -1 – 15 = c -16 = c So: the equation is y=3x-16

18. EXAMPLE: FIND THE EQUATION OF THE LINE ON THIS GRAPH...  From this graph, we can determine two points: the x-intercept (3,0) and the y-intercept (0,6)  Since the y-intercept is 6, we already know our ‘c’ value in the equation which is 6  We now need to find the ‘m’ value or the gradient using: (x 1, y 1 ) and (x 2, y 2 ) where:  x 1 is 3, x 2 is 0  y 1 is 0, y 2 is 6  So, we use the formula: m = (6-0)/(0-3) m = 6/-3 m = -2  Now we know our m and c values, we can plug those into the equation y=mx+c to determine the equation of the line: y = -2x + 6

19. EXAMPLE: FIND THE EQUATION OF THE LINE ON THIS GRAPH...  From this graph, we can determine two points: the x and y intercept (0,0) and we can read the point (2,1)  Since the y-intercept is 0, we already know our ‘c’ value in the equation which is 0 (meaning we don’t even need to have a ‘c’ value)  We now need to find the ‘m’ value or the gradient using: (x 1, y 1 ) and (x 2, y 2 ) where:  x 1 is 0, x 2 is 2  y 1 is 0, y 2 is 1  So, we use the formula: m = (1-0)/(2-0) m = 1/2  Now we know our m and c values, we can plug those into the equation y=mx (remember no need for +c in this one as c=0) to determine the equation of the line: y = 1/2x + 0 or y = 1/2x

20. ANOTHER METHOD  If we are just given two points and we don’t know or are unable to determine the y-intercept, we need to firstly find the gradient using the two points and then find the y-intercept

21. EXAMPLE: FIND THE EQUATION OF THE LINE PASSING THROUGH THE POINTS (-2,5) AND (1,-1)  Find the gradient using where x 1 is -2 and x 2 is 1 y 1 is 5 and y 2 is -1 [we get this from the points given (-2, 5) and (1,-1)]  Using the formula for ‘m’, this gives us m = (-1 – 5)/(1 - -2) m = (-6)/(3) m = -2  So now we have y = -2x + c We need to find the ‘c’ or the y-intercept. To do this, choose one of the points on the graph (-2,5) or (1,-1) [it doesn’t matter which you choose] and substitute this into the equation and then solve to find ‘c’ y = -2x + c Let’s use (1,-1)  y is -1 and x is 1 so we substitute these in: -1 = -2(1) + c -1 = -2 + c -1 + 2 = c 1 = c  Now we can write our formula: y = -2x + 1

22. DISTANCE BETWEEN TWO POINTS (3C)

23. WHAT THIS MEANS  So far, we have:  Looked at various methods of sketching linear graphs  Looked at methods of finding the equation of a graph, depending on what information we are given (e.g. the gradient, the y-intercept, one or two points on the line)  Today we are looking at how to find how long the line is between two points (or, the distance between two points along the line)  We use Pythagoras’ theorem to find the distance between two points on a graph  Remember: c 2 = a 2 + b 2  ‘c’ is always the hypotenuse (this is the distance between the points)  ‘a’ and ‘b’ are the rise and run between the two points

24. AN OVERVIEW OF WHAT WE NEED TO DO (x 2, y 2 ) (x 1, y 1 ) A B C

25. YOU CAN USE THE FORMULA:

26. EXAMPLE

27. SOMETIMES…  Some more complex problems require you to find the distance between several points in order to ‘prove’ or ‘determine’ something about the shape.  If a question asks you to ‘prove this is an isosceles triangle’, or ‘prove that points A and B are vertices of an isosceles triangle’ you need to find the lengths of the sides (using the points given) and if two of the side lengths of the triangle are the same, then it is an isosceles triangle  If a question asks you to ‘prove’ or ‘determine’ whether a triangle is equilateral, it means you need to find the lengths of the three sides and if they are all the same length, it is an equilateral triangle  If a question gives you four points to find the distances between. If you are finding four side lengths, it means you are finding a quadrilateral, or a shape with 4 sides (e.g. a square if all the sides are equal, a rectangle if opposite pairs of sides are equal)

28. EXAMPLE: PROVE THAT THE POINTS A(1, 1), B(3, -1) AND C(-1, -3) ARE THE VERTICES OF AN ISOSCELES TRIANGLE.

29. MIDPOINT OF A SEGMENT (3D)

30. SO FAR…  We know that to find the distance between two points on a linear graph, we use a modified form of Pythagoras’ theorem  Now we are looking at how to find the midpoint of a segment of a linear graph  This means finding the coordinates of the point (x,y) that are halfway between points A and B 

31. LET ME EXPLAIN…  In this diagram, point A is (x 1, y 1 )  Point B is (x 2, y 2 )  Point M is the midpoint between these lines (x,y) – this is the point we are trying to find  The = signs means that the distance between A and M is the same as the distance from M to B  We find M by using the formula given A B

32. FIND THE COORDINATES OF THE MIDPOINT OF THE LINE SEGMENT JOINING (-2, 5) AND (7, 1).

33. THE COORDINATES OF THE MIDPOINT, M, OF THE LINE SEGMENT AB ARE (7, 2). IF THE COORDINATES OF A ARE (1, - 4), FIND THE COORDINATES OF B.

34. FOR WORDED PROBLEMS  It is always useful to draw a sketch first  For example: Find the coordinates of the centre of a square with vertices A(0,0), B(2,4), C(6,2) and D(4,-2) B(2.4) A(0,0) D(4,-2) C(6,2) For this question, we would need to find the midpoints of BD or AC to find the centre of this square