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The point ( t , 2 t ) lies on the line 4 x − 3 y + 6 = 0 . Find the value of t .

1. The point ( t , 2 t ) lies on the line 4 x − 3 y + 6 = 0 . Find the value of t. Substitute the coordinates of ( t , 2 t ) in for x and y :. 4 x − 3 y + 6 = 0. 4( t ) − 3(2 t ) + 6 = 0. 4 t − 6 t + 6 = 0. −2 t + 6 = 0. 6 = 2 t. = t. 3 = t. 2.

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The point ( t , 2 t ) lies on the line 4 x − 3 y + 6 = 0 . Find the value of t .

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  1. 1. The point (t, 2t) lies on the line 4x − 3y + 6 = 0. Find the value of t. Substitute the coordinates of (t, 2t) in for x and y: 4x − 3y + 6 = 0 4(t) − 3(2t) + 6 = 0 4t − 6t + 6 = 0 −2t + 6 = 0 6 = 2t = t 3 = t

  2. 2. Investigate whether the lines l: 2x − 3y + 7 = 0 and m: 6x + 4y − 5 = 0 are perpendicular. l: 2x − 3y + 7 = 0 m: 6x + 4y − 5 = 0 a = 2, b = −3 a = 6, b = 4 Slope = Slope = = = SlopeL = SlopeL = If the lines are perpendicular, ml × mm = −1 ml × mm = = −1 Therefore, lm

  3. 3. Find the equation of the line k, which passes through the point (3, −2) and is perpendicular to the line passing through (3, 3) and (5, 7). Find the slope of the line passing through (3,3) and (5,7) Slope = (3, 3) (5, 7) (x1, y1) (x2, y2) Slope = = = 2

  4. 3. Find the equation of the line k, which passes through the point (3, −2) and is perpendicular to the line passing through (3, 3) and (5, 7). Since the lines are perpendicular: m1 × m2 = −1 2 × m2 = −1 m2 = (x1, y1) = (3, −2)m = (y − y1) = m(x − x1) (y − (−2)) = (y + 2) = 2(y + 2) = −1(x − 3) 2y + 4 = − x + 3 x + 2y + 1 = 0

  5. 4. A (−2, 3) and B (2, 6) are two points. (i) Plot the points A and B on a coordinate diagram.

  6. 4. A (−2, 3) and B (2, 6) are two points. (ii) Find the equation of AB (−2, 3) (2, 6) (x1, y1) = (−2, 3) Slope = (x1, y1) (x2, y2) Slope = Slope = = 0 = 3x − 4y + 18 =

  7. 4. A (−2, 3) and B (2, 6) are two points. (iii) k is the line 3x + 2y − 9 = 0. Show that k passes through the midpoint of [AB]. A (−2, 3) B (2, 6) (x1, y1) (x2, y2) Midpoint = = = =

  8. 4. A (−2, 3) and B (2, 6) are two points. (iii) k is the line 3x + 2y − 9 = 0. Show that k passes through the midpoint of [AB]. Substitute the coordinates of the midpoint into the equation of the line, k: k: 3x + 2y − 9 = 0 0 + 9 − 9 = 0 0 = 0  Midpoint is on k

  9. 4. A (−2, 3) and B (2, 6) are two points. (iv) Investigate if k is perpendicular to AB. k: 3x + 2y − 9 = 0 Slope = a = 3, b = 2 Slope = =

  10. 4. A (−2, 3) and B (2, 6) are two points. (iv) Investigate if k is perpendicular to AB. If the lines are perpendicular lines then m1 × m2 = −1 Therefore, the lines are not perpendicular.

  11. 5. The point (3, 2) is on the line k: ax + y − 8 = 0. Find the value of a. (i) Substitute the coordinates of (3, 2) in for x and y: k: ax + y − 8 = 0 a(3) + 2 − 8 = 0 3a − 6 = 0 3a = 6 a = 2

  12. 5. The point (3, 2) is on the line k: ax + y − 8 = 0. Find the points where the line k intersects the x and y axes. (ii) At the y-axis, x = 0: At the x-axis, y = 0: 2x + y − 8 = 0 2x + y − 8 = 0 2(0) + y − 8 = 0 2x + 0 − 8 = 0 0 + y − 8 = 0 2x = 8 y = 8 x = 4 y-intercept = (0, 8) x-intercept = (4, 0)

  13. 5. The point (3, 2) is on the line k: ax + y − 8 = 0. Hence sketch the line k. (iii)

  14. 6. A (0, 0), B (−4, 1) and C (2, 8) are the vertices of a triangle. Prove that the triangle is right-angled at the point A. (i) If the triangle is right-angled then [AB] is perpendicular to [AC], so m1 × m2 = − 1.

  15. 6. A (0, 0), B (−4, 1) and C (2, 8) are the vertices of a triangle. Prove that the triangle is right-angled at the point A. (i) Slope of [AB]: Slope of [AC]: (0, 0) (−4, 1) (0, 0) (2, 8) (x1, y1) (x2, y2) (x1, y1) (x2, y2) Slope = Slope = mAB= =

  16. 6. A (0, 0), B (−4, 1) and C (2, 8) are the vertices of a triangle. Prove that the triangle is right-angled at the point A. (i) Investigate if: mAB × mAC = −1 −1 = −1 Therefore ABAC, showing the triangle is a right-angled triangle at the point A.

  17. 6. A (0, 0), B (−4, 1) and C (2, 8) are the vertices of a triangle. Find the area of the triangle ABC. (ii) Area = (0, 0) (−4, 1) (2, 8) (x1, y1) (x2, y2) = = = = = 17 units2

  18. 7. A (−3, 4), B (5, 6) and C (1, −2) are three points. Pis the midpoint of [AB] and Qis the midpoint of [AC]. Find the coordinates of P and the coordinates of Q. (i) A (−3, 4) B (5, 6) Midpoint = (x1, y1) (x2, y2)

  19. 7. A (−3, 4), B (5, 6) and C (1, −2) are three points. Pis the midpoint of [AB] and Qis the midpoint of [AC]. Find the coordinates of P and the coordinates of Q. (i) A (−3, 4) C (1, −2) Midpoint = (x1, y1) (x2, y2)

  20. 7. A (−3, 4), B (5, 6) and C (1, −2) are three points. Pis the midpoint of [AB] and Qis the midpoint of [AC]. Plot A, B, C, P and Q on a coordinate diagram and show the line segments [BC] and [PQ] on your diagram. (ii)

  21. 7. A (−3, 4), B (5, 6) and C (1, −2) are three points. Pis the midpoint of [AB] and Qis the midpoint of [AC]. Investigate whether PQ is parallel to BC. (iii) If the lines are parallel their slope will be equal. P (1, 5) Q (−1, 1) B (5, 6) C (1, −2) (x1, y1) (x2, y2) (x1, y1) (x2, y2)

  22. 7. A (−3, 4), B (5, 6) and C (1, −2) are three points. Pis the midpoint of [AB] and Qis the midpoint of [AC]. Investigate whether PQ is parallel to BC. (iii) Parallel lines have equal slopes. SincemPQ = mBC , PQ || BC

  23. 8. A (2, 3), B (1, −2) and C (12, 1) are the vertices of a triangle ABC. Investigate if the triangle is right-angled. (i) Right-angled implies two sides are perpendicular to each other A (2, 3) B (1, −2) A (2, 3) C (12, 1) (x1, y1) (x2, y2) (x1, y1) (x2, y2)

  24. 8. A (2, 3), B (1, −2) and C (12, 1) are the vertices of a triangle ABC. Investigate if the triangle is right-angled. (i) If the lines are perpendicular, Therefore, the triangle is right-angled at the point A.

  25. 8. A (2, 3), B (1, −2) and C (12, 1) are the vertices of a triangle ABC. The point (3, k) is on the line AB. Find the value of k. (ii) Find the equation of the line AB: (x1, y1) = (2, 3) m = 5 (y – y1) = m(x – x1) (y – 3) = 5(x – 2) y – 3 = 5x – 10 0 = 5x – y – 7

  26. 8. A (2, 3), B (1, −2) and C (12, 1) are the vertices of a triangle ABC. The point (3, k) is on the line AB. Find the value of k. (ii) Substitute the coordinates of (3, k) in for x and y: 0 = 5x – y – 7 0 = 5(3) – k – 7 0 = 15 – k – 7 0 = 8 – k k = 8

  27. 9. The line l intersect the x-axis at (-4, 0) and the y-axis at (0, 3). Find the slope of l. (i) (−4, 0) (0, 3) (x1, y1) (x2, y2)

  28. 9. The line l intersect the x-axis at (-4, 0) and the y-axis at (0, 3). Find the equation of l. (ii) (x1, y1) = (−4, 0) m= (y – y1) = m(x – x1)

  29. 9. The line m passes through (0, 0) and is perpendicular to l. Show the lines l and m on a coordinate diagram. (iii) Perpendicular lines m1m2 = −1 mlmm = −1 mm = −1 mm =

  30. 9. The line m passes through (0, 0) and is perpendicular to l. Find the equation of m (iv) (x1, y1) = (0, 0) m = (y – y1) = m(x – x1) (y – 0) = 3(y – 0) = −4(x – 0) 3y = −4x 4x + 3y = 0

  31. 9. The line m passes through (0, 0) and is perpendicular to l. The point (t, 4) is on the line m. Find the value of t. (v) Substitute the coordinates of (t, 4) into the line m: 4x + 3y = 0 4(t) + 3(4) = 0 4t + 12 = 0 4t = −12 t = −3

  32. 10. The line k: 3x + 2y − 12 = 0 intersects the x-axis at the point Pand the y-axis at the point Q. Find the coordinates of the points P and Q. (i) At y-axis, x = 0: At x-axis, y = 0: 3x + 2y – 12 = 0 3x + 2y – 12 = 0 3(0) + 2y – 12 = 0 3x + 2(0) −12 = 0 2y = 12 3x = 12 y = 6 x = 4 y-intercept, Q = (0, 6) x-intercept, P = (4, 0)

  33. 10. The line k: 3x + 2y − 12 = 0 intersects the x-axis at the point Pand the y-axis at the point Q. Graph the line k. (ii)

  34. 10. The line k: 3x + 2y − 12 = 0 intersects the x-axis at the point Pand the y-axis at the point Q. Hence, or otherwise, find the area of the triangle formed by the x-axis, y-axis and the line k. (iii) (0, 0) (4, 0) (0, 6) (x1, y1) (x2, y2) Area = = = = = 12 units2

  35. 11. Given the line m: 3x − 4y + 16 = 0: (i) Find the slope of m 3x – 4y + 16 = 0 a = 3, b = −4

  36. 11. Given the line m: 3x − 4y + 16 = 0: (ii) The line n is perpendicular to m and cuts the x-axis at the point (3, 0). Find the equation of n. (x1, y1) = (3, 0) mn = m nmmmn = −1 mn = −1 (y – y1) = m(x – x1) (y – 0) = mn = 3(y – 0) = −4(x – 3) 3y = −4x + 12 4x + 3y – 12 = 0

  37. 11. Given the line m: 3x − 4y + 16 = 0: (iii) Find the coordinates of the point of intersection of the lines m and n.  3x – 4y = −16 Let x = 0: ( 3) ( 4) 4x + 3y = 12 3x – 4y = −16 9x – 12y = −48 3(0) – 4y = −16 16x + 12y = 48 −4y = −16 25x = 0 y = 4 x = 0 Point of intersection = (0, 4)

  38. 12. Given the line l: 2x – 5y + 16 = 0, find: (i) the coordinates of A, where l intersects the x-axis At the x-axis,y = 0: 2x – 5y + 16 = 0 2x – 5(0) + 16 = 0 2x = −16 x = −8 A: (−8, 0)

  39. 12. Given the line l: 2x – 5y + 16 = 0, find: (ii) the equation of the line k, of slope 2 and passing through B (0, −4) (x1, y1) = (0, −4) mk = 2 (y – y1) = m(x – x1) (y – (−4)) = 2(x – 0) y + 4 = 2x 0 = 2x – y – 4

  40. 12. Given the line l: 2x – 5y + 16 = 0, find: (iii) the point C, where l ∩ k = {C}. ( −1)  2x – 5y = −16 Let y = 5: 2x – y = 4 2x – y = 4 –2x + 5y = 16 2x – 5 = 4 2x – y = 4 2x = 9 4y = 20 x = y = 5 C = = (4·5, 5)

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