{"group":{"id":1,"name":"Community","lockable":false,"created_at":"2012-01-18T18:02:15.000Z","updated_at":"2026-05-06T00:09:22.000Z","description":"Problems submitted by members of the MATLAB Central community.","is_default":true,"created_by":161519,"badge_id":null,"featured":false,"trending":false,"solution_count_in_trending_period":0,"trending_last_calculated":"2026-05-06T00:00:00.000Z","image_id":null,"published":true,"community_created":false,"status_id":2,"is_default_group_for_player":false,"deleted_by":null,"deleted_at":null,"restored_by":null,"restored_at":null,"description_opc":null,"description_html":null,"published_at":null},"problems":[{"id":43642,"title":"Euclidean distance from a point to a polynomial","description":"A not uncommon problem in the area of computational geometry is to find the closest point to a straight line from a given point, or the distance from a point to a line. As you might expect, there is a simple formula for those things.\r\n\r\nAs an extension, I decided one day to write a tool that would compute the distance from a point to a general polynomial function in the (x,y) plane. That is your problem here:\r\n\r\nGiven a point (x0,y0), and a polynomial in the form y=P(x) where the function P is defined by the coefficients of a polynomial, you need to compute the minimum \u003chttps://en.wikipedia.org/wiki/Euclidean_distance Euclidean distance\u003e to that polynomial. So you need to find and return the minimum distance in the (x,y) plane between the point (x0,y0), and the function y=P(x).\r\n\r\nThe function P will be passed in as the coefficients of a polynomial in standard MATLAB form, thus with the highest order coefficient first in a vector, like that generated by polyfit, and used by polyval. (P might be as simple as a constant function.) The point in question will be passed in as a vector of length 2, thus [x0,y0].\r\n\r\nAs test case for you to check your code, the distance from the point (-2,-5) to the curve y=x^2/2+3*x-5 should be:\r\n\r\n  x0y0 = [-2 -5];\r\n  P = [0.5 3 -5];\r\n  D = distance2polynomial(P,xy)\r\n  D =\r\n          1.89013819497707\r\n\r\n(Be careful plotting these curves in case you want to plot your solution. The command \"axis equal\" is a good idea.)\r\n\r\nThe symbolic TB tells me the distance is 1.8901381949770695260066523338279..., but I'll allow some slop in your solution, since you may have chosen a different algorithm than the one I chose. You should expect to provide at least 13 correct significant digits in the solution.\r\n\r\nDisclaimer: I'm not really sure why anyone needs such a code, which is why I've not posted my solution on the FEX. Anyway, my solution is a pretty one that I thought might make a fun Cody problem, and I wanted to see how others might approach the problem. I expect that my reference solution will score poorly for Cody purposes, since it is carefully coded, complete with error checks, and returns more than just the minimum distance.","description_html":"\u003cp\u003eA not uncommon problem in the area of computational geometry is to find the closest point to a straight line from a given point, or the distance from a point to a line. As you might expect, there is a simple formula for those things.\u003c/p\u003e\u003cp\u003eAs an extension, I decided one day to write a tool that would compute the distance from a point to a general polynomial function in the (x,y) plane. That is your problem here:\u003c/p\u003e\u003cp\u003eGiven a point (x0,y0), and a polynomial in the form y=P(x) where the function P is defined by the coefficients of a polynomial, you need to compute the minimum \u003ca href = \"https://en.wikipedia.org/wiki/Euclidean_distance\"\u003eEuclidean distance\u003c/a\u003e to that polynomial. So you need to find and return the minimum distance in the (x,y) plane between the point (x0,y0), and the function y=P(x).\u003c/p\u003e\u003cp\u003eThe function P will be passed in as the coefficients of a polynomial in standard MATLAB form, thus with the highest order coefficient first in a vector, like that generated by polyfit, and used by polyval. (P might be as simple as a constant function.) The point in question will be passed in as a vector of length 2, thus [x0,y0].\u003c/p\u003e\u003cp\u003eAs test case for you to check your code, the distance from the point (-2,-5) to the curve y=x^2/2+3*x-5 should be:\u003c/p\u003e\u003cpre class=\"language-matlab\"\u003ex0y0 = [-2 -5];\r\nP = [0.5 3 -5];\r\nD = distance2polynomial(P,xy)\r\nD =\r\n        1.89013819497707\r\n\u003c/pre\u003e\u003cp\u003e(Be careful plotting these curves in case you want to plot your solution. The command \"axis equal\" is a good idea.)\u003c/p\u003e\u003cp\u003eThe symbolic TB tells me the distance is 1.8901381949770695260066523338279..., but I'll allow some slop in your solution, since you may have chosen a different algorithm than the one I chose. You should expect to provide at least 13 correct significant digits in the solution.\u003c/p\u003e\u003cp\u003eDisclaimer: I'm not really sure why anyone needs such a code, which is why I've not posted my solution on the FEX. Anyway, my solution is a pretty one that I thought might make a fun Cody problem, and I wanted to see how others might approach the problem. I expect that my reference solution will score poorly for Cody purposes, since it is carefully coded, complete with error checks, and returns more than just the minimum distance.\u003c/p\u003e","function_template":"function D = distance2polynomial(P,x0y0)\r\n  % compute the minimum Euclidean distance between a point and a polynomial\r\n  D = rand;\r\nend\r\n","test_suite":"%%\r\nx0y0 = [-2 5];\r\nP = [0.5 3 -5];\r\ny_correct = 4.3093988461280149175163000679048;\r\ntol = 5e-13;\r\nassert(abs(distance2polynomial(P,x0y0)-y_correct) \u003c tol)\r\n\r\n%%\r\nx0y0 = [pi, pi];\r\nP = [10];\r\ny_correct = 6.8584073464102067615373566167205;\r\ntol = 7e-13;\r\nassert(abs(distance2polynomial(P,x0y0)-y_correct) \u003c tol)\r\n\r\n%%\r\nx0y0 = [0.25,50];\r\nP = [1 2 3 4 5];\r\ny_correct = 1.6470039192886012020234097061626;\r\ntol = 5e-13;\r\nassert(abs(distance2polynomial(P,x0y0)-y_correct) \u003c tol)\r\n\r\n%%\r\nx0y0 = [-3 -3];\r\nP = [-2 1];\r\ny_correct = 4.4721359549995793928183473374626;\r\ntol = 5e-13;\r\nassert(abs(distance2polynomial(P,x0y0)-y_correct) \u003c tol)\r\n\r\n%%\r\nx0y0 = [0 5];\r\nP = [1 0 1];\r\ny_correct = 1.9364916731037084425896326998912;\r\ntol = 2e-13;\r\nassert(abs(distance2polynomial(P,x0y0)-y_correct) \u003c tol)\r\n\r\n%%\r\nx0y0 = [-2 -5];\r\nP = [0.5 3 -5];\r\ny_correct = 1.8901381949770695260066523338279;\r\ntol = 2e-13;\r\n(abs(distance2polynomial(P,x0y0)-y_correct) \u003c tol)\r\n\r\n","published":true,"deleted":false,"likes_count":3,"comments_count":2,"created_by":544,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":35,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":37,"created_at":"2016-10-28T21:00:14.000Z","updated_at":"2026-05-08T12:02:16.000Z","published_at":"2016-10-28T21:08:10.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"targetMode\":\"\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"targetMode\":\"\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\\n\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eA not uncommon problem in the area of computational geometry is to find the closest point to a straight line from a given point, or the distance from a point to a line. As you might expect, there is a simple formula for those things.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eAs an extension, I decided one day to write a tool that would compute the distance from a point to a general polynomial function in the (x,y) plane. That is your problem here:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eGiven a point (x0,y0), and a polynomial in the form y=P(x) where the function P is defined by the coefficients of a polynomial, you need to compute the minimum\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:hyperlink w:docLocation=\\\"https://en.wikipedia.org/wiki/Euclidean_distance\\\"\u003e\u003cw:r\u003e\u003cw:t\u003eEuclidean distance\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:hyperlink\u003e\u003cw:r\u003e\u003cw:t\u003e to that polynomial. So you need to find and return the minimum distance in the (x,y) plane between the point (x0,y0), and the function y=P(x).\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eThe function P will be passed in as the coefficients of a polynomial in standard MATLAB form, thus with the highest order coefficient first in a vector, like that generated by polyfit, and used by polyval. (P might be as simple as a constant function.) The point in question will be passed in as a vector of length 2, thus [x0,y0].\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eAs test case for you to check your code, the distance from the point (-2,-5) to the curve y=x^2/2+3*x-5 should be:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"code\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003e\u003c![CDATA[x0y0 = [-2 -5];\\nP = [0.5 3 -5];\\nD = distance2polynomial(P,xy)\\nD =\\n        1.89013819497707]]\u003e\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003e(Be careful plotting these curves in case you want to plot your solution. The command \\\"axis equal\\\" is a good idea.)\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eThe symbolic TB tells me the distance is 1.8901381949770695260066523338279..., but I'll allow some slop in your solution, since you may have chosen a different algorithm than the one I chose. You should expect to provide at least 13 correct significant digits in the solution.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eDisclaimer: I'm not really sure why anyone needs such a code, which is why I've not posted my solution on the FEX. Anyway, my solution is a pretty one that I thought might make a fun Cody problem, and I wanted to see how others might approach the problem. I expect that my reference solution will score poorly for Cody purposes, since it is carefully coded, complete with error checks, and returns more than just the minimum distance.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003c/w:body\u003e\u003c/w:document\u003e\"},{\"partUri\":\"/matlab/output.xml\",\"contentType\":\"text/xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\" standalone=\\\"no\\\" ?\u003e\u003cembeddedOutputs\u003e\u003cmetaData\u003e\u003cevaluationState\u003emanual\u003c/evaluationState\u003e\u003clayoutState\u003ecode\u003c/layoutState\u003e\u003coutputStatus\u003eready\u003c/outputStatus\u003e\u003c/metaData\u003e\u003coutputArray type=\\\"array\\\"/\u003e\u003cregionArray type=\\\"array\\\"/\u003e\u003c/embeddedOutputs\u003e\"}]}"}],"problem_search":{"errors":[],"problems":[{"id":43642,"title":"Euclidean distance from a point to a polynomial","description":"A not uncommon problem in the area of computational geometry is to find the closest point to a straight line from a given point, or the distance from a point to a line. As you might expect, there is a simple formula for those things.\r\n\r\nAs an extension, I decided one day to write a tool that would compute the distance from a point to a general polynomial function in the (x,y) plane. That is your problem here:\r\n\r\nGiven a point (x0,y0), and a polynomial in the form y=P(x) where the function P is defined by the coefficients of a polynomial, you need to compute the minimum \u003chttps://en.wikipedia.org/wiki/Euclidean_distance Euclidean distance\u003e to that polynomial. So you need to find and return the minimum distance in the (x,y) plane between the point (x0,y0), and the function y=P(x).\r\n\r\nThe function P will be passed in as the coefficients of a polynomial in standard MATLAB form, thus with the highest order coefficient first in a vector, like that generated by polyfit, and used by polyval. (P might be as simple as a constant function.) The point in question will be passed in as a vector of length 2, thus [x0,y0].\r\n\r\nAs test case for you to check your code, the distance from the point (-2,-5) to the curve y=x^2/2+3*x-5 should be:\r\n\r\n  x0y0 = [-2 -5];\r\n  P = [0.5 3 -5];\r\n  D = distance2polynomial(P,xy)\r\n  D =\r\n          1.89013819497707\r\n\r\n(Be careful plotting these curves in case you want to plot your solution. The command \"axis equal\" is a good idea.)\r\n\r\nThe symbolic TB tells me the distance is 1.8901381949770695260066523338279..., but I'll allow some slop in your solution, since you may have chosen a different algorithm than the one I chose. You should expect to provide at least 13 correct significant digits in the solution.\r\n\r\nDisclaimer: I'm not really sure why anyone needs such a code, which is why I've not posted my solution on the FEX. Anyway, my solution is a pretty one that I thought might make a fun Cody problem, and I wanted to see how others might approach the problem. I expect that my reference solution will score poorly for Cody purposes, since it is carefully coded, complete with error checks, and returns more than just the minimum distance.","description_html":"\u003cp\u003eA not uncommon problem in the area of computational geometry is to find the closest point to a straight line from a given point, or the distance from a point to a line. As you might expect, there is a simple formula for those things.\u003c/p\u003e\u003cp\u003eAs an extension, I decided one day to write a tool that would compute the distance from a point to a general polynomial function in the (x,y) plane. That is your problem here:\u003c/p\u003e\u003cp\u003eGiven a point (x0,y0), and a polynomial in the form y=P(x) where the function P is defined by the coefficients of a polynomial, you need to compute the minimum \u003ca href = \"https://en.wikipedia.org/wiki/Euclidean_distance\"\u003eEuclidean distance\u003c/a\u003e to that polynomial. So you need to find and return the minimum distance in the (x,y) plane between the point (x0,y0), and the function y=P(x).\u003c/p\u003e\u003cp\u003eThe function P will be passed in as the coefficients of a polynomial in standard MATLAB form, thus with the highest order coefficient first in a vector, like that generated by polyfit, and used by polyval. (P might be as simple as a constant function.) The point in question will be passed in as a vector of length 2, thus [x0,y0].\u003c/p\u003e\u003cp\u003eAs test case for you to check your code, the distance from the point (-2,-5) to the curve y=x^2/2+3*x-5 should be:\u003c/p\u003e\u003cpre class=\"language-matlab\"\u003ex0y0 = [-2 -5];\r\nP = [0.5 3 -5];\r\nD = distance2polynomial(P,xy)\r\nD =\r\n        1.89013819497707\r\n\u003c/pre\u003e\u003cp\u003e(Be careful plotting these curves in case you want to plot your solution. The command \"axis equal\" is a good idea.)\u003c/p\u003e\u003cp\u003eThe symbolic TB tells me the distance is 1.8901381949770695260066523338279..., but I'll allow some slop in your solution, since you may have chosen a different algorithm than the one I chose. You should expect to provide at least 13 correct significant digits in the solution.\u003c/p\u003e\u003cp\u003eDisclaimer: I'm not really sure why anyone needs such a code, which is why I've not posted my solution on the FEX. Anyway, my solution is a pretty one that I thought might make a fun Cody problem, and I wanted to see how others might approach the problem. I expect that my reference solution will score poorly for Cody purposes, since it is carefully coded, complete with error checks, and returns more than just the minimum distance.\u003c/p\u003e","function_template":"function D = distance2polynomial(P,x0y0)\r\n  % compute the minimum Euclidean distance between a point and a polynomial\r\n  D = rand;\r\nend\r\n","test_suite":"%%\r\nx0y0 = [-2 5];\r\nP = [0.5 3 -5];\r\ny_correct = 4.3093988461280149175163000679048;\r\ntol = 5e-13;\r\nassert(abs(distance2polynomial(P,x0y0)-y_correct) \u003c tol)\r\n\r\n%%\r\nx0y0 = [pi, pi];\r\nP = [10];\r\ny_correct = 6.8584073464102067615373566167205;\r\ntol = 7e-13;\r\nassert(abs(distance2polynomial(P,x0y0)-y_correct) \u003c tol)\r\n\r\n%%\r\nx0y0 = [0.25,50];\r\nP = [1 2 3 4 5];\r\ny_correct = 1.6470039192886012020234097061626;\r\ntol = 5e-13;\r\nassert(abs(distance2polynomial(P,x0y0)-y_correct) \u003c tol)\r\n\r\n%%\r\nx0y0 = [-3 -3];\r\nP = [-2 1];\r\ny_correct = 4.4721359549995793928183473374626;\r\ntol = 5e-13;\r\nassert(abs(distance2polynomial(P,x0y0)-y_correct) \u003c tol)\r\n\r\n%%\r\nx0y0 = [0 5];\r\nP = [1 0 1];\r\ny_correct = 1.9364916731037084425896326998912;\r\ntol = 2e-13;\r\nassert(abs(distance2polynomial(P,x0y0)-y_correct) \u003c tol)\r\n\r\n%%\r\nx0y0 = [-2 -5];\r\nP = [0.5 3 -5];\r\ny_correct = 1.8901381949770695260066523338279;\r\ntol = 2e-13;\r\n(abs(distance2polynomial(P,x0y0)-y_correct) \u003c tol)\r\n\r\n","published":true,"deleted":false,"likes_count":3,"comments_count":2,"created_by":544,"edited_by":null,"edited_at":null,"deleted_by":null,"deleted_at":null,"solvers_count":35,"test_suite_updated_at":null,"rescore_all_solutions":false,"group_id":37,"created_at":"2016-10-28T21:00:14.000Z","updated_at":"2026-05-08T12:02:16.000Z","published_at":"2016-10-28T21:08:10.000Z","restored_at":null,"restored_by":null,"spam":false,"simulink":false,"admin_reviewed":false,"description_opc":"{\"relationships\":[{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/document\",\"targetMode\":\"\",\"relationshipId\":\"rId1\",\"target\":\"/matlab/document.xml\"},{\"relationshipType\":\"http://schemas.mathworks.com/matlab/code/2013/relationships/output\",\"targetMode\":\"\",\"relationshipId\":\"rId2\",\"target\":\"/matlab/output.xml\"}],\"parts\":[{\"partUri\":\"/matlab/document.xml\",\"relationship\":[],\"contentType\":\"application/vnd.mathworks.matlab.code.document+xml\",\"content\":\"\u003c?xml version=\\\"1.0\\\" encoding=\\\"UTF-8\\\"?\u003e\\n\u003cw:document xmlns:w=\\\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\\\"\u003e\u003cw:body\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eA not uncommon problem in the area of computational geometry is to find the closest point to a straight line from a given point, or the distance from a point to a line. As you might expect, there is a simple formula for those things.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eAs an extension, I decided one day to write a tool that would compute the distance from a point to a general polynomial function in the (x,y) plane. That is your problem here:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eGiven a point (x0,y0), and a polynomial in the form y=P(x) where the function P is defined by the coefficients of a polynomial, you need to compute the minimum\u003c/w:t\u003e\u003c/w:r\u003e\u003cw:r\u003e\u003cw:t\u003e \u003c/w:t\u003e\u003c/w:r\u003e\u003cw:hyperlink w:docLocation=\\\"https://en.wikipedia.org/wiki/Euclidean_distance\\\"\u003e\u003cw:r\u003e\u003cw:t\u003eEuclidean distance\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:hyperlink\u003e\u003cw:r\u003e\u003cw:t\u003e to that polynomial. So you need to find and return the minimum distance in the (x,y) plane between the point (x0,y0), and the function y=P(x).\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eThe function P will be passed in as the coefficients of a polynomial in standard MATLAB form, thus with the highest order coefficient first in a vector, like that generated by polyfit, and used by polyval. (P might be as simple as a constant function.) The point in question will be passed in as a vector of length 2, thus [x0,y0].\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eAs test case for you to check your code, the distance from the point (-2,-5) to the curve y=x^2/2+3*x-5 should be:\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"code\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003e\u003c![CDATA[x0y0 = [-2 -5];\\nP = [0.5 3 -5];\\nD = distance2polynomial(P,xy)\\nD =\\n        1.89013819497707]]\u003e\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003e(Be careful plotting these curves in case you want to plot your solution. The command \\\"axis equal\\\" is a good idea.)\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eThe symbolic TB tells me the distance is 1.8901381949770695260066523338279..., but I'll allow some slop in your solution, since you may have chosen a different algorithm than the one I chose. You should expect to provide at least 13 correct significant digits in the solution.\u003c/w:t\u003e\u003c/w:r\u003e\u003c/w:p\u003e\u003cw:p\u003e\u003cw:pPr\u003e\u003cw:pStyle w:val=\\\"text\\\"/\u003e\u003c/w:pPr\u003e\u003cw:r\u003e\u003cw:t\u003eDisclaimer: I'm not really sure why anyone needs such a code, which is why I've not posted my solution on the FEX. Anyway, my solution is a pretty one that I thought might make a fun Cody problem, and I wanted to see how others might approach the problem. 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