By Eben Matlis

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To use the distance formula on this problem, we need to show that if we square then add the lengths of the two legs (the sides that are not CHAPTER 3 The xy Coordinate Plane the hypotenuse), this will equal the square of the hypotenuse. While not really necessary, we should plot the points to see which two sides are the legs and which side is the hypotenuse. Fig. 3-16. From the graph we can see that a is the distance from ðÀ2ó 13 2 Þ to ð1ó 2Þ; b is the distance from ð1ó 2Þ to ð4ó 4Þ; and c is the distance from ðÀ2ó 13 2Þ to ð4ó 4Þ.

31 CHAPTER 3 The xy Coordinate Plane 32 4. ð2ó 2Þ Right 2, up 2 Fig. 3-5. 5. ðÀ6ó 4Þ Left 6, up 4 Fig. 3-6. 6. ð0ó 3Þ No horizontal movement, up 3 Fig. 3-7. CHAPTER 3 The xy Coordinate Plane 7. 33 ðÀ4ó 0Þ Left 4, no vertical movement Fig. 3-8. 8. ð0ó 0Þ No horizontal movement, no vertical movement. These are the coordinates of the origin. Fig. 3-9. The Distance Between Two Points At times we need to find the distance between two points. If the points are on the same vertical line (the x-coordinates are the same), the distance between the points is the absolute value of the difference between the y-coordinates.

A) x ¼ 8 b) x ¼ 6 and x ¼ 8 c) x ¼ 8 and x ¼ À8 What is the solution for jxj > 4? a) ð4ó À 4Þ b) ðÀ4ó 4Þ c) ð4ó 1Þ d) x ¼ 6 d) ðÀ1ó À 4Þ [ ð4ó 1Þ CHAPTER 2 28 3. 4. 5. 6. Absolute Value Solve for x: 12 jx À 6j À 4 ¼ À1. a) x ¼ 12 b) x ¼ 8 and x ¼ 4 no solution c) x ¼ 12 and x ¼ 0 What is the solution for j2x À 3j a) ðÀ1ó 2Š b) ðÀ1ó 1Š [ ½2ó 1Þ 1? c) ðÀ1ó 1Š [ ½À1ó 2Š What is the solution for jÀ 12 x þ 5j > 1? a) ðÀ1ó 8Þ [ ð12ó 1Þ b) ð8ó 12Þ c) ðÀ1ó 12Þ [ ð8ó 1Þ Solve for x: j3x À 1j ¼ 2. a) x ¼ 1 b) x ¼ 1 and x ¼ À 13 1 and x ¼ 3 SOLUTIONS 1.

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