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Q: Determing coordinate points of triangles ( Answered,   9 Comments )
Question  
Subject: Determing coordinate points of triangles
Category: Science > Math
Asked by: groves222-ga
List Price: $10.00
Posted: 27 May 2006 19:57 PDT
Expires: 26 Jun 2006 19:57 PDT
Question ID: 732965
I need the steps involved to determine the 3 coordinate points (x,y)
of a triangle as if it were plotted on a graph.

For example, triangle ABC has the following given information:

Point A is positioned at (1,1)
Point B at (5,2)
Angle BAC is 25 degrees
Angle ACB is 90 degrees

What is the coordinate of point C?

Please see this diagram for a better illustration:
http://www.swiftwx.com/triangle.gif

*** Please provide detailed steps so i can reproduce this in a
programming language. I am not a math major, i need specific steps
here.
Answer  
Subject: Re: Determing coordinate points of triangles
Answered By: maniac-ga on 06 Jun 2006 18:43 PDT
 
Hello Groves222,

There are several good references explaining the concepts as well as
source code that solves parts of the problems in different ways.
Perhaps the simplest is to solve the equations in slope / intercept
form.
NOTE: you have to check for + / - infinity in the slope calculations
and handle the results appropriately.   I have a note at the end to
address at least one work around for that condition.

First, let's walk through the known information and start deriving
some factors helpful in the calculation of the intercept.

A is at (1,1)
B is at (5,2)

The line A-B can be represented in slope / intercept form as
  Y = M * X + B
when computed as described at
http://www-static.cc.gatech.edu/gvu/multimedia/nsfmmedia/graphics/elabor/math/mathfaq_lines.html
where it starts by computing the slope (M)
  M = (By - Ay) / (Bx - Ax) = (2-1)/(5-1) = 0.25
and generates the intercept (B) by
  B = Ay - (Ax * M) = 1 - (1*0.25) = 0.75
or line A-B in slope / intercept form is
  Y = 0.25 * X + 0.75

The line A-C has a slope (M) of
  M = 0.25 + sin(25 degrees) = 0.672618262
and has intercept (B) of
  B = 1 - (1*0.672618262) - 0.327381738
or line A-C in slope / intercept form is
  Y = 0.672618262 * X + 0.327381738

The line B-C which is perpendicular to A-B has slope (M) of
  M[BC] = -(1 / M[AB]) = -(1/0.25) = -4
computing the intercept (B) as before, we get
  B = 2 - (5 * (-4)) = 22
or line B-C in slope / intercept form is
  Y = (-4) * X + 22

You can compute point C from this information using a formula
  Cx = (B[BC]-B[AC])/(M[BC]-M[AC])
  Cx = (22 - 0.327381738) / (-4 - 0.672618262)
  Cx = 4.638217172
and
  Cy = M[AC]*Cx + B[AC]
  Cy = 0.672618262 * 4.638217172 + 0.327381738
  Cy = 3.44713131

When you have a line with slope (M) of inifinity, there is no
intercept (B) for that line. For example, when B-C is vertical, you
can simply solve for point C using the known X using a simple
adjustment to the intercept method. There is a similar fix needed if
A-B is vertical.

Additional explanations of this type of solution is at
http://id.mind.net/~zona/mmts/intersections/intersectionOfTwoLines1/intersectionOfTwoLines1.html
and with some additional links and "source code" at
  http://www.pdas.com/lineint.htm
or another explanation and source code at
  http://www.geog.ubc.ca/courses/klink/gis.notes/ncgia/u32.html#SEC32.3.5

There is also a nice graphical application at
  http://www.analyzemath.com/Slope/Slope.html
where you can examine some of these relationships graphically
(computes slope / intercept for one or two lines).

Search phrases used to locate the answer included:
  compute slope intercept line
  calculate intersection two lines
  intersection two lines source code
as well as similar phrases can be used to find additional information.

Please make a clarification request if some part of the answer is
unclear or incomplete. I would be glad to help you further. Good luck
with your work.

  --Maniac

Clarification of Answer by maniac-ga on 07 Jun 2006 16:35 PDT
Hello Groves222,

I need to make a correction to the answer as follows:

The line A-C has a slope (M) of
  M = TAN(25+ATAN(M[AB])) = TAN(39.03624347 degrees) = 0.810831944
and has intercept (B) of
  B = 1 - (1*0.810831944) - 0.189168056
or line A-C in slope / intercept form is
  Y = 0.810831944 * X + 0.189168056

This revised calculation properly captures the rotation of 25 degrees
from A-B to A-C.

As a result of this change, and using the corrected values the answer
should now read

You can compute point C from this information using a formula
  Cx = (B[BC]-B[AC])/(M[BC]-M[AC])
  Cx = (22 - 0.189168056) / (-4 - 0.810831944)
  Cx = 4.533692342
and
  Cy = M[AC]*Cx + B[AC]
  Cy = 0.810831944 * 4.533692342 + 0.189168056
  Cy = 3.865230633

I am not quite sure the minor differences in the answer I am providing
and from the comments. Make a clarification request if you want a copy
of the spreadsheet I used to calculate the results and perform some
cross checks to make sure the results are OK.

  --Maniac
Comments  
Subject: Re: Determing coordinate points of triangles
From: frde-ga on 28 May 2006 10:29 PDT
 
Your diagram is wrong
- A and B are known - you have put C where B is

You can calculate the length of A-B it is Sqrt( 1 + 16 )
( from a triangle A-D-B with D at (5,1)

Using Cos and Sin you can get the lengths of the other two sides
You then need to get the angle BAD add it to CAB 
From that you have another triangle A-C-E where E is below C and level with A

This is a bit complicated to explain without diagrams
Subject: Re: Determing coordinate points of triangles
From: groves222-ga on 28 May 2006 12:42 PDT
 
Ok, i updated the diagram.

Can you please walk through the steps with the example above?
Subject: Re: Determing coordinate points of triangles
From: ansel001-ga on 29 May 2006 02:26 PDT
 
You still need to fix your diagram.  Angle C should be 90.  You made
angle B the right angle instead.
Subject: Re: Determing coordinate points of triangles
From: klbr-ga on 29 May 2006 12:43 PDT
 
Question is not clear.

Is Angle ACB is 90 degrees? (like mentioned in the question) or 

Is Angle ABC is 90 degrees? (like mentioned in the diagram)
Subject: Re: Determing coordinate points of triangles
From: dgfinck-ga on 02 Jun 2006 21:16 PDT
 
Is the poster still interested?  Is it desired generate a process to
solve it in general for any locations of the points A & B, and any
angle BAC?  Or just this specific case?  Will the 90deg angle remain
right?  Or is a solution desired for an arbitrary (2nd) angle?

df
Subject: Re: Determing coordinate points of triangles
From: berkeleychocolate-ga on 05 Jun 2006 18:21 PDT
 
Here's a fairly neat way to solve this problem: Equate points of the
plane with complex numbers. So (x,y) = x + y i. Say C = (x,y). Let r
be the distance from A to C. Then since  ABC is a right triangle with
hypotenuse sqrt(17), r = sqrt(17)*cos25. So the vector from A to C can
be obtained by rotating the unit vector in the direction from A to B
by 25 degrees and multiplying it by r. That is, C = A + this vector =
(1+i) + r*(cos25+isin25)*(4+i)/sqrt(17) = (1+i) + cos(25)*(cos25 +
isin25)*(4+i) =  (1+i) + cos25 * ( (4cos25 - sin25) + i*(cos25+4sin25)
). So x = 1 + cos25 * (4cos25 - sin25) and y = 1 + cos25 * (cos25 +
sin 25).
Subject: Re: Determing coordinate points of triangles
From: brix24-ga on 07 Jun 2006 02:53 PDT
 
I get values for the x and y coordinates of point C that differ from
both maniac-ga?s and berkeleychocolate-ga?s. The values I get are:

x(c)				y(c)
3.90255299781359	3.35348269108123
and
4.66859744093257	0.289304918605314

Both of these points seem to satisfy the condition of a right triangle
with AB as the hypotenuse, that is,

[x(c)-x(a)]^2 + [y(c)-y)a)]^2 + [x(c)-x(b)]^2 + [y(c)-y(b)]^2 = 17

In confirmation of the above x and y values, I get 17 when I post
either of the following into Excel (perhaps Google spreadsheets
soon?):
=(3.90255299781359-1)^2 + (3.35348269108123-1)^2 +
(3.90255299781359-5)^2 + (3.35348269108123-2)^2

=(4.66859744093257-1)^2 + (0.289304918605314-1)^2
+(4.66859744093257-5)^2 + (0.289304918605314-2)^2

----
Google search strategies:

slope perpendicular lines

triangle coordinates

The first search was to affirm (and correct) what I remembered about
perpendicular lines. The second search led to drawing the correct line
to solve the problem with a diagram.

-----
Initial formulas that I got:
x(c)=x(a) + [x(b)-x(a)][cos alpha]^2 ? (tan alpha)[y(b)-y(a)](cos alpha)^2
y(c)=y(a) + [y(b)-y(a)][cos alpha]^2 + (tan alpha)[x(b)-x(a)] (cos alpha)^2

and

x(c)=x(a) + [x(b)-x(a)][cos alpha]^2 + (tan alpha)[y(b)-y(a)](cos alpha)^2
y(c)=y(a) + [y(b)-y(a)][cos alpha]^2 - (tan alpha)[x(b)-x(a)] (cos alpha)^2

These reduce to:
x(c)=x(a) + (cos alpha) * {[x(b)-x(a)]*[cos alpha] ? [y(b)-y(a)]*(sin alpha)}
y(c)=y(a) + (cos alpha) * {[y(b)-y(a)]*[cos alpha] + [x(b)-x(a)]*(sin alpha)}

and

x(c)=x(a) + (cos alpha) * {[x(b)-x(a)]*[cos alpha] + [y(b)-y(a)]*(sin alpha)}
y(c)=y(a) + (cos alpha) * {[y(b)-y(a)]*[cos alpha] - [x(b)-x(a)]*(sin alpha)}

Note: I haven?t checked out the generality of these formulas, that is,
whether the signs of the [x(b)-x(a)] and [y(b)-y(a)] terms take care
of the left/right, above/below relationships of points A and B in the
general case.

--------
Difference from berkeleychocolate-ga?s results: I can explain the Y
value difference I get if berkeleychocolate?s result is simply missing
a ?4? before the final sine.

Difference from maniac?s result: I?m not sure why we differ; I got lost at

> The line A-C has a slope (M) of
>  M = 0.25 + sin(25 degrees).
Subject: Re: Determing coordinate points of triangles
From: berkeleychocolate-ga on 07 Jun 2006 12:38 PDT
 
Correction: x is as I gave it and y is 1 + cos25*(cos25+4sin25). This
gives agreement with brix24. I just lost a factor of 4 in the last
step. One could rotate by -25 degrees and get the other solution.
Subject: Re: Determing coordinate points of triangles
From: brix24-ga on 15 Jun 2006 06:27 PDT
 
groves22-ga,

Maniac's revised results fit the triangle that you show in your figure
(AB perpendicular to BC), ABC = 90 degrees.

Berkeleychocolate's and my results are for the triangle given by the
data (AC is perpendicular to BC; ACB = 90 degrees).

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