Distance between Two Points.. working out troubles

[Finalshine]

Im doing distance between two points as its set out

 

 

----------------------------------

/ 2 2

d = / (x2-x1) + (y2-y1) =

 

 

 

and so i write the equation im doing out and follow the set rules to working it out....

 

 

x1,y1 x2,y2

( 4 , 5 ) ( 1 , 1 )

 

 

---------------------------------------

/ 2 2

d= / (1-4) + (1-5) =

 

 

 

 

 

Now my text Book says the Answer is 5.. but i don't no how im getting 31.1.. im following what im meant to be doing but still getting it wrong somehow..

 

any help would be great..

 

 

 

[Unity+]

Im doing distance between two points as its set out

 

 

----------------------------------

/ 2 2

d = / (x2-x1) + (y2-y1) =

 

 

 

and so i write the equation im doing out and follow the set rules to working it out....

 

 

x1,y1 x2,y2

( 4 , 5 ) ( 1 , 1 )

 

 

---------------------------------------

/ 2 2

d= / (1-4) + (1-5) =

 

 

 

 

 

Now my text Book says the Answer is 5.. but i don't no how im getting 31.1.. im following what im meant to be doing but still getting it wrong somehow..

 

any help would be great..

 

 

 

The equation for finding the distance between two points is .

From the coordinates, I also get 5. It is hard to tell how you got you answer because of the way you laid it out. Could you use latex?

Edited March 10, 2014 by Unity+

[Spyman]

First off, this looks like homework and if so should have been placed in the Homework Help section.

 

Secondly, are you saying that you have two coordinates (4,5) and (1,1) where you want to calculate the distance between them?

 

If you draw this out in a coordinate system you will find that you get a right triangle and can use the Pythagorean theorem.

 

It looks like you have the equation ok but make a mistake somewhere in your calculation.

 

[math]d=\sqrt{(1-4)^{2}+(1-5)^{2}}=\sqrt{(-3)^{2}+(-4)^{2}}=\sqrt{(9)+(16)}=\sqrt{25}=5[/math]

[Finalshine]

i figured out where i have gone wrong, with you showing me how to do it then, somehow i was adding and not subtracting as my teacher has said anything that is showed is the opposite to what you are working out.. like add is to subtract, and multiply is to divide an back and forth but got this order mixed up..

[imatfaal]

Not sure about how your teacher phrased it but I think it would be worth you working through so you understand it completely.

 

Think of it as distances moved from the origin (0,0) in the x-axis (along the page) and y-axis (up and down the page). (4,5) is 4 along and 5 up and (1,1) is 1 along and 1 up. To get to (4,5) from (1,1) you need to walk another 3 in direction the x-axis and another 4 in the direction of the y-axis - that is to say in math terms change in x = x_2 - x_1 and change in y = y_2-y_1.

 

Draw a nice coordinate system (with positive and negative x and y) on a piece of graph/squared paper, plot pairs of points and work out what the change in x and change in y must be.

 

For starters work out the change in x and change in y for moving from (4,5) to (1,1) - you already know the distance so you can check you have it right and that you know how to use the pythagoras formula.

[pzkpfw]

... and the beauty of squaring in this stuff is that it doesn't matter if the difference between the two x's works out as positive or negative (and the same for the difference between the y's). All you need is the difference, to get the sides of the triangle.

 

That is, you could look at it as being from (1,1) to (4,5) or (4,5) to (1,1) - and get the same answer (in terms of distance, if not direction).

 

That is (1-4)^2 = (-3)^2 = 9

Also (4-1)^2 = 3^2 = 9

 

And (1-5)^2 = (-4)^2 = 16

Also (5-1)^2 = 4^2 = 16

 

Or - if you look at a rectangle with sides of length 4 and 3, it doesn't matter which diagonal you look at - both diagonals are the same length.

 

And finally: the 3-4-5 triangle is a common thing, as it's so "nice" that the lengths "square and square root" to simple integers. That triangle is used in building as a simple way to find a right angle. Say you need to go out 90 degrees from a point on a wall. Get another point 3 metres along the wall. Measure 5 metres from that point and 4 metres from the first point - and you easily make a right-angle triangle.

 

 

(I find with math it helps to look at something from multiple directions, and see that they all make sense together.)

Edited March 10, 2014 by pzkpfw