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g r a p h i n g

g r a p h i n g. Year 10 Maths for Methods. I n t r o du c t i o n - l i n e a r g r a p h s. A linear graph is a STRAIGHT LINE graph . The intercepts are points where the line crosses the x-axis and y-axis: The y-intercept is the point that the line

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g r a p h i n g

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  1. graphing Year 10 Maths for Methods

  2. Introduction - linear graphs A linear graph is a STRAIGHT LINE graph. The intercepts are points where the line crosses the x-axis and y-axis: The y-intercept is the point that the line crosses the y-axis. At this point x = 0. The x-interceptis the point that the line crosses the x-axis. At this point y = 0.

  3. Introduction - linear graphs gradient y-intercept The gradient of a graph refers to the slope / steepness of the graph • The y-intercept of a graph refers to: • Where the graph crosses the y-axis • At this point x = 0 The equation to represent a linear equation is given by: y = m + c

  4. gradient

  5. gradient + As a line gets steeper, the gradient gets larger We say a vertical line has a gradient that is UNDEFINED + +

  6. gradient In the same way, as a line with a negative gradient gets steeper, the gradient becomes more negative We say a horizontal line has a gradient that is ZERO - - -

  7. GRAdient& y-intercepT Match each graph to the equation Gradient = 2 Y-Int = 6 Gradient = -3 Y-Int = 12 C B A Gradient = 10 Y-Int = -9 State the gradient and the y-intercept of the following: a) b) c)

  8. GRAdient& y-intercepT c) y +3x = 1 3y – 2x = 12 + 4 to both sides ÷ both sides by 2 - 3 to both sides + 2 to both sides ÷ both sides by 3 Stating the gradient and y-intercept when given an equation. When given in the form first rearrange to the form eg. Rearrange the following equations to y = m + c form: a) 2y - 4x = 6

  9. GRAPHInGSTRAightlines y-intercept gradient Gradient = 2 = Using the gradient-intercept method a)

  10. GRAPHInGSTRAightlines y-intercept gradient Gradient = Using the gradient-intercept method b)

  11. GRAPHInGSTRAightlines y-intercept gradient Gradient = -3 = Using the gradient-intercept method c)

  12. WorksheEt-PAGE 1 NOW DO

  13. GRAPHInGSTRAightlines Let x = 0, solve for y: 2y – (4 0) = 6 2y – 0 = 6 2y = 6 y = = 3 Let y = 0, solve for x: (2 0) – 4x = 6 0 – 4x = 6 - 4x = 6 x = = - 1.5 (0, 3) (-1.5, 0) Using the x and y intercept method Use when equation is in form Find the x and y interceptsand draw a line through the points. eg. 2y - 4x = 6

  14. GRAPHInGSTRAightlines Let x = 0, solve for y: 2y – (4 0) = 6 2y – 0 = 6 2y = 6 y = = 3 (0, 3) Let y = 0, solve for x: (2 0) – 4x = 6 0 – 4x = 6 - 4x = 6 x = = - 1.5 (-1.5, 0) Using the x and y intercepts eg. 2y - 4x = 6

  15. GRAPHInGSTRAightlines Let x = 0, solve for y: 3y – (2 0) = 12 3y – 0 = 12 3y = 12 y = = 4 (0, 4) Let y = 0, solve for x: (3 0) – 2x = 12 0 – 2x = 12 - 2x = 12 x = = - 6 (-6, 0) Using the x and y intercepts eg2. 3y – 2x = 12

  16. GRAPHInGSTRAightlines Let x = 0, solve for y: y + (3 0) = 9 y + 0 = 9 y = 9 ( 0, 9 ) Let y = 0, solve for x: 0 + 3x = 9 x = = 3 ( 3, 0 ) Using the x and y intercepts eg3. y + 3x = 9

  17. GRAPHInGSTRAightlines Let x = 0, solve for y: y = (4 x 0) + 8 y = 0 + 8 y = 8 ( 0, 8 ) Let y = 0, solve for x: 0 = 4x + 8 -8 = 4x x = x = -2 ( -2, 0 ) Using the x and y intercepts eg4. y = 4x + 8

  18. Now lets graph using the classpad WorksheEt-PAGE 2 NOW DO

  19. real lIfeapplications A plumber charges a call out fee of $100 plus an additional $50 per hour. How could we model his Charges (C) per hour (h), using a linear equation? A pool initially has 2000 litres of water in it before being filled at a rate of 10 litres per minute. How could we model the amount of water in the pool (W) per minute (m), using a linear equation?

  20. real lIfeapplications 20000 Litres 100 Litres per minute At the end of the swimming season the pool is emptied. The emptying of the pool is given by How much water was in the pool before being emptied? At what rate is water being drained?

  21. FINDING IFA POiNTIS On A LINE Step 1: Substitute the x value given in the point in question, in this case x = 1 into the equation. Step 2: Solve the equation. (IF the y value is equal to the y value given in the question, in this case y = 4 then the point IS on the line) Once solved, this examples gives us y = 10 so the point is NOT on the line the equation y = 4x + 6 y = 4x + 6 y = (4 1) + 6 y = 4 + 6 y = 10 (1, 4) If we are given a linear equation and a point (set of co-ordinates), we can find if that point is on the graph of the equation. Look at the following example… eg. For the equation y = 4x + 6, is the point ( 1, 4) on it’s line?

  22. FINDING IFA POiNTIS On A LINE Proof: y = -2x + 3 y = -(2 4) + 3 y = -8 + 3 y = -5 Step 1: Substitute the x value given in the point in question, in this case x = 4 into the equation. Step 2: Solve the equation. If the y value is equal to the y value given in the question, in this case y = -5 then the point IS on the line. Once solved, this examples gives us y = -5 so the point IS on the line the equation y = -2x + 3 (4, -5) eg2. For the equation y = -2x + 3, is the point ( 4, -5) on it’s line?

  23. exercisE1F NOW DO Q8bchk, Q4 (LHS), Q5acf, Q7gjl, Q9, Q13, Q14, Q15, Q17ac

  24. Linear graphs m = How could we find the gradient of this line? gradient = m = State the coordinates of any 2 points on the graph Label the points ) and ) Substitute the values into the gradient equation and solve.

  25. Linear graphs What would the equation of the line be? Read the y-intercept off the graph m = You try: Find the gradient of this line? gradient = m = State the coordinates of any 2 points on the graph Label the points ) and ) Substitute the values into the gradient equation and solve.

  26. FINDING THE EQUATIoNOF A LINE y = m + c W H E N W E A R E G I V E N T H E G R A P H eg1. Find the equation of the line: y = m+ cgives y = 1+ 6 y = + 6 m = gradient = m = y-intercept = y-intercept gradient 6

  27. FINDING THE EQUATIoNOF A LINE y = m + c W H E N W E A R E G I V E N T H E G R A P H Eg2. Find the equation of the line: y = m+ cgives y = 2+ 3 m = gradient = m = y-intercept = y-intercept gradient 3

  28. FINDING THE EQUATIoNOF A LINE y = m + c W H E N W E A R E G I V E N T H E G R A P H Eg3. Find the equation of the line: y = m+ cgives y = 3+ 2 m = gradient = m = y-intercept = y-intercept gradient 2

  29. FINDING THE EQUATIoNOF A LINE y = m + c W H E N W E A R E G I V E N T H E G R A P H Eg4. Find the equation of the line: m = y = 0+ 2gives y = 2 y-intercept gradient gradient = m = y-intercept = 0 2

  30. FINDING THE EQUATIoNOF A LINE W H E N W E A R E G I V E N T H E G R A P H Eg5. Find the equation of the line. m = 3 y-intercept gradient gradient = m = y-intercept = undefined no y-intercept y = m + c

  31. FINDING THE EQUATIoNOF A LINE W H E N W E A R E G I V E N T W OPOIN T S Find the equation of the line that passes the points (2, 7) and (4, 13) Find the gradient * Label the points ) and ) * Sub the values into the gradient equation and solve 2. Write the equation for a straight line , inserting the gradient into ‘m’ 3. Choose either point and sub into the x and y of the equation, then solve c. m =

  32. FINDING THE EQUATIoNOF A LINE W H E N W E A R E G I V E N T W OPOIN T S You try: Find the equation of the line that passes the points (3, -4) and (5, 8) Find the gradient * Label the points ) and ) * Sub the values into the gradient equation and solve 2. Write the equation for a straight line , inserting the gradient into ‘m’ 3. Choose either point and sub into the x and y of the equation, then solve c. m =

  33. exercisE1G NOW DO Q2ace, Q3bh, Q4bd, Q5bde, Q6b, 12

  34. length of aline Eg1. Find the exact distance between the two points (2, 4) and (5, 10) The length of a line segment (or distance between two points) is found using:

  35. length of aline Eg2. Find the exact distance between the two points (-3, 4) and (2, -2) The length of a line segment (or distance between two points) is found using:

  36. length of aline Eg3. A line has a length of has end points (a, 10) and (4, 12). Solve the values of a and b. Two solutions as has two possible solutions: and

  37. length of aline Eg4. A line has a length of has end points (3, 2) and (7, b). Solve the values of a and b. Two solutions as has two possible solutions: or

  38. midpointof aline Eg. Find the midpoint between the points (2, 5) and (6, 11) The midpoint of a line segment (between two points) is found using:

  39. midpointof aline Eg2. Find the midpoint between the points (-4, 0) and (10, 9)

  40. midpointof aline Eg3. The midpoint between the points (a, 7) and (8, b) is (3, 15) Find the values of a and b.

  41. midpointof aline Eg4. The midpoint between the points (-2, b) and (a, 21) is (6, 12) Find the values of a and b.

  42. exercisE1H NOW DO Q4ace, Q5ace, Q7a, Q8a, Q13

  43. ParalLellines • Never intersect (touch each other) • Share the same gradient • Have different y-intercepts • The lines pictured are parallel lines. • If their gradient is 2, then the equation for the lines both takes the form • The equation for top line would be: The equation for lower line would be:

  44. Perpendicular lines • Lines intersect (touch each other) forming four 90° angles • Their gradients are different, but share a relationship • given by: • OR • The lines pictured are perpendicular lines. • If the gradient of line 1 is 2, then the gradient line 2 is: Where is the Gradient of line 1 and is the Gradient of line 2

  45. Find the equation of the line that is: Parallel to and passes through the point (0, 9) b) Parallel to and passes through the point (1, 10) 1. Parallel so gradients are the same Substitute m = 3 into The point given (0, 9) is the y-intercept (as So substitute c = 9 into the equation. 1. Parallel so gradients are the same Substitute m = -4 into The point given (1, 10) is NOT the y-intercept (as So substitute (1, 10) the equation: x becomes 1, y becomes 10 3. Solve ‘c’ 4. Re-write equation with m = -4 and c = 14 in

  46. Find the equation of the line that is: Perpendicular to and passes through the point (0, 3) b) Perpendicular to and passes through the point (10, -6) 1. Perpendicular so gradients have relationship Substitute = 4 and solve 2. Substitute into 3. The point given (0, 3) is the y-intercept (as So substitute c = 3 into the equation. 1. Perpendicular so gradients have relationship Substitute and solve 2. Substitute into 3. The point given (10, -6) is NOT the y-intercept (as So substitute (10, -6) into the equation: x becomes 10, y becomes -6 4. Solve ‘c’ 5. Re-write equation with m = and c = in

  47. exercisE1I NOW DO Q1ac, Q2bc, Q5acegij, Q11ac, Q13

  48. SIMULTAneousEQUATIOnS Simultaneous equations involve: • Finding a solution to a set of two equations that each have the same 2 unknowns variables • Both equations share the same solution for each variable. We can solve simultaneous equations in a number of ways, including: • Graphing two equations and finding their point of intersection • The Substitution method • The Elimination method

  49. SIMULTAneousEQUATIOnS G R A P H I N G T H E N F I N D I N G T H E P O I N T O F I N T E R S E C T I O N Solve the simultaneous equations 1. Graph them on the same axis 2. Find the point where they intersect – the co-ordinates at this point is the solution. The graphs intersect at (3, 1) x = 3, y = 1

  50. SIMULTAneousEQUATIOnS y = 3x - 3 y = -x + 1 G R A P H I N G T H E N F I N D I N G T H E P O I N T O F I N T E R S E C T I O N eg1. Solve the simultaneous equations: 1. Graph them on the same axis 2. Find the point where they intersect Intersect at ( 1, 0 ) So, x = 1, y = 0

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