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Definitions and Postulates

Definitions and Postulates. Measure the length of ST to the nearest tenth of a centimeter. Align one mark of a metric ruler with S . Then estimate the coordinate of T . For example, if you align S with 2 , T appears to align with 5.4. ST = 5.4 – 2 = 3.4.

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Definitions and Postulates

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  1. Definitions and Postulates

  2. Measure the length of STto the nearest tenth of a centimeter. Align one mark of a metric ruler with S. Then estimate the coordinate of T. For example, if you align Swith 2, T appears to align with 5.4. ST= 5.4 – 2 = 3.4 The length of STis about 3.4 centimeters. ANSWER EXAMPLE 1 Apply the Segment Addition Postulate SOLUTION Use Ruler Postulate.

  3. Maps The cities shown on the map lie approximately in a straight line. Use the given distances to find the distance from Lubbock, Texas, to St. Louis, Missouri. The distance from Lubbock to St. Louis is about 740 miles. ANSWER EXAMPLE 2 Apply the the Segment Addition Postulate SOLUTION Because Tulsa, Oklahoma, lies between Lubbock and St. Louis, you can apply the Segment Addition Postulate. LS = LT + TS = 380 + 360 = 740

  4. 1 8 1. ANSWER 5 MN1 in 8 2. ANSWER 3 PQ1 in 8 for Examples 1 and 2 GUIDED PRACTICE Use a ruler to measure the length of the segment to the nearest inch. Use ruler postulate Use rulerpostulate

  5. In Exercises 3 and 4, use the diagram shown. 3. Use the Segment Addition Postulate to find XZ. ANSWER xz = 73 for Examples 1 and 2 GUIDED PRACTICE SOLUTION xz = xy + yz Segment addition postulate = 23 + 50 Substitute 23 for xy and 50 for yz = 73 Add

  6. In the diagram, WY = 30. Can you use the Segment Addition Postulate to find the distance between points Wand Z? 4. ANSWER NO; Because w is not between x and z. for Examples 1 and 2 GUIDED PRACTICE

  7. Use the diagram to find GH. FH = FG+ GH 36 21+GH = 15 GH = EXAMPLE 3 Find a length SOLUTION Use the Segment Addition Postulate to write an equation. Then solve the equation to find GH. Segment Addition Postulate. Substitute 36 for FHand 21 for FG. Subtract 21 from each side.

  8. To find the length of a horizontal segment, find the absolute value of the difference of the x-coordinates of the endpoints. JK = 2 – (– 3) = 5 EXAMPLE 4 Compare segments for congruence Plot J(– 3, 4), K(2, 4), L(1, 3), and M(1, – 2) in a coordinate plane. Then determine whether JKand LMare congruent. SOLUTION Use Ruler Postulate.

  9. LM = – 2 – 3 = 5 ~ JKand LMhave the same length. So, JK LM. = ANSWER EXAMPLE 4 Compare segments for congruence To find the length of a vertical segment, find the absolute value of the difference of the y-coordinates of the endpoints. Use Ruler Postulate.

  10. 5. Use the diagram at the right to find WX. 144 = 37 + wx = wx 107 ANSWER WX = 107 for Examples 3 and 4 GUIDED PRACTICE Use the segment addition postulate to write an equation. Then solve the equation to find WX SOLUTION vx = vw + wx Segment addition postulate Substitute 37 for vw and144for vx Subtract 37 from each side

  11. Plot the points A(– 2, 4),B(3, 4), C(0, 2), and D(0, – 2) in a coordinate plane. Then determine whether ABand CDare congruent. Length of AB is not equal to the length of CD, so they are not congruent ANSWER for Examples 3 and 4 GUIDED PRACTICE 6.

  12. a. FIND MIDPOINTThe endpoints ofRSare R(1,–3) and S(4, 2). Find the coordinates of the midpoint M. EXAMPLE 3 Use the Midpoint Formula

  13. 1 , – , M M = 2 5 2 The coordinates of the midpoint Mare 1 5 – , 2 2 ANSWER – 3 + 2 1 + 4 2 2 EXAMPLE 3 Use the Midpoint Formula SOLUTION a. FIND MIDPOINTUse the Midpoint Formula.

  14. FIND ENDPOINTLet (x, y) be the coordinates of endpoint K. Use the Midpoint Formula. b. FIND ENDPOINTThe midpoint of JKis M(2, 1). One endpoint is J(1, 4). Find the coordinates of endpoint K. STEP 1 Find x. STEP 2 Find y. 4+ y 1+ x 1 2 = = 2 2 ANSWER The coordinates of endpoint Kare (3, – 2). EXAMPLE 3 Use the Midpoint Formula 4 + y = 2 1 + x = 4 y =–2 x =3

  15. 3. The endpoints of ABare A(1, 2) andB(7, 8).Find the coordinates of the midpoint M. () M (4, 5) = M 1 + 7 2 + 8 ANSWER The Coordinates of the midpoint M are(4,5). , 2 2 for Example 3 GUIDED PRACTICE SOLUTION Use the midpoint formula.

  16. STEP 1 Find x. STEP 2 Find y. 4+ x – 1 = 4+ y – 2 2 = 2 ANSWER The coordinates of endpoint V is (– 6, – 8) for Example 3 GUIDED PRACTICE 4. The midpoint of VWis M(– 1, – 2). One endpoint is W(4, 4). Find the coordinates of endpoint V. SOLUTION Let (x, y) be the coordinates of endpoint V. Use the Midpoint Formula. 4 + x = – 2 4 + y = – 4 x = –6 y = –8

  17. Use the Distance Formula. You may find it helpful to draw a diagram. EXAMPLE 4 Standardized Test Practice SOLUTION

  18. RS = 2 2 = (x– x) + (y–y) 2 1 2 1 = 2 2 [(4 – 2)] + [(–1) –3] = 2 2 (2) + (–4 ) = = 4+16 4.47 20 ANSWER The correct answer is C. EXAMPLE 4 Standardized Test Practice Distance Formula Substitute. Subtract. Evaluate powers. Add. Use a calculator to approximate the square root.

  19. 5. In Example 4, does it matter which ordered pair you choose to substitute for (x , y ) and which ordered pair you choose to substitute for (x , y )? Explain. 1 1 2 2 ANSWER No, when squaring the difference in the coordinate you get the same answer as long as you choose the x and y value from the some period for Example 4 GUIDED PRACTICE

  20. 2 2 (x – x ) + (y – y ) 2 1 2 1 = AB 6. What is the approximate length of AB, with endpoints A(–3, 2) and B(1, –4)? 2 2 = [2 –(–3)] + (–4 –1) = 2 2 (5) + (5 ) 6.1 units 7.2 units 8.5 units 10.0 units for Example 4 GUIDED PRACTICE SOLUTION Use the Distance Formula. You may find it helpful to draw a diagram. Distance Formula Substitute. Subtract.

  21. ANSWER The correct answer is B = 25+25 50 = = 7.2 for Example 4 GUIDED PRACTICE Evaluate powers. Add. Use a calculator to approximate the square root.

  22. Skateboard In the skateboard design, VWbisects XYat point T, and XT=39.9cm. Find XY. Point Tis the midpoint of XY . So, XT = TY = 39.9 cm. EXAMPLE 1 Find segment lengths SOLUTION XY = XT + TY Segment Addition Postulate = 39.9 + 39.9 Substitute. = 79.8cm Add.

  23. ALGEBRA Point Mis the midpoint of VW. Find the length of VM . STEP 1 Write and solve an equation. Use the fact that that VM = MW. EXAMPLE 2 Use algebra with segment lengths SOLUTION VM= MW Write equation. 4x–1= 3x + 3 Substitute. x – 1 = 3 Subtract 3xfrom each side. x = 4 Add 1 to each side.

  24. STEP 2 Evaluate the expression for VMwhen x = 4. So, the length of VMis 15. Check: Because VM = MW, the length of MWshould be 15. If you evaluate the expression for MW, you should find that MW = 15. MW = 3x + 3 = 3(4) +3 = 15 EXAMPLE 2 Use algebra with segment lengths VM = 4x – 1 = 4(4) – 1 = 15

  25. M is midpoint and line MN bisects the line PQ at M. SoMNis the segment bisector of PQ. So PM = MQ =1 PQ = PM + MQ = 1 1 3 + 7 7 7 3 8 8 8 4 = for Examples 1 and 2 GUIDED PRACTICE In Exercises 1 and 2, identify the segment bisectorof PQ . Then find PQ. 1. SOLUTION Segment addition postulate. Substitute Add.

  26. In Exercises 1 and 2, identify the segment bisector of PQ . Then find PQ. for Examples 1 and 2 GUIDED PRACTICE 2. SOLUTION M is midpoint and line l bisects the line PQ of M. So lis the segment bisector of PQ. So PM = MQ

  27. STEP 1 Write and solve an equation Evaluate the expression for PQwhen x = + 4 x = 18 7 STEP 2 3 5 11 7 18 Substitute for x. 7 18 18 PQ = 7 7 = for Examples 1 and 2 GUIDED PRACTICE PM = MQ Write equation. 5x – 7 = 11–2x Substitute. 7x = 18 Add 2xand 7 each side. Divide each side by 7. PQ = 5x – 7 + 11 – 2x = 3x + 4 Simplify.

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