https://community.graphisoft.com/t5/Libraries-objects/A-math-question-a-tough-one/m-p/97463#M31693 It's fairly simple if I understand correctly. You can get the distance (let's call it "D1") between the known points and use the proportion.<BR /> <PRE>D1 = SQR((X2-X1)^2 + (Y2-Y1)^2) ! per Pythagoras ! since (X3-X1)/(X2-X1) = D/D1 then it follows that... X3 = (D/D1)*(X2-X1) + X1 ! and... Y3 = (D/D1)*(Y2-Y1) + Y1</PRE> Sun, 18 Feb 2007 15:14:05 GMT Anonymous 2007-02-18T15:14:05Z https://community.graphisoft.com/t5/Libraries-objects/A-math-question-a-tough-one/m-p/97461#M31691 <DIV class="actalk-migrated-content"><T>I am working on a lib part where I need to figure out the point coordinates defined by the distance and vector. See attached image... I am not after the pretty obvious one using cos(a) and sin(a) but a solution based rather on vector translations for a reason that is too long to be explained in this topic, anyway... <BR /> X1,Y1 known <BR /> X2,Y2 known <BR /> D known <BR /> X3,Y3 ???<BR /> <BR /> thanks</T></DIV><BR /><IMG src="https://community.graphisoft.com/t5/image/serverpage/image-id/37497iA3A0AF2AD90FF3A1/image-size/large?v=v2&amp;px=999" border="0" alt="VECTORQUESTION.jpg" title="VECTORQUESTION.jpg" /> Sun, 18 Feb 2007 11:15:51 GMT https://community.graphisoft.com/t5/Libraries-objects/A-math-question-a-tough-one/m-p/97461#M31691 Rob 2007-02-18T11:15:51Z https://community.graphisoft.com/t5/Libraries-objects/A-math-question-a-tough-one/m-p/97462#M31692 It's not very clear what you are trying to do... what do you mena by "vector translations"... do you mean ROT2? Sun, 18 Feb 2007 12:09:51 GMT https://community.graphisoft.com/t5/Libraries-objects/A-math-question-a-tough-one/m-p/97462#M31692 TomWaltz 2007-02-18T12:09:51Z https://community.graphisoft.com/t5/Libraries-objects/A-math-question-a-tough-one/m-p/97463#M31693 It's fairly simple if I understand correctly. You can get the distance (let's call it "D1") between the known points and use the proportion.<BR /> <PRE>D1 = SQR((X2-X1)^2 + (Y2-Y1)^2) ! per Pythagoras ! since (X3-X1)/(X2-X1) = D/D1 then it follows that... X3 = (D/D1)*(X2-X1) + X1 ! and... Y3 = (D/D1)*(Y2-Y1) + Y1</PRE> Sun, 18 Feb 2007 15:14:05 GMT https://community.graphisoft.com/t5/Libraries-objects/A-math-question-a-tough-one/m-p/97463#M31693 Anonymous 2007-02-18T15:14:05Z https://community.graphisoft.com/t5/Libraries-objects/A-math-question-a-tough-one/m-p/97464#M31694 Matthew, you have nailed it down! that's exactly what I was after...I knew I am very close but I just could not think of that last step... perfect<BR /> thanks very much. Sun, 18 Feb 2007 21:27:07 GMT https://community.graphisoft.com/t5/Libraries-objects/A-math-question-a-tough-one/m-p/97464#M31694 Rob 2007-02-18T21:27:07Z