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Introduction

Introduction. In this introduction, you will learn how to - calculate the length of a line segment. y. B. C. A. D. x. x. x. x. x. Revision – reading coordinate points on an axes. Reading a set of coordinates : - read off values on the x-axis read off values on the y-axis

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Introduction

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  1. Introduction In this introduction, you will learn how to - calculate the length of a line segment

  2. y B C A D x x x x x Revision – reading coordinate points on an axes • Reading a set of coordinates : • - read off values on the x-axis • read off values on the y-axis • put the two values in the order • (x, y) and you get a coordinate pair 3 2 1 1 2 3 -3 -2 -1 -1 -2 -3

  3. y A F E D B C x x x x x x x Revision – reading coordinate points on an axes 3 2 1 3 3 (3, 3) 2 -2 (2, -2) 1 2 3 -3 -2 -1 -1 -2 -1 (-2, -1) -2 -1.5 1 (-1.5, 1) -3 2 1 (2, 1) Use the spacebar to see the next set

  4. (7, 10) B x Join the two points Produce a right-angled triangle Mark a point C (1, 2) x C y Length of Line Segment Here, we want to find the length of the line segment AB 12 We read off the coordinates of the two points 10 8 6 4 x 2 A x 2 4 6 8 10 12 Next we will determine the length of AC and BC

  5. (7, 10) B x (1, 2) x C(7, 2) y Length of Line Segment Here, we want to find the length of the line segment AC 12 Looking at the x-values of both points A and C, x-value of point A = 1 x-value of point C = 7 Therefore length AC = (7 – 1) = 6 10 8 6 4 Looking at the y-values of both points B and C, y-value of point B = 10 y-value of point C = 2 Therefore length BC = (10 – 2) = 8 x 2 A x 2 4 6 8 10 12 By pythagorus’ theorem, AB2 = AC2 + BC2

  6. (7, 10) B x (1, 2) x C(7, 2) AC = 6 units, BC = 8 units y Length of Line Segment By pythagorus’ theorem, AB2 = AC2 + BC2 12 10 8 6 4 units x 2 A x 2 4 6 8 10 12

  7. (x2, y2) B x (x1, y1) x C(x2, y1) Length of Line Segment – IN GENERAL y Replacing A with coordinates (x1, y1) and B with coordinates (x2, y2) Therefore point C will have (x2, y1) 12 y2 10 Looking at the x-values of both points A and C, x-value of point A = x1 x-value of point C = x2 Therefore length AC = (x2 – x1) 8 6 4 y1 x 2 Looking at the y-values of both points B and C, y-value of point B = y2 y-value of point C = y1 Therefore length BC = (y2 – y1) A x 2 4 6 8 10 12 x1 x2

  8. (x2, y2) B x (x1, y1) x C(x2, y1) Length of Line Segment – IN GENERAL y Using Pythagorus’ theorem, 12 AB2 = AC2 + BC2 y2 10 8 6 4 y1 x 2 A x 2 4 6 8 10 12 x1 x2

  9. Applying Pythagorus’ theorem : Distance = (to 3 sig. fig.) Example 1 : Find the distance between the points (2, 8) and (3, 16) Let A be (2, 8)  (x1, y1) B be (3, 16)  (x2, y2)

  10. Applying Pythagorus’ theorem : Distance = units (to 3 sig. fig.) Example 2 : Find the distance between the points (-3, 8) and (-8, 14) Let A be (–3 , 5)  (x1, y1) B be (–8, 14)  (x2, y2)

  11. Complete the following exercise in foolscap papers and hand up when you return to school. • Find the distance between the following pairs of points: • (a) (2, 3) and (9, 7) • (b) (–1 , 4) and (8, 5) • (c) (2, –5) and (6, 12) • (d) (–10, 2) and (–4, –7) • If the distance between the points A(k, 0) and B(0, k) is 10, find the possible values of k. • The coordinates of two points are A(–2, 6) and B(9, 3). Find the coordinates of the point C such • that AC = BC. • 4. The distance between the points (1, 2t) and (1 – t , 1) is , • find the possible values of t. For all Sec 4 E and 5 N(A), please also complete the following: Ten Years’ Series Page 67 Q1 to 7. End of Lesson

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