

Subject:
Determing coordinate points of triangles
Category: Science > Math Asked by: groves222ga 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. 

Subject:
Re: Determing coordinate points of triangles
Answered By: maniacga 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 AB can be represented in slope / intercept form as Y = M * X + B when computed as described at http://wwwstatic.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) = (21)/(51) = 0.25 and generates the intercept (B) by B = Ay  (Ax * M) = 1  (1*0.25) = 0.75 or line AB in slope / intercept form is Y = 0.25 * X + 0.75 The line AC 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 AC in slope / intercept form is Y = 0.672618262 * X + 0.327381738 The line BC which is perpendicular to AB 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 BC 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 BC 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 AB 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  


Subject:
Re: Determing coordinate points of triangles
From: frdega 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 AB it is Sqrt( 1 + 16 ) ( from a triangle ADB 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 ACE 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: groves222ga 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: ansel001ga on 29 May 2006 02:26 PDT 
You still need to fix your diagram. Angle C should be 90º. You made angle B the right angle instead. 
Subject:
Re: Determing coordinate points of triangles
From: klbrga 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: dgfinckga 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: berkeleychocolatega 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: brix24ga on 07 Jun 2006 02:53 PDT 
I get values for the x and y coordinates of point C that differ from both maniacga?s and berkeleychocolatega?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?): =(3.902552997813591)^2 + (3.353482691081231)^2 + (3.902552997813595)^2 + (3.353482691081232)^2 =(4.668597440932571)^2 + (0.2893049186053141)^2 +(4.668597440932575)^2 + (0.2893049186053142)^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 berkeleychocolatega?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 AC has a slope (M) of > M = 0.25 + sin(25 degrees). 
Subject:
Re: Determing coordinate points of triangles
From: berkeleychocolatega 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: brix24ga on 15 Jun 2006 06:27 PDT 
groves22ga, 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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