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example 1 use the number line to find the coordinate of the midpoint of…

Question

example 1
use the number line to find the coordinate of the midpoint of each segment.
j k l m n p
-7-6-5-4-3-2-1 0 1 2 3 4 5 6

  1. $overline{rm}$
  2. $overline{jp}$
  3. $overline{ln}$
  4. $overline{mp}$
  5. $overline{lp}$
  6. $overline{jn}$

use the number line to find the coordinate of the midpoint of each segment.
e f g h j k l
-6 -4 -2 0 2 4 6 8 10

  1. $overline{fk}$
  2. $overline{hk}$
  3. $overline{ef}$
  4. $overline{fg}$
  5. $overline{jl}$
  6. $overline{el}$

use tools use the number line to find the coordinate of the midpoint of each segment.
a b c d e
-6 -4 -2 0 2 4 6 8 10 12

  1. $overline{de}$
  2. $overline{bc}$
  3. $overline{bd}$
  4. $overline{ad}$

example 2

  1. home improvement callie wants to build a fence halfway between her house and her neighbor’s house. how far away from callie’s house should the fence be built?
  1. drawing calvino’s home is located at the midpoint between fast pizza and pizza now. fast pizza is a quarter mile away from calvino’s home. how far away is pizza now from calvino’s home? how far apart are the two pizzerias?

Explanation:

Response

Let's solve these midpoint problems one by one. We'll use the midpoint formula for a number line, which is the average of the two endpoints' coordinates, i.e., \( \text{Midpoint} = \frac{x_1 + x_2}{2} \), where \( x_1 \) and \( x_2 \) are the coordinates of the endpoints.

Problem 1: \( \overline{KM} \)
  • From the first number line, \( K \) is at \( -5 \) and \( M \) is at \( 1 \).
  • Midpoint formula: \( \frac{-5 + 1}{2} = \frac{-4}{2} = -2 \)
Problem 2: \( \overline{JP} \)
  • \( J \) is at \( -7 \) and \( P \) is at \( 6 \).
  • Midpoint: \( \frac{-7 + 6}{2} = \frac{-1}{2} = -0.5 \)
Problem 3: \( \overline{LN} \)
  • \( L \) is at \( -2 \) and \( N \) is at \( 4 \).
  • Midpoint: \( \frac{-2 + 4}{2} = \frac{2}{2} = 1 \)
Problem 4: \( \overline{MP} \)
  • \( M \) is at \( 1 \) and \( P \) is at \( 6 \).
  • Midpoint: \( \frac{1 + 6}{2} = \frac{7}{2} = 3.5 \)
Problem 5: \( \overline{LP} \)
  • \( L \) is at \( -2 \) and \( P \) is at \( 6 \).
  • Midpoint: \( \frac{-2 + 6}{2} = \frac{4}{2} = 2 \)
Problem 6: \( \overline{JN} \)
  • \( J \) is at \( -7 \) and \( N \) is at \( 4 \).
  • Midpoint: \( \frac{-7 + 4}{2} = \frac{-3}{2} = -1.5 \)
Problem 7: \( \overline{FK} \)
  • From the second number line, \( F \) is at \( -4 \) and \( K \) is at \( 10 \).
  • Midpoint: \( \frac{-4 + 10}{2} = \frac{6}{2} = 3 \)
Problem 8: \( \overline{HK} \)
  • \( H \) is at \( 4 \) and \( K \) is at \( 10 \).
  • Midpoint: \( \frac{4 + 10}{2} = \frac{14}{2} = 7 \)
Problem 9: \( \overline{EF} \)
  • \( E \) is at \( -6 \) and \( F \) is at \( -4 \).
  • Midpoint: \( \frac{-6 + (-4)}{2} = \frac{-10}{2} = -5 \)
Problem 10: \( \overline{FG} \)
  • \( F \) is at \( -4 \) and \( G \) is at \( 0 \).
  • Midpoint: \( \frac{-4 + 0}{2} = \frac{-4}{2} = -2 \)
Problem 11: \( \overline{JL} \)
  • \( J \) is at \( 6 \) and \( L \) is at \( 12 \) (Wait, no, looking at the second number line: \( J \) is at \( 6 \), \( L \) is at \( 12 \)? Wait, the second number line has \( E(-6), F(-4), G(0), H(4), J(6), K(10), L(12) \)? Wait, maybe I misread. Wait, the second number line: \( E \) at -6, \( F \) at -4, \( G \) at 0, \( H \) at 4, \( J \) at 6, \( K \) at 10, \( L \) at 12? Wait, no, the labels: \( E, F, G, H, J, K, L \) with coordinates -6, -4, 0, 2, 4, 6, 8? Wait, maybe the original number line is: \( E(-6), F(-4), G(0), H(2), J(4), K(6), L(8) \)? Wait, the user's image: the second number line has marks at -6, -4, -2, 0, 2, 4, 6, 8, 10. So \( E \) at -6, \( F \) at -4, \( G \) at 0, \( H \) at 2, \( J \) at 4, \( K \) at 6, \( L \) at 8? Wait, maybe I made a mistake. Let's recheck. The second number line: "E F G H J K L" with coordinates -6, -4, 0, 2, 4, 6, 8? Wait, the user's image: the second number line has: E at -6, F at -4, G at 0, H at 2, J at 4, K at 6, L at 8? Wait, the problem 7 is \( \overline{FK} \): F is at -4, K is at 6. Then midpoint is \( \frac{-4 + 6}{2} = 1 \). Wait, maybe my initial reading was wrong. Let's correct:

Looking at the second number line (the one with E, F, G, H, J, K, L):

  • E: -6
  • F: -4
  • G: 0
  • H: 2
  • J: 4
  • K: 6
  • L: 8

So problem 7: \( \overline{FK} \): F(-4), K(6). Midpoint: \( \frac{-4 + 6}{2} = 1 \)

Problem 8: \( \overline{HK} \): H(2), K(6). Midpoint: \( \frac{2 + 6}{2} = 4 \)

Problem 9: \( \overline{EF} \): E(-6), F(-4). Midpoint: \( \frac{-6 + (-4)}{2} = -5 \)

Problem 10: \( \overline{FG} \): F(-4), G(0). Midpoint: \( \frac{-4 + 0}{2} = -2 \)

Problem 11: \( \overline{JL} \): J(4), L(8). Midpoint: \( \frac{4 + 8}{2} = 6 \)

Problem 12: \( \overline{EL} \): E(-6), L(8). Midpoint: \( \frac{-6 + 8}{2} = 1 \)…

Answer:

Let's solve these midpoint problems one by one. We'll use the midpoint formula for a number line, which is the average of the two endpoints' coordinates, i.e., \( \text{Midpoint} = \frac{x_1 + x_2}{2} \), where \( x_1 \) and \( x_2 \) are the coordinates of the endpoints.

Problem 1: \( \overline{KM} \)
  • From the first number line, \( K \) is at \( -5 \) and \( M \) is at \( 1 \).
  • Midpoint formula: \( \frac{-5 + 1}{2} = \frac{-4}{2} = -2 \)
Problem 2: \( \overline{JP} \)
  • \( J \) is at \( -7 \) and \( P \) is at \( 6 \).
  • Midpoint: \( \frac{-7 + 6}{2} = \frac{-1}{2} = -0.5 \)
Problem 3: \( \overline{LN} \)
  • \( L \) is at \( -2 \) and \( N \) is at \( 4 \).
  • Midpoint: \( \frac{-2 + 4}{2} = \frac{2}{2} = 1 \)
Problem 4: \( \overline{MP} \)
  • \( M \) is at \( 1 \) and \( P \) is at \( 6 \).
  • Midpoint: \( \frac{1 + 6}{2} = \frac{7}{2} = 3.5 \)
Problem 5: \( \overline{LP} \)
  • \( L \) is at \( -2 \) and \( P \) is at \( 6 \).
  • Midpoint: \( \frac{-2 + 6}{2} = \frac{4}{2} = 2 \)
Problem 6: \( \overline{JN} \)
  • \( J \) is at \( -7 \) and \( N \) is at \( 4 \).
  • Midpoint: \( \frac{-7 + 4}{2} = \frac{-3}{2} = -1.5 \)
Problem 7: \( \overline{FK} \)
  • From the second number line, \( F \) is at \( -4 \) and \( K \) is at \( 10 \).
  • Midpoint: \( \frac{-4 + 10}{2} = \frac{6}{2} = 3 \)
Problem 8: \( \overline{HK} \)
  • \( H \) is at \( 4 \) and \( K \) is at \( 10 \).
  • Midpoint: \( \frac{4 + 10}{2} = \frac{14}{2} = 7 \)
Problem 9: \( \overline{EF} \)
  • \( E \) is at \( -6 \) and \( F \) is at \( -4 \).
  • Midpoint: \( \frac{-6 + (-4)}{2} = \frac{-10}{2} = -5 \)
Problem 10: \( \overline{FG} \)
  • \( F \) is at \( -4 \) and \( G \) is at \( 0 \).
  • Midpoint: \( \frac{-4 + 0}{2} = \frac{-4}{2} = -2 \)
Problem 11: \( \overline{JL} \)
  • \( J \) is at \( 6 \) and \( L \) is at \( 12 \) (Wait, no, looking at the second number line: \( J \) is at \( 6 \), \( L \) is at \( 12 \)? Wait, the second number line has \( E(-6), F(-4), G(0), H(4), J(6), K(10), L(12) \)? Wait, maybe I misread. Wait, the second number line: \( E \) at -6, \( F \) at -4, \( G \) at 0, \( H \) at 4, \( J \) at 6, \( K \) at 10, \( L \) at 12? Wait, no, the labels: \( E, F, G, H, J, K, L \) with coordinates -6, -4, 0, 2, 4, 6, 8? Wait, maybe the original number line is: \( E(-6), F(-4), G(0), H(2), J(4), K(6), L(8) \)? Wait, the user's image: the second number line has marks at -6, -4, -2, 0, 2, 4, 6, 8, 10. So \( E \) at -6, \( F \) at -4, \( G \) at 0, \( H \) at 2, \( J \) at 4, \( K \) at 6, \( L \) at 8? Wait, maybe I made a mistake. Let's recheck. The second number line: "E F G H J K L" with coordinates -6, -4, 0, 2, 4, 6, 8? Wait, the user's image: the second number line has: E at -6, F at -4, G at 0, H at 2, J at 4, K at 6, L at 8? Wait, the problem 7 is \( \overline{FK} \): F is at -4, K is at 6. Then midpoint is \( \frac{-4 + 6}{2} = 1 \). Wait, maybe my initial reading was wrong. Let's correct:

Looking at the second number line (the one with E, F, G, H, J, K, L):

  • E: -6
  • F: -4
  • G: 0
  • H: 2
  • J: 4
  • K: 6
  • L: 8

So problem 7: \( \overline{FK} \): F(-4), K(6). Midpoint: \( \frac{-4 + 6}{2} = 1 \)

Problem 8: \( \overline{HK} \): H(2), K(6). Midpoint: \( \frac{2 + 6}{2} = 4 \)

Problem 9: \( \overline{EF} \): E(-6), F(-4). Midpoint: \( \frac{-6 + (-4)}{2} = -5 \)

Problem 10: \( \overline{FG} \): F(-4), G(0). Midpoint: \( \frac{-4 + 0}{2} = -2 \)

Problem 11: \( \overline{JL} \): J(4), L(8). Midpoint: \( \frac{4 + 8}{2} = 6 \)

Problem 12: \( \overline{EL} \): E(-6), L(8). Midpoint: \( \frac{-6 + 8}{2} = 1 \)

Third Number Line (A, B, C, D, E):
  • A: -4
  • B: -2
  • C: 3
  • D: 7
  • E: 11

Problem 13: \( \overline{DE} \): D(7), E(11). Midpoint: \( \frac{7 + 11}{2} = 9 \)

Problem 14: \( \overline{BC} \): B(-2), C(3). Midpoint: \( \frac{-2 + 3}{2} = 0.5 \)

Problem 15: \( \overline{BD} \): B(-2), D(7). Midpoint: \( \frac{-2 + 7}{2} = 2.5 \)

Problem 16: \( \overline{AD} \): A(-4), D(7). Midpoint: \( \frac{-4 + 7}{2} = 1.5 \)

Problem 17: Home Improvement

Callie's house and neighbor's house: distance between them is 20 yd? Wait, the diagram shows 20 yd from Callie's house to neighbor's. Wait, the problem says "build a fence halfway between her house and her neighbor’s house". So midpoint. If Callie's house is at 0 and neighbor's at 20, midpoint is \( \frac{0 + 20}{2} = 10 \) yd from Callie's house.

Problem 18: Calvino's Home

Calvino’s home is the midpoint between Fast Pizza and Pizza Now. Fast Pizza is a quarter mile from Calvino’s home. So distance from Calvino’s to Pizza Now is also a quarter mile (since midpoint). Distance between the two pizzerias: \( \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \) mile.

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Let's summarize the answers:

  1. \( \overline{KM} \): \(-2\)
  2. \( \overline{JP} \): \(-0.5\)
  3. \( \overline{LN} \): \(1\)
  4. \( \overline{MP} \): \(3.5\)
  5. \( \overline{LP} \): \(2\)
  6. \( \overline{JN} \): \(-1.5\)
  7. \( \overline{FK} \): \(1\) (corrected)
  8. \( \overline{HK} \): \(4\) (corrected)
  9. \( \overline{EF} \): \(-5\)
  10. \( \overline{FG} \): \(-2\)
  11. \( \overline{JL} \): \(6\) (corrected)
  12. \( \overline{EL} \): \(1\) (corrected)
  13. \( \overline{DE} \): \(9\)
  14. \( \overline{BC} \): \(0.5\)
  15. \( \overline{BD} \): \(2.5\)
  16. \( \overline{AD} \): \(1.5\)
  17. Fence distance: \(10\) yd
  18. Pizza Now distance: \(\frac{1}{4}\) mile; Pizzerias distance: \(\frac{1}{2}\) mile

(Note: Some answers were corrected based on reinterpreting the number line coordinates. If the initial number line coordinates were different, adjust accordingly.)