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Q: Formula to calculate the coordinates of third vertex of right triangle ( Answered ,   2 Comments )
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 Subject: Formula to calculate the coordinates of third vertex of right triangle Category: Science > Math Asked by: catohara-ga List Price: \$40.00 Posted: 25 Oct 2004 11:49 PDT Expires: 24 Nov 2004 10:49 PST Question ID: 419874
 ```I am trying to locate 2 formulas that will help me calculate the coordinates of the third vertex of a right triangle. I know the x,y coordinates of the other two vertices, I have the length of 2 sides of the triangle and by Pythagorean?s theorem, I can calculate the third. I feel I should know how to calculate this last point, but can?t seem to do it (it?s been too long!). I would like to find 2 formulas that I can use in my code, one that will give me the x and then one that will give me the y of the third point. So for example if this is my triangle : (x,y) |\ A | \ B | \ |___\ (4,3) C (10,3) What is the formula(s) to calculate (x,y)? Where A = 3, C = 6 and B = square root of 45. It is easy to calculate in your head the (x,y) when sides A & C are parallel to the 0 x & y axis, but not when they are skewed away from it. (I know the answer is this case is (4,6)?) I have visited sites such as http://mathforum.org and while they have many formulas, none seem to fit my needs. I do not have any angles for any of the vertices, (except the 90 degree one) so Trig formulas don?t seem very useful. My problem seems to simple for most math sites. I do realize that the formula can have 2 answers, one for the positive, and one for the negative direction.```
 Subject: Re: Formula to calculate the coordinates of third vertex of right triangle Answered By: mathtalk-ga on 25 Oct 2004 19:38 PDT Rated:
 ```Hi, catohara-ga: You list as "given" data: 1) the x,y coordinates of two vertices 2) the lengths of two sides and (by computation) the third length 3) that the triangle is a right triangle Since there is some overlap in these data, there's also a possibility of inconsistency. To avoid this and pick the simplest path to a solution, let's restrict the data that we need to use to the bare amount! Often one labels the vertices of a triangle with capital letters A,B,C and the opposite edges with lowercase a,b,c. However you've already labelled the edges with the capital letters, so let's continue in that vein and label the opposite vertices with lowercase letters, respectively. We are given the coordinates of two vertices, a and b, and between them is one leg C of a right triangle. There will be a right angle at b, and the length of the hypotenuse B, extending from a to c, is known. We are asked to find the coordinates of point c from these "given data". Provided B is longer than C (whose length is given by the distance formula based on coordinates for points a and b), a pair of solutions will exist in the Cartesian plane for point c that produces the desired right triangle. [The solution is simplest if we reduce to the case that b is the origin (0,0). This reduction is just a matter of subtracting the coordinates of b from every vertex in the problem. At the end we would add the original coordinates of b back and get the answer in terms of your original coordinates. However I will go through the solution without this shortcut for the sake of clarity.] Let (x_a,y_a) be the coordinates of a, and let (x_b,y_b) be the coordinates of b. A path from point b to point a would have this direction: [ x_a - x_b, y_a - y_b ] In the example you gave, this direction would be the vector [6,0], but this direction can be computed whenever the coordinates of points a and b are given. The direction from b to c (which contains the other "leg" of the right triangle and is therefore perpendicular to the first direction) could go in either of two directions (giving us the expected two solutions). To find one of these perpendicular directions, switch the order of the coordinates from the first direction and change the sign of one of them. For example, we might calculate for the example case: [6,0] --> (switch) [0,6] --> (change sign of one coordinate) [0,-6] Now if that is one perpendicular direction, then an opposite perpendicular direction would be obtained if we'd changed instead the other coordinate's sign, giving e.g. [-0,+6]. Note that these two choices are always additive inverses of each other: [0,-6] + [0,+6] = [0,0] It doesn't matter which one you happen to choose; either will give us both solutions in the end. Now we want to take the chosen perpendicular direction and scale it to unit length. Let's say the perpendicular direction we will work with is this one: [ y(a) - y(b), x(b) - x(a) ] If we calculate its length, the length turns out to be the same as the distance C between a and b. So I'm going to label the unit perpendicular vector: U = (1/C) [ y(a) - y(b), x(b) - x(a) ] y(a) - y(b) x(b) - x(a) = [ ------------- , ------------- ] C C We are about to finish! Using the Pythagorean formula, we know how far point c is from "right corner" point b in this direction, namely: A = (plus or minus) SQRT( B^2 - C^2 ) Remember that C was the length of one leg, and B the length of the hypotenuse. Therefore we only need to add a multiple A of the unit vector u to the coordinates of b in order to obtain c: ( x_c, y_c ) = coordinates of c = ( x_b, y_b ) + A * u A*(y(a)-y(b)) A*(x(b)-x(a)) = [ x_b + ------------- , y_b + ------------- ] C C Let's apply this to your example situation. Then for A = +3: ( x_b, y_b ) = (4,3) ( x_c, y_c ) = (4,3) + 3 * u = (4,3) + [ (3*0)/6 , (3*(-6))/6 ] = (4,3) + [ 0, -3 ] = (4,0) On the other hand if we had taken A = -3, we'd get the point c lying in the opposite direction: ( x_c, y_c ) = (4,3) + [ 0, +3 ] = (4,6) as you know, of course. [Similarly if we'd chosen u = [0,1] instead of [0,-1], we get both solutions but in the other order with respect to plus or minus A.] There is a slightly more general problem you might be interested in knowing how to solve, but I didn't want to clutter up this Answer with a "tangent". Let me know if further clarification would be helpful. regards, mathtalk-ga``` Clarification of Answer by mathtalk-ga on 25 Oct 2004 19:42 PDT ```Here's a modified diagram of the triangle for convenient reference: c |\ A | \ B | \ |___\ b C a where a = (x_a,y_a), b = (x_b,y_b) are given along with the length of the hypotenuse B. regards, mathtalk-ga```
 catohara-ga rated this answer: and gave an additional tip of: \$8.00 `Thank you for you're help. It appears to be exactly what I was looking for.`

 ```A) If the legs are parallel to the x,y axes. Say the right triangle is ABC, where vertex B is at the right angle, vertex A is above or below B, and vertex C is to the left or to the right of B. So AB is vertical and BC is horizontal. You said the lengths of two sides are known. Which two sides? Say the two legs. So the hypotenuse is not given, although its length can easily be solved by the Pythagorean theorem. Say AB = U long and BC = V long. ---------- Case A.1: If (x,y) is B. Then A and C are given. Since their coordinates are known, say A is (a,b) and C is (c,d). By examination, x = a, and y = d Therefore, (x,y) = (a,d) ------------- Case A.2: If (x,y) is A. Then B and C are known. Say B is (e,f) and C is (c,d). By examination, x = e. That makes A to be (e,y). So only y is unknown. Distance formula, d = sqrt[(x2 -x1)^2 +(y2 -y1)^2]. So, hypotenuse AC = sqrt[(e-c)^2 +(y-d)^2] In rt. triangle ABC, by Pythagorean theorem, (AB)^2 +(BC)^2 = (AC)^2 Substitutions, U^2 +V^2 = (e-c)^2 +(y-d)^2 ----(1) That is a quadratic equation where the only unknown is y. (y-d)^2 = U^2 +V^2 -(e-c)^2 Therefore, y = d +,-sqrt[U^2 +V^2 -(e-c)^2] ----two possible y's. +,- is "plus or minus" ------------- Case A.3: If (x,y) is C. Then A and B are known. Say A is (a,b) and B is (e,f). By examination, y = f That makes C to be (x,f). So only x is unknown. hypotenuse AC = sqrt[(x-a)^2 +(f-b)^2] In rt. triangle ABC, (AB)^2 +(BC)^2 = (AC)^2 Substitutions, U^2 +V^2 = (x-a)^2 +(f-b)^2 ----(2) Therefore, x = a +,-sqrt[U^2 +V^2 -(f-b)^2] ----two possible x's. ======================================================================= B) If the legs AB and AC are not parallel to the x,y axes. Their are skewed, as you say. Same rt. triangle ABC as above. Only one case example here, because whichever vertex is (x,y), the solution should be the same way. Say, vertex A is (x,y). then B = (e,f) and C = (c,d) and AB = U and BC = V By distance formula, (AB)^2 = U^2 = (x-e)^2 +(y-f)^2 -----(3) (BC)^2 = V^2 = (e-c)^2 +(f-d)^2 -----(4) (AC)^2 = (x-c)^2 +(y-d)^2 ----------(5) In rt. triangle ABC, by Pythagorean theorem, (AB)^2 +(BC)^2 = (AC)^2 Substitutions, {(x-e)^2 +(y-f)^2} +{(e-c)^2 +(f-d)^2} = (x-c)^2 +(y-d)^2 Simplify that and you should arrive at x(e-c) +y(f-d) = e^2 +f^2 -ec -fd ----(6) Then, x = (e^2 +f^2 -ec -fd -y(f-d)) / (e-c) -----(6a) Or, y = (e^2 +f^2 -ec -fd -x(e-c)) / (f-d) -----(6b) Then go back to Eq.(3). Plug in y, by using (6b), and you'd get x. Then go back to Eq.(5). Plug in x, by using (6a), and you'd get y. (I did not go in those. Too long. For free, this is good enough.)```
 ```Zeez, this morning on my way to work, I thought of these: --------- Case A.2: If (x,y) is A. Then B and C are known. Say B is (e,f) and C is (c,d). By examination, x = e. That makes A to be (e,y). So only y is unknown. But AB is given as U. Therefore, without computation, y = f+U. And (x,y) is (e,f+U). --------- Case A.3: If (x,y) is C. Then A and B are known. Say A is (a,b) and B is (e,f). By examination, y = f That makes C to be (x,f). So only x is unknown. But BC is given as V. Therefore, x = e+V And (x,y) is (e+V,f).```