The Distance Between Vertices

  • Thread starter LLS
  • Start date
I understand now. In summary, the conversation discusses a problem involving finding the distance between the midpoint of two vertices and a third vertex in a triangle. The correct calculation is done using the distance formula, resulting in an answer of √26, which can be left in surd form or approximated to a certain degree of accuracy.
  • #1
LLS
40
0
[SOLVED] The Distance Between Vertices

Homework Statement



A triangle has vertices at points A, B and C, which are located at (1,0), (-3,0), and (0,5) respectively. What is the distance from the midpoint of AB to point C?

Homework Equations




The Attempt at a Solution



I plotted the triangle. Side AC = 5

The midpoint of AB is at (-1,0)

The distance from the midpoint of AB to C = 5. It's the same as AC.

I think that I am missing something.
 
Physics news on Phys.org
  • #2
LLS said:
Side AC = 5
No it isn't.
LLS said:
The midpoint of AB is at (-1,0)
Correct :approve:
LLS said:
The distance from the midpoint of AB to C = 5
No it isn't.
LLS said:
It's the same as AC.
Correct :approve:
 
  • #3
You correctly chose that the midpoint of AB is (-1, 0); but AC is not equal to 5.

Point C is at (0, 5). Use the distance formula to find the length from point C (-1, 0) to the point (0, 5).
 
  • #4
symbolipoint said:
Use the distance formula to find the length from point C (-1, 0) to the point (0, 5).
I'm sure symbolipoint means,

Use the distance formula to find the length from the midpoint of AB (-1, 0) to the point C (0, 5).
 
  • #5
symbolipoint said:
You correctly chose that the midpoint of AB is (-1, 0); but AC is not equal to 5.

Point C is at (0, 5). Use the distance formula to find the length from point C (-1, 0) to the point (0, 5).

c^2 = (5-0)^2 + (1-0)^2

c^2 = 25 + 1 = 26

c = 5.097

I still get that the distance = 5.

What did I do wrong?
 
  • #6
LLS said:
c^2 = (5-0)^2 + (1-0)^2

c^2 = 25 + 1 = 26

c = 5.097

I still get that the distance = 5.

What did I do wrong?

How is [itex]\sqrt{26} = 5.097... =5[/itex] ?

Now you're just making things up.
 
  • #7
Hootenanny said:
How is [itex]\sqrt{26} = 5.097... =5[/itex] ?

Now you're just making things up.

It's what the calculator indicated:

5.097019514 to be exact

I'm lost. Did I do the distance formula calc correctly?
 
  • #8
LLS said:
It's what the calculator indicated:

5.097019514 to be exact

I'm lost.

All I'm say is that 5.097019514... is NOT the same as 5! Therefore, you distance from the midpoint of AB to C is NOT 5.
 
  • #9
Hootenanny said:
All I'm say is that 5.097019514... is NOT the same as 5! Therefore, you distance from the midpoint of AB to C is NOT 5.

I rounded it. I didn't round up to 5.1. I rounded down to 5.

Is 5.097019514 the correct answer?

Have I done any other calcs correctly?

Please steer me towards a calc that works. I don't want to give up.
 
  • #10
its best if you do not plug in numbers in until the very end. for instance leave sqrt(5) as sqrt(5) and do not put it down as 5.097...
 
  • #11
LLS said:
I rounded it. I didn't round up to 5.1. I rounded down to 5.

Is 5.097019514 the correct answer?

Have I done any other calcs correctly?

Please steer me towards a calc that works. I don't want to give up.
Your calculation is correct, your final answer is not. If you are rounding then you should explicitly state the degree of accuracy to which you have quoted the answer. However an exact answer is always preferable, you should leave it in surd form unless instructed otherwise.
 
  • #12
Hootenanny said:
Your calculation is correct, your final answer is not. If you are rounding then you should explicitly state the degree of accuracy to which you have quoted the answer. However an exact answer is always preferable, you should leave it in surd form unless instructed otherwise.

Was it the correct calculation? I may have a correct calculation but if the formula was wrong or the values the answer is wrong.

I'm really confused as to what the answer is.
 
  • #13
LLS said:
Was it the correct calculation? I may have a correct calculation but if the formula was wrong or the values the answer is wrong.

I'm really confused as to what the answer is.
The formula is correct and the answer is [itex]\sqrt{26}[/itex] as you calculated, there is no need to write it in decimal form (unless stated otherwise). An answer in this form is exact, as opposed to an approximated decimal.

However, if you do wish to write it in decimal form, you must state explicitly the degree of accuracy which you are rounding it to. Simply writing AB=5 is incorrect, but writing AB=5 (1sf) is acceptable.
 
  • #14
Hootenanny said:
The formula is correct and the answer is [itex]\sqrt{26}[/itex] as you calculated, there is no need to write it in decimal form (unless stated otherwise). An answer in this form is exact, as opposed to an approximated decimal.

However, if you do wish to write it in decimal form, you must state explicitly the degree of accuracy which you are rounding it to. Simply writing AB=5 is incorrect, but writing AB=5 (1sf) is acceptable.

Thank you
 

What is "The Distance Between Vertices"?

"The Distance Between Vertices" is a concept in mathematics and computer science that refers to the measure of the shortest path or the number of edges between two vertices in a graph or network.

How is "The Distance Between Vertices" calculated?

The distance between vertices is calculated using algorithms such as Dijkstra's algorithm, which finds the shortest path between two vertices in a graph by considering the weight of each edge. Other algorithms, such as Floyd-Warshall or Bellman-Ford, can also be used depending on the type of graph and the desired outcome.

What is the significance of "The Distance Between Vertices" in real-world applications?

"The Distance Between Vertices" is important in various real-world applications such as navigation systems, social networks, transportation networks, and internet routing. It helps in determining the most efficient or optimal path between two points in a network.

Can the distance between vertices be negative?

No, the distance between vertices cannot be negative. It is always a positive value that represents the number of edges between two vertices in a graph.

Is "The Distance Between Vertices" the same as the length of a path?

No, the distance between vertices is not always the same as the length of a path. The distance between vertices refers to the minimum number of edges between two vertices, while the length of a path is the sum of the weights of all the edges in that path. In some cases, the shortest path may not be the path with the lowest total weight.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
4
Views
3K
  • Precalculus Mathematics Homework Help
Replies
4
Views
920
  • Precalculus Mathematics Homework Help
Replies
5
Views
2K
  • Precalculus Mathematics Homework Help
Replies
2
Views
924
  • Precalculus Mathematics Homework Help
Replies
16
Views
2K
  • Precalculus Mathematics Homework Help
Replies
11
Views
2K
Replies
1
Views
838
  • Introductory Physics Homework Help
Replies
7
Views
256
  • Precalculus Mathematics Homework Help
Replies
8
Views
2K
  • Precalculus Mathematics Homework Help
Replies
20
Views
2K
Back
Top