How find End point of line with help of length & Angle

[sathyboopathiraja]

Dear All,

I know the start point_x ,start point_y , length , angle from this how can i get the end point of the line.

Is there any way to find end point of the line? If yes means please guide me!......

With Regards,
Boopathiraja.N.

Thanks in advance to all!.

[toraj58]

it is very easy:

tan(theta) = (y2-y1) / (x2-x1)

and

L = sqrt((y2-y1)^2 - (x2-x1)^2)

solve these two equations in respect to y2 and x2 and find them.

inorder to code the solution for n-equation with n-variables i suggest that you use Cramer's rule

see here:

http://en.wikipedia.org/wiki/Cramer's_rule

[sathyboopathiraja]

Dear toraj,

i am not familiar on mathmatics so please can you give any solved equation. I am not familiar on the maths, thats why i am asking once again help.

I hope once again you will guide me!...

With regards,
Boopathiraja.N

[toraj58]

would you please wait till tomorrow.
i will write the code for you tomorrow.

if i had scanner here then i would scratch the steps for you with figures to help you to understand it.

but if i write the code for you; you may not understand it.

would you please tell me how much do you know math(algebra and geometric and trigonometry) then i will the explain obscure parts for you. if you don't understand it at all.....i should paint figures (maybe in paint brush) to demstrate it to you.

so please wait till tomorrow and till then see here:

http://images.google.com/images?sour...N&hl=en&tab=wi

and

http://en.wikipedia.org/wiki/Trigonometry

and

http://images.google.com/images?um=1...eta%29%3Dy%2Fx

[sathyboopathiraja]

Dear toraj,

I like your attitude & your spirit and also really say lot of thanks for ur help. but i am waiting for your valuable information. once again thanks mr.toraj for your effort.

With regards,
Boopathiraja.N.

[toraj58]

hi my friend,

i contemplated more and found a simpler approach that don't need to solve 2-equations/ 2-variables.

it is like this:

assume that:
L = Length of line
(x1, y1) = start point
(x2, y2) = end point
theta = angle

then:

sin(theta) = (y2 - y1) / L
so: y2 = [L * sin(theta)] + y1

and

cos(theta) = (x2 - x1) / L
so: x2 = [L * cos(theta)] + x1

so it is very easy to code this one.

[sathyboopathiraja]

Dear Mr.Toraj,

Mine sincere thanks for ur effort mr.toraj. it is working as too fine.

[Personal Info About you]

i don't know now you are in which position, but one day definetly you will achieve your goal fro this no doubt because you are having the good attitude. god will always help to you, i really pray the god to shine your future as bright.

if i say anything as wrong really sorry.

Finally this equation working as too fine.

With Regards,
Boopathiraja.

[All prizes to God]

[toraj58]

you are good man.
it is my pleasure to help you.

thanks.

[rkrishnan_g]

you are good man.
it is my pleasure to help you.

thanks.

Mr. Toraj and Mr. Boopathiraja,

Both of yours conversation was very good. Actually i was a techniq to find the same thing which you both discussed and now i am noting the formula given by Toraj.

Toraj: I am also thanking you for the formula which you posted in this forum. You are really expert in Maths having very Good Attitude and postive approach as Boopathiraja wrote about you.

Boopathiraja: By your posting only i got to know this techniq from this forum. So Thank You. And I can really find you are Thankfull man. After getting the techniq from Mr. Toraj you didnt finish up in one word of thanks but you wrote your real Wishes and Bless for him. Very Nice !!

And Finally i thank you both for your time and efforts.

Wish you both Good Time ahead !!

Radhakrishnan.