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Do Now: Pass out calculators. Complete EOC Review Week # 17. Have your homework out ready to check.

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Presentation on theme: "Do Now: Pass out calculators. Complete EOC Review Week # 17. Have your homework out ready to check."— Presentation transcript:

1 Do Now: Pass out calculators. Complete EOC Review Week # 17. Have your homework out ready to check.

2 Do Now: Pass out calculators. Complete the problem below. Write an explanation next to each multiple choice answer.

3 Objective: To apply the distance and midpoint formulas.

4 Distance Formula:

5 EXAMPLE 1 Find the distance between two points Find the distance between (– 1, 3) and (5, 2) Let ( x 1, y 1 ) = ( –1, 3) and ( x 2, y 2 ) = ( 5, 2 ). d ( x 2 – x 1 ) 2 + ( y 2 – y 1 ) 2 = Distance formula = (5 – (–1)) 2 + ( 2 – 3) 2 Substitute. = 6 2 + ( – 1) 2 = 37 Simplify. ANSWER The distance between the points is units. 37

6 Extra Example… Find the distance between (-3, 1) and (2, 3).

7 GUIDED PRACTICE for Examples 1 and 2 Find the distance between the points. 1. (3, 0), (3, 6) ANSWER The distance between the points is 6 units. 2. (–2, 1), (2, 5) ANSWER The distance between the points is 4 units. 2 3. (6, –2), (–4, 7) ANSWER The distance between the points is units. 181

8 EXAMPLE 2 Find a missing coordinate The distance between (3, – 5) and (7, b) is 5 units. Find the value of b. SOLUTION Use the distance formula with d = 5. Let ( x 1, y 1 ) = ( 3, –5) and ( x 2, y 2 ) = ( 7, b ). Then solve for b d ( x 2 – x 1 ) 2 + ( y 2 – y 1 ) 2 = Distance formula Substitute. = (7 – 3) 2 + ( b – (– 5)) 2 5 Multiply. = 16 + b 2 + 10b + 25 5

9 EXAMPLE 2 Find a missing coordinate Simplify. = b 2 + 10b + 41 5 Square each side. b 2 + 10b + 41 25 = Write in standard form. b 2 + 10b + 16 0 = Factor. (b + 2)(b + 8) 0 = b + 2 = 0 or b + 8 = 0 Zero-product property b = – 2 or b = – 8 Solve for b. ANSWER The value of b is – 2 or – 8

10 GUIDED PRACTICE for Examples 1 and 2 4. The distance between (1, a) and (4, 2) is 3 units. Find the value of a. ANSWER The value of a is 2.

11

12 EXAMPLE 3 Standardized Test Practice SOLUTION Let ( x 1, y 1 ) = ( –1, – 2) and ( x 2, y 2 ) = ( 3, – 4 ). ( ) x 1 + x 2 y 1 + y 2 2 2, = ( ) – 1 + 3 – 2 + (– 4) 2 2, Substitute = (1,– 3) Simplify ANSWER The correct answer is B.

13 GUIDED PRACTICE for Examples 3 and 4 5. Find the midpoint of the line segment with endpoints (4, 3) and (2, 5). (3, 4) ANSWER

14 Pg. 748 # 39, 41

15 Exit Ticket: 1.Can you find the distance between two points? Find the distance between (4, 8), (4, 7) 2. Can you find the missing coordinate using the distance formula? (13, -3), (b, 2); d = 13 3. Can you find the midpoint of the line segment with the given endpoint? (6, -3), (4, -7)


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