1 / 50

The coordinates of P and Q are (1, 5) and (9, −3) respectively, find

1. The coordinates of P and Q are (1, 5) and (9, −3) respectively, find. ( i ). the equation of the line PQ. P (1, 5) Q (9, −3). ( x 1 , y 1 ) ( x 2 , y 2 ). ( x 1 , y 1 ) = (1, 5) m = −1. ( y – y 1 ) = m ( x – x 1 ). ( y – 5) = −1( x – 1).

amunoz
Download Presentation

The coordinates of P and Q are (1, 5) and (9, −3) respectively, find

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 1. The coordinates of PandQ are (1, 5) and (9, −3) respectively, find (i) the equation of the line PQ P (1, 5) Q (9, −3) (x1, y1) (x2, y2) (x1, y1) = (1, 5) m = −1 (y – y1) = m(x – x1) (y – 5) = −1(x – 1) y – 5 = − x + 1 x + y – 6 = 0

  2. 1. The coordinates of PandQ are (1, 5) and (9, −3) respectively, find (ii) the equation of the line k, which passes through the origin O and is perpendicular to PQ Perpendicular lines: m1m2 = −1 −1 × mk = −1 mk = 1 (x1, y1) = (0, 0) mk = 1 (y – y1) = m(x – x1) (y – 0) = 1(x – 0) y = x 0 = x − y

  3. 1. The coordinates of PandQ are (1, 5) and (9, −3) respectively, find (iii) the coordinates of the point A, where PQ ∩ k = {A} x + y = 6 Let x = 3: x – y = 0 x – y = 0 2x = 6 3 – y = 0 x = 3 3 = y Point of intersection= (3, 3)

  4. 1. The coordinates of PandQ are (1, 5) and (9, −3) respectively, find (iv) the coordinates of R, where QOPR is a parallelogram. • Plot the three given points O, P and Q. • The vertices of the parallelogram are in the order of QOPR, as you go around the shape. This means that the translation which maps O to Q must be the same as the one which maps P to R. • To go from O (0, 0) to Q (9, −3), add 9 to the x coordinate and subtract 3 from the y coordinate. • Apply this same translation to P (1, 5) results in R: (10, 2)

  5. 2. l is the line 5x − 2y + c = 0 (i) (3, 4) is a point on l. Find the value of c. Substitute the coordinates of (3, 4) into the equation: 5x − 2y + c = 0 5(3) − 2(4) + c = 0 15 − 8 + c = 0 7 + c = 0 c = −7

  6. 2. l is the line 5x − 2y + c = 0 (ii) The line k is parallel to l and passes through the point (−2, 0). Find the equation of k. l: 5x − 2y − 7 = 0 a = 5, b = −2 Parallel lines have equal slopes, therefore

  7. 2. l is the line 5x − 2y + c = 0 (ii) The line k is parallel to l and passes through the point (−2, 0). Find the equation of k. (y − y1) = m(x − x1) (y − 0) = (x − (−2)) (y − 0) = (x + 2) 2(y − 0) = 5(x + 2) 2y = 5x + 10 k: 0 = 5x − 2y + 10

  8. 2. l is the line 5x − 2y + c = 0 (iii) The lines l and k together with the line x = 2 and the y-axis, form parallelogram.Find the coordinates of the vertices of the parallelogram. l: 5x − 2y − 7 = 0 Let x = 2: At the y-axis, x = 0 5x − 2y − 7 = 0 5x − 2y − 7 = 0 5(2) − 2y − 7 = 0 5(0) − 2y = 7 10 − 7 = 2y y = = y

  9. 2. l is the line 5x − 2y + c = 0 (iii) The lines l and k together with the line x = 2 and the y-axis, form parallelogram.Find the coordinates of the vertices of the parallelogram. k: 5x − 2y + 10 = 0 At the y-axis, x = 0: Let x = 2: 5x − 2y + 10 = 0 5x − 2y + 10 = 0 5(0) − 2y + 10 = 0 5(2) − 2y + 10 = 0 10 + 10 = 2y 10 = 2y 10 = y 5 = y (2, 10) (0, 5)

  10. 2. l is the line 5x − 2y + c = 0 (iii) The lines l and k together with the line x = 2 and the y-axis, form parallelogram.Find the coordinates of the vertices of the parallelogram.

  11. 3. A (6, 4) and B (4, 3) are two points on the coordinated plane. (i) Find the equation of the line AB. A (6, 4) B (4, 3) (x1,y1) (x2, y2) (x1, y1) = (6, 4) m = (y − y1) = m(x − x1) (y − 4) = 2(y − 4) = 1(x − 6) 2y − 8 = x − 6 0 = x − 2y + 2

  12. 3. A (6, 4) and B (4, 3) are two points on the coordinated plane. (ii) The line AB crosses the y-axis at the point C. Find the coordinates of C. Crossing y-axis, x = 0: 0 = x − 2y + 2 0 = 0 − 2y + 2 2y = 2 y = 1 C: (0, 1)

  13. 3. A (6, 4) and B (4, 3) are two points on the coordinated plane. (iii) Find the ratio , giving your answer in the form , where p, q ∈ . A: (6, 4) B: (4, 3) B: (4, 3) C: (0, 1) (x1,y1) (x2, y2) (x1,y1) (x2, y2)

  14. 4. The line lhas equation 2x − 3y + 12 = 0. A (0, 4) and B (−6, 0) are two points. (i) The line l has equation 2x − 3y + 12 = 0. A (0, 4) and B (−6, 0) are two points. Substitute the point B = (−6, 0): Substitute the point A = (0, 4): 2x − 3y + 12 = 0 2x − 3y + 12 = 0 2(0) − 3(4) + 12 = 0 2(−6) −3(0) + 12 = 0 0 − 12 + 12 = 0 −12 − 0 + 12 = 0 0 = 0 0 = 0 Therefore, A is on the line. Therefore, B is on the line.

  15. 4. The line lhas equation 2x − 3y + 12 = 0. A (0, 4) and B (−6, 0) are two points. (ii) Find the slope of l. l: 2x − 3y + 12 = 0 a = 2, b = −3

  16. 4. The line lhas equation 2x − 3y + 12 = 0. A (0, 4) and B (−6, 0) are two points. (iii) The line k is perpendicular to l and it contains the point B. Find the equation of k. For perpendicular lines: m1 m2= −1 (x1,y1) = (−6, 0) mk = (y − y1) = m (x − x1) mlmk= −1 mk= −1 (y − 0) = mk = (y − 0) = 2(y − 0) = −3(x + 6) 2y = −3x – 18 3x + 2y + 18 = 0

  17. 4. The line lhas equation 2x − 3y + 12 = 0. A (0, 4) and B (−6, 0) are two points. (iv) k intersects the y-axis at the point C. Find the coordinates of C. At the y-axis, x = 0: 3x + 2y + 18 = 0 3(0) + 2y + 18 = 0 2y = −18 y = −9 C: (0, −9)

  18. 4. The line lhas equation 2x − 3y + 12 = 0. A (0, 4) and B (−6, 0) are two points. (v) • Using the same axis and scale, plot the points A, B and C and graph the lines l and k.

  19. 4. The line lhas equation 2x − 3y + 12 = 0. A (0, 4) and B (−6, 0) are two points. (vi) • D is another point such that ABCD is a rectangle. Find the coordinates of D. • The vertices of the rectangle are in the order of ABCD, as you go around the shape. This means that the translation which maps B to A must be the same as the one which maps C to D. • To go from B (−6, 0) to A (0, 4), add 6 to the x-coordinate and add 4 to the y-coordinate. • Apply this same translation to C (0, −9) results in D: (6, −5)

  20. 4. The line lhas equation 2x − 3y + 12 = 0. A (0, 4) and B (−6, 0) are two points. (vii) • Calculate the area of ABCD. One possible solution: Move one point to (0, 0) and move all other points under the same translation: (−6, 0) (−6 + 6, 0) (0, 0) (0, −9) (0 + 6, −9) (6, −9) (0, 4) (0 + 6, 4) (6, 4) (6, −5) (6 + 6, −5) (12, −5)

  21. 4. The line lhas equation 2x − 3y + 12 = 0. A (0, 4) and B (−6, 0) are two points. (vii) • Calculate the area of ABCD. (0, 0) (6, −9) (12, −5) Area1 (x1, y1) (x2, y2) = 39 units2

  22. 4. The line lhas equation 2x − 3y + 12 = 0. A (0, 4) and B (−6, 0) are two points. (vii) • Calculate the area of ABCD. (0, 0) (6, 4) (12, −5) Area2 (x1, y1) (x2, y2) = 39 units2

  23. 4. The line lhas equation 2x − 3y + 12 = 0. A (0, 4) and B (−6, 0) are two points. (vii) • Calculate the area of ABCD. Total area = Area1 + Area2 = 39 + 39 = 78 units2

  24. 5. An electrician charges a call-out fee of €45 and then an additional €1·20 for every minute that he is working on a repair job in a house. (i) Write an equation to show the relationship between the total fee which the electrician will charge (F) and the number of minutes he is working on the job (T). y = starting value + growth rate(x) Starting value = 45 Growth rate = €1·20 = 1·2 F = 45 + 1·2T

  25. 5. An electrician charges a call-out fee of €45 and then an additional €1·20 for every minute that he is working on a repair job in a house. (ii) Graph this linear relationship for 0 ≤ T ≤ 120. T = 120 T = 0 F = 45 + 12(120) F = 45 + 12(0) = 45 + 144 F = €45 (0, 45) F = €189 (120, 189)

  26. 5. An electrician charges a call-out fee of €45 and then an additional €1·20 for every minute that he is working on a repair job in a house. (iii) Use your graph to find the cost of a 40-minute repair job. • Go to 40 on the horizontal axis. • Draw a line vertically upwards until you touch the graph. • Draw a line horizontally across until you touch the vertical axis. From the graph, when T = 40 minutes, F = €93

  27. 5. An electrician charges a call-out fee of €45 and then an additional €1·20 for every minute that he is working on a repair job in a house. (iv) Use your graph to find how long the electrician was working on a job, for which he charged €135. • Go to 135 on the vertical axis. • Draw a line horizontally across until you touch the graph. • Draw a line vertically downwards until you touch the horizontal axis. From the graph, when F = €135, T = 75 minutes.

  28. 5. An electrician charges a call-out fee of €45 and then an additional €1·20 for every minute that he is working on a repair job in a house. (v) The point (60, 117) is on this graph. Write a sentence that describes the meaning of these coordinates. A 60 minute job would be charged a fee of €117.

  29. 5. An electrician charges a call-out fee of €45 and then an additional €1·20 for every minute that he is working on a repair job in a house. (vi) If the electrician changed his rate per minute to €1.50, what effect would this have on the graph? Greater rate of change. It would make the slope of the graph steeper.

  30. 5. An electrician charges a call-out fee of €45 and then an additional €1·20 for every minute that he is working on a repair job in a house. (vii) If the electrician changed his call-out fee to €20, what effect would this have on the graph? The graph would be parallel to the current line, but with a lower starting value. It would start at €20 on the y-axis, instead of €45.

  31. 6. A (2, 3), B (−4, 1) and C (−2, 5) are three points. Prove that AC ⊥ BC (i) If AC ⊥ BC then m1 × m2 = −1. B: (−4, 1) C: (−2, 5) A: (2, 3) C: (−2, 5) (x1, y1) (x2, y2) (x1, y1) (x2, y2)

  32. 6. A (2, 3), B (−4, 1) and C (−2, 5) are three points. Prove that AC ⊥ BC (i) If two lines are perpendicular, then m1 × m2 = −1 Therefore, AC ⊥ BC

  33. 6. A (2, 3), B (−4, 1) and C (−2, 5) are three points. Prove that |AC| = |BC| (ii) B: (−4, 1) C(−2, 5) A: (2, 3) C: (−2, 5) (x1, y1) (x2, y2) (x1, y1) (x2, y2) , therefore |AC| = |BC|

  34. 6. A (2, 3), B (−4, 1) and C (−2, 5) are three points. • Calculate the area of the triangle BAC. (iii) Move the point A (2, 3) to (0, 0) and move B and C by the same translation:

  35. 6. A (2, 3), B (−4, 1) and C (−2, 5) are three points. • Calculate the area of the triangle BAC. (iii) Area1 = 10 units2

  36. 6. A (2, 3), B (−4, 1) and C (−2, 5) are three points. • The diagonals of the square BAHG intersect at C. Find the coordinates of H and the coordinates of G. (iv) To find the coordinates of H: Go from B into C and out the same amount on the far side

  37. 6. A (2, 3), B (−4, 1) and C (−2, 5) are three points. • The diagonals of the square BAHG intersect at C. Find the coordinates of H and the coordinates of G. (iv) To find the coordinates of G: Go from A into C and out the same amount on the far side

  38. 6. A (2, 3), B (−4, 1) and C (−2, 5) are three points. Find the equation of the line BC and show that H lies on this line. (v) B: (−4, 1) C:(−2, 5) (x1, y1) (x2, y2) (x1, y1) = (−4, 1) m = 2 (y − y1) = m(x − x1) (y − 1) = 2(x − (−4)) (y − 1) = 2(x + 4) y − 1 = 2x + 8 0 = 2x − y + 9 Slope of BC = 2

  39. 6. A (2, 3), B (−4, 1) and C (−2, 5) are three points. Find the equation of the line BC and show that H lies on this line. (v) Substitute the coordinates of H: (0, 9) into the equation: 0 = 2x − y + 9 0 = 2(0) − 9 + 9 0 = 0 − 9 + 9 0 = 0 Therefore H is on the line.

  40. 7. The line kmakes equal intercepts on the axes at the points Pand Q, as shown. (i) Write down the slope of k. Rise = Run = a

  41. 7. The line kmakes equal intercepts on the axes at the points Pand Q, as shown. (ii) The point (2, 7) is on k. Find the equation of k. (x1, y1) = (2, 7) m = 1 (y − y1) = m(x − x1) y − 7 = 1(x − 2) y − 7 = x − 2 k: 0 = x − y + 5

  42. 7. The line kmakes equal intercepts on the axes at the points Pand Q, as shown. (iii) The line n is perpendicular to k and contains the point (2, 1). Find the equation of n. Perpendicular lines, therefore mk × mn = −1 1 × mn = −1 mn = −1 (x1, y1) = (2, 1) (y − y1) = m(x − x1) y − 1 = −1(x − 2) y − 1 = − x + 2 n: x + y − 3 = 0

  43. 7. The line kmakes equal intercepts on the axes at the points Pand Q, as shown. (iv) Find the coordinates of the point R, such that k ∩ n = {R}. Let x = −1: x + y = 3  x − y = −5 −1 + y = 3  x + y = 3 y = 4 2x = −2 x = −1 Therefore, k ∩ n = R = (−1, 4)

  44. 8. The points P, Qand Rare the vertices of a triangle, as shown. QRis the line 2x + y + 3 = 0. Pin the point (4, 4) and Qis the point (−2, 1). (i) Find the slope of the line PQ. P: (4, 4) Q: (−2, 1) (x1, y1) (x2, y2)

  45. 8. The points P, Qand Rare the vertices of a triangle, as shown. QRis the line 2x + y + 3 = 0. Pin the point (4, 4) and Qis the point (−2, 1). (ii) State whether or not PQ and QR are perpendicular to each other. Give a reason for your answer. QR: 2x + y + 3 = 0 a = 2, b = 1

  46. 8. The points P, Qand Rare the vertices of a triangle, as shown. QRis the line 2x + y + 3 = 0. Pin the point (4, 4) and Qis the point (−2, 1). (ii) State whether or not PQ and QR are perpendicular to each other. Give a reason for your answer. If the lines are perpendicular lines, then m1 × m2 = −1 mPQ × mQR = −1 × −2 = −1 −1 = −1 Therefore PQQR

  47. 8. The points P, Qand Rare the vertices of a triangle, as shown. QRis the line 2x + y + 3 = 0. Pin the point (4, 4) and Qis the point (−2, 1). (iii) QR intersects the y-axis at the point R. Find the coordinates of the point R. At the y-axis, x = 0: 2x + y + 3 = 0 2(0) + y + 3 = 0 y + 3 = 0 y= −3 R: (0, −3)

  48. 8. The points P, Qand Rare the vertices of a triangle, as shown. QRis the line 2x + y + 3 = 0. Pin the point (4, 4) and Qis the point (−2, 1). (iv) • Find the area of the triangle PQR. Move the point R to (0, 0) and then move the points P and Q under the same translation:

  49. 8. The points P, Qand Rare the vertices of a triangle, as shown. QRis the line 2x + y + 3 = 0. Pin the point (4, 4) and Qis the point (−2, 1). (iv) • Find the area of the triangle PQR. Area1 = 15 units2

More Related