Editing Computational geometry
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==Translation along a line== | ==Translation along a line== | ||
− | Given some line <math>Ax + By + C = 0</math> and a point <math>(x_0, y_0)</math> known to be on the line, we wish to locate another point on the line that is distance <math>h</math> away from <math>(x_0, y_0)</math>. We can do this by observing that the general point is given by <math>(x_0 - Bt, y_0 + At)</math>, which is distance <math>\sqrt{(Bt)^2 + (At)^2} = t \sqrt{A^2+B^2} | + | Given some line <math>Ax + By + C = 0</math> and a point <math>(x_0, y_0)</math> known to be on the line, we wish to locate another point on the line that is distance <math>h</math> away from <math>(x_0, y_0)</math>. We can do this by observing that the general point is given by <math>(x_0 - Bt, y_0 + At)</math>, which is distance <math>\sqrt{(Bt)^2 + (At)^2} = t \sqrt{A^2+B^2}</math> away from <math>(x_0, y_0)</math>. Therefore, we want <math>t = \frac{h}{A^2+B^2}</math>, and finally obtain <math>\left(x_0 - \frac{Bh}{A^2+B^2}, y_0 + \frac{Ah}{A^2+B^2}\right)</math>. Note that unless <math>h=0</math>, there is another possible answer, which is simply "on the other side": <math>\left(x_0 + \frac{Bh}{A^2+B^2}, y_0 - \frac{Ah}{A^2+B^2}\right)</math>. |
==Dropping a perpendicular== | ==Dropping a perpendicular== |