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Math help (line segment)


bort
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Hello does anyone know how to find the end point of a line segment if you know the starting X,Y coordinate and the distance along the line?

 

Here is a picture to describe. I know the starting (A) and ending (B) points of the overall line segment. Now I want to find the new end point of a "sub" line segment that starts at A and follows the same direction for a specified distance.

 

If you can help I would appreciate it very much.

 

math_help.jpg

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Given that you have two coordinates (2,5) and (8,2)

we assign the values y2 and y1 to the y values

and same for x values.

y2 = 2

y1 = 5

x2 = 8

x1 = 2

 

 

Apply the slope formula:

 

(y2-y1) / (x2-x1 ) = slope(m)

 

(2 - 5)/(8 - 2)

(-3) / (6)

 

-3 over 6, slope of -0.5

 

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

 

Now we want our new coordinates, we know that the x value in the new point is 4 (initial distance is 2 already, so add 2 more)

 

so our new coordiantes are (4,y2) and (2,5)

 

so we substitute these values back in, we know the slope is -0.5

 

so we setup our formula this way, since we know slope (we can replace our slope variable with -0.5)

 

-0.5 = (y2 - 5) / (4 - 2)

 

simplify

 

-.5 = (y2 - 5) / (2)

 

multiply both sides by two to nullify 2

 

-1 = (y2 - 5)

 

add five to nullify the (-5) on the right side)

 

-1 + 5 = 4

 

From this answer, we can verify that the y2 coordinate is 4

 

to verify the slope and thus verify the coordinate

 

(4 - 5) / (4 - 2)

 

simplify

 

(-1)/(2)

 

= -.5

 

therefore your coordinate is (4,4)

Edited by enigma#
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Wow... I was gonna suggest using the given points to find the angle that the line makes with the X-axis and using that angle and the subsequent value of sine and cosine to find the proper x and y values for a given hypotenuse of 2.

 

...

 

Basic math is too basic now. DX ffffffffuuuuuuuuuuuu Physics

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Wow... I was gonna suggest using the given points to find the angle that the line makes with the X-axis and using that angle and the subsequent value of sine and cosine to find the proper x and y values for a given hypotenuse of 2.

 

...

 

Basic math is too basic now. DX ffffffffuuuuuuuuuuuu Physics

 

O.o lol

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Given that you have two coordinates (2,5) and (8,2)

we assign the values y2 and y1 to the y values

and same for x values.

y2 = 2

y1 = 5

x2 = 8

x1 = 2

 

 

Apply the slope formula:

 

(y2-y1) / (x2-x1 ) = slope(m)

 

(2 - 5)/(8 - 2)

(-3) / (6)

 

-3 over 6, slope of -0.5

 

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

 

Now we want our new coordinates, we know that the x value in the new point is 4 (initial distance is 2 already, so add 2 more)

 

so our new coordiantes are (4,y2) and (2,5)

 

so we substitute these values back in, we know the slope is -0.5

 

so we setup our formula this way, since we know slope (we can replace our slope variable with -0.5)

 

-0.5 = (y2 - 5) / (4 - 2)

 

simplify

 

-.5 = (y2 - 5) / (2)

 

multiply both sides by two to nullify 2

 

-1 = (y2 - 5)

 

add five to nullify the (-5) on the right side)

 

-1 + 5 = 4

 

From this answer, we can verify that the y2 coordinate is 4

 

to verify the slope and thus verify the coordinate

 

(4 - 5) / (4 - 2)

 

simplify

 

(-1)/(2)

 

= -.5

 

therefore your coordinate is (4,4)

Yeah this, I knew that

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OK so, i think i'm on to something... Since I know the two points A and B, I can pretend that AB is the hypotenuse of a big triangle that is 3 units high and 6 units long.

 

I can also get the length of the hypotenuse AB by doing that silly distance formula distance-formula4.jpg.

 

THEN, since the distance from point A to the mystery point is known (i said it was 2 in the picture), I can basically scale the big triangle down proportionately. Example: If the big hypotenuse was SqrRt(45) or 6.7082039325, then the little hypotenuse of 2 is 3.35410196625 times smaller. So make the legs of 3 and 6 become 3.35410196625 times smaller as well.

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