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Distance between two coordinates

distance between two landmarks..Find distance between two stations given travel times of trainsDistance between two 3D vectorsDistance between two 3D linesFind the distance between two pointsStraight distance between two townsFinding x-coordinate of a parabola's vertex given the y-coordinates of two points and the distance between themDistance Between Two AirplanesDistance between two stationsMinimum distance between two boats travelling in different velocities

1

0

$begingroup$

I'm working through the exercises on online cnx book Algebra and Trigonometry here.

If the coordinates on a map for San Francisco are $(53,17)$ and the coordinates of San Jose are $(76, -12)$, find the distance between the two cities.

Throughout this chapter I have been learning about Cartesian coordinate system, distance formula with Pythagoras' theorem and the mid point formula.

The solution provided is $37$.

With what I have learned in the chapter I cannot arrive at $37$. If there were $3$ points and the distance was a hypotenuse I could use the Pythagoras formula. But there are only two pairs of coordinates provided.

Given I know that San Francisco is north of San Jose, I tried calculating the distance between the $Y$ coordinates:

distance = $y_2 - y_1 = 17- -12 = 29$.

How can I arrive at $37$?

Edit. Adding my sketch to demonstrate my confusion:

I cannot see how the Pythagoras theorem applies since I do not know the length of sides A and B?

algebra-precalculus

share|cite|improve this question

edited Mar 14 at 16:22

Doug Fir

asked Mar 14 at 16:00

Doug FirDoug Fir

4188

$endgroup$

add a comment | 

1

0

$begingroup$

I'm working through the exercises on online cnx book Algebra and Trigonometry here.

If the coordinates on a map for San Francisco are $(53,17)$ and the coordinates of San Jose are $(76, -12)$, find the distance between the two cities.

Throughout this chapter I have been learning about Cartesian coordinate system, distance formula with Pythagoras' theorem and the mid point formula.

The solution provided is $37$.

With what I have learned in the chapter I cannot arrive at $37$. If there were $3$ points and the distance was a hypotenuse I could use the Pythagoras formula. But there are only two pairs of coordinates provided.

Given I know that San Francisco is north of San Jose, I tried calculating the distance between the $Y$ coordinates:

distance = $y_2 - y_1 = 17- -12 = 29$.

How can I arrive at $37$?

Edit. Adding my sketch to demonstrate my confusion:

I cannot see how the Pythagoras theorem applies since I do not know the length of sides A and B?

algebra-precalculus

share|cite|improve this question

edited Mar 14 at 16:22

Doug Fir

asked Mar 14 at 16:00

Doug FirDoug Fir

4188

$endgroup$

add a comment | 

$begingroup$

I'm working through the exercises on online cnx book Algebra and Trigonometry here.

If the coordinates on a map for San Francisco are $(53,17)$ and the coordinates of San Jose are $(76, -12)$, find the distance between the two cities.

Throughout this chapter I have been learning about Cartesian coordinate system, distance formula with Pythagoras' theorem and the mid point formula.

The solution provided is $37$.

With what I have learned in the chapter I cannot arrive at $37$. If there were $3$ points and the distance was a hypotenuse I could use the Pythagoras formula. But there are only two pairs of coordinates provided.

Given I know that San Francisco is north of San Jose, I tried calculating the distance between the $Y$ coordinates:

distance = $y_2 - y_1 = 17- -12 = 29$.

How can I arrive at $37$?

Edit. Adding my sketch to demonstrate my confusion:

I cannot see how the Pythagoras theorem applies since I do not know the length of sides A and B?

algebra-precalculus

share|cite|improve this question

edited Mar 14 at 16:22

Doug Fir

asked Mar 14 at 16:00

Doug FirDoug Fir

4188

$endgroup$

share|cite|improve this question

edited Mar 14 at 16:22

Doug Fir

asked Mar 14 at 16:00

Doug FirDoug Fir

4188

share|cite|improve this question

edited Mar 14 at 16:22

Doug Fir

asked Mar 14 at 16:00

Doug FirDoug Fir

4188

asked Mar 14 at 16:00

Doug FirDoug Fir

4188

asked Mar 14 at 16:00

Doug FirDoug Fir

4188

3 Answers
3

active

oldest

votes

1

$begingroup$

The distance is $sqrt23^2+29^2 approx 37$ according to Pythagoras, the $x$-distance is $23$ and $y$-distance is $29$ as you concluded. So did you use Pythagoras correctly?

share|cite|improve this answer

answered Mar 14 at 16:05

MaxMax

9071318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm confused because I drew a right angled triangle on paper. One end of the hypotenuse is San Francisco, San Jose at the other. To use the Pythagoras formula I thought that I needed to have the distances of both the smaller sides? I can see you using the theorem and arriving at the correct answer, but how would this look if I was drawing the points on a triangle on paper?
  $endgroup$
  – Doug Fir
  Mar 14 at 16:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DougFir The hypotenuse is always opposite to the right angle ($90$ degrees), did you count for this in your picture? Btw, you did calculate one of the smaller distances, for $y$, correctly. Now do this again for $x$. Then 'plug' it in Pythagoras.
  $endgroup$
  – Max
  Mar 14 at 16:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Added a sketch to my post to show where I'm confused. How can I calculate the distance of the hypotenuse if I don't know A and B?
  $endgroup$
  – Doug Fir
  Mar 14 at 16:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @DougFir You do know A and B; A is the $x$-distance and B is the $y$-distance, as you calculated. Does this help you reach the final conclusion?
  $endgroup$
  – Max
  Mar 14 at 16:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, it clicked when you put it that way. Thank you!
  $endgroup$
  – Doug Fir
  Mar 14 at 16:47
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

The distance $d$ between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $d = sqrt(x_2-x_1)^2 + (y_2-y_1)^2 $. Can you apply this?

(The formula is a simple consequence of the Pythagorean theorem).

share|cite|improve this answer

answered Mar 14 at 16:05

DeepakDeepak

17.5k11539

$endgroup$

add a comment | 

1

$begingroup$

Find the distances between the two cities' x and y coordinates and apply the Pythagorean theorem:

$$
Delta x=76-53=23\
Delta y=17+12=29\
Distance=sqrtDelta x^2+Delta y^2=sqrt23^2+29^2approx 37 length units
$$

You could use something called the absolute value to find the distances, but in this case it is not really necessary as you could easily see that the distance between the x points $53$ and $76$ is equal to $76-53$ and the distance between the y points $17$ and $-12$ is equal to $17 + 12$.

share|cite|improve this answer

edited Mar 14 at 16:47

answered Mar 14 at 16:42

Michael RybkinMichael Rybkin

3,939421

$endgroup$

add a comment | 

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1

$begingroup$

The distance is $sqrt23^2+29^2 approx 37$ according to Pythagoras, the $x$-distance is $23$ and $y$-distance is $29$ as you concluded. So did you use Pythagoras correctly?

share|cite|improve this answer

answered Mar 14 at 16:05

MaxMax

9071318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm confused because I drew a right angled triangle on paper. One end of the hypotenuse is San Francisco, San Jose at the other. To use the Pythagoras formula I thought that I needed to have the distances of both the smaller sides? I can see you using the theorem and arriving at the correct answer, but how would this look if I was drawing the points on a triangle on paper?
  $endgroup$
  – Doug Fir
  Mar 14 at 16:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DougFir The hypotenuse is always opposite to the right angle ($90$ degrees), did you count for this in your picture? Btw, you did calculate one of the smaller distances, for $y$, correctly. Now do this again for $x$. Then 'plug' it in Pythagoras.
  $endgroup$
  – Max
  Mar 14 at 16:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Added a sketch to my post to show where I'm confused. How can I calculate the distance of the hypotenuse if I don't know A and B?
  $endgroup$
  – Doug Fir
  Mar 14 at 16:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @DougFir You do know A and B; A is the $x$-distance and B is the $y$-distance, as you calculated. Does this help you reach the final conclusion?
  $endgroup$
  – Max
  Mar 14 at 16:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, it clicked when you put it that way. Thank you!
  $endgroup$
  – Doug Fir
  Mar 14 at 16:47
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

The distance is $sqrt23^2+29^2 approx 37$ according to Pythagoras, the $x$-distance is $23$ and $y$-distance is $29$ as you concluded. So did you use Pythagoras correctly?

share|cite|improve this answer

answered Mar 14 at 16:05

MaxMax

9071318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm confused because I drew a right angled triangle on paper. One end of the hypotenuse is San Francisco, San Jose at the other. To use the Pythagoras formula I thought that I needed to have the distances of both the smaller sides? I can see you using the theorem and arriving at the correct answer, but how would this look if I was drawing the points on a triangle on paper?
  $endgroup$
  – Doug Fir
  Mar 14 at 16:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DougFir The hypotenuse is always opposite to the right angle ($90$ degrees), did you count for this in your picture? Btw, you did calculate one of the smaller distances, for $y$, correctly. Now do this again for $x$. Then 'plug' it in Pythagoras.
  $endgroup$
  – Max
  Mar 14 at 16:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Added a sketch to my post to show where I'm confused. How can I calculate the distance of the hypotenuse if I don't know A and B?
  $endgroup$
  – Doug Fir
  Mar 14 at 16:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @DougFir You do know A and B; A is the $x$-distance and B is the $y$-distance, as you calculated. Does this help you reach the final conclusion?
  $endgroup$
  – Max
  Mar 14 at 16:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, it clicked when you put it that way. Thank you!
  $endgroup$
  – Doug Fir
  Mar 14 at 16:47
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

The distance is $sqrt23^2+29^2 approx 37$ according to Pythagoras, the $x$-distance is $23$ and $y$-distance is $29$ as you concluded. So did you use Pythagoras correctly?

share|cite|improve this answer

answered Mar 14 at 16:05

MaxMax

9071318

$endgroup$

share|cite|improve this answer

answered Mar 14 at 16:05

MaxMax

9071318

answered Mar 14 at 16:05

MaxMax

9071318

answered Mar 14 at 16:05

MaxMax

9071318

1

$begingroup$

The distance $d$ between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $d = sqrt(x_2-x_1)^2 + (y_2-y_1)^2 $. Can you apply this?

(The formula is a simple consequence of the Pythagorean theorem).

share|cite|improve this answer

answered Mar 14 at 16:05

DeepakDeepak

17.5k11539

$endgroup$

add a comment | 

1

$begingroup$

The distance $d$ between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $d = sqrt(x_2-x_1)^2 + (y_2-y_1)^2 $. Can you apply this?

(The formula is a simple consequence of the Pythagorean theorem).

share|cite|improve this answer

answered Mar 14 at 16:05

DeepakDeepak

17.5k11539

$endgroup$

add a comment | 

$begingroup$

The distance $d$ between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $d = sqrt(x_2-x_1)^2 + (y_2-y_1)^2 $. Can you apply this?

(The formula is a simple consequence of the Pythagorean theorem).

share|cite|improve this answer

answered Mar 14 at 16:05

DeepakDeepak

17.5k11539

$endgroup$

share|cite|improve this answer

answered Mar 14 at 16:05

DeepakDeepak

17.5k11539

answered Mar 14 at 16:05

DeepakDeepak

17.5k11539

answered Mar 14 at 16:05

DeepakDeepak

17.5k11539

1

$begingroup$

Find the distances between the two cities' x and y coordinates and apply the Pythagorean theorem:

$$
Delta x=76-53=23\
Delta y=17+12=29\
Distance=sqrtDelta x^2+Delta y^2=sqrt23^2+29^2approx 37 length units
$$

You could use something called the absolute value to find the distances, but in this case it is not really necessary as you could easily see that the distance between the x points $53$ and $76$ is equal to $76-53$ and the distance between the y points $17$ and $-12$ is equal to $17 + 12$.

share|cite|improve this answer

edited Mar 14 at 16:47

answered Mar 14 at 16:42

Michael RybkinMichael Rybkin

3,939421

$endgroup$

add a comment | 

1

$begingroup$

Find the distances between the two cities' x and y coordinates and apply the Pythagorean theorem:

$$
Delta x=76-53=23\
Delta y=17+12=29\
Distance=sqrtDelta x^2+Delta y^2=sqrt23^2+29^2approx 37 length units
$$

You could use something called the absolute value to find the distances, but in this case it is not really necessary as you could easily see that the distance between the x points $53$ and $76$ is equal to $76-53$ and the distance between the y points $17$ and $-12$ is equal to $17 + 12$.

share|cite|improve this answer

edited Mar 14 at 16:47

answered Mar 14 at 16:42

Michael RybkinMichael Rybkin

3,939421

$endgroup$

add a comment | 

$begingroup$

Find the distances between the two cities' x and y coordinates and apply the Pythagorean theorem:

$$
Delta x=76-53=23\
Delta y=17+12=29\
Distance=sqrtDelta x^2+Delta y^2=sqrt23^2+29^2approx 37 length units
$$

You could use something called the absolute value to find the distances, but in this case it is not really necessary as you could easily see that the distance between the x points $53$ and $76$ is equal to $76-53$ and the distance between the y points $17$ and $-12$ is equal to $17 + 12$.

share|cite|improve this answer

edited Mar 14 at 16:47

answered Mar 14 at 16:42

Michael RybkinMichael Rybkin

3,939421

$endgroup$

share|cite|improve this answer

edited Mar 14 at 16:47

answered Mar 14 at 16:42

Michael RybkinMichael Rybkin

3,939421

answered Mar 14 at 16:42

Michael RybkinMichael Rybkin

3,939421

answered Mar 14 at 16:42

Michael RybkinMichael Rybkin

3,939421