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clipka <ano### [at] anonymous org> wrote:
> Am 15.02.2013 15:24, schrieb Trevor G Quayle:
> > Or to really simplify it, we can combine all of the above and get formulae to
> > solve directly for x and y from the points:
> >
> > xi =
> >
[(x21-x22)*(x11*y12-x12*y11)+(x11-x12)*(x22*y21-x21*y22)]/[(x21-x22)*(y12-y11)+(x11-x12)(y21-y22)]
>
> That's a clean, nice & safe term. Note how the denominator is fully
> symmetric with respect to both x vs. y as well as line 1 vs. line 2,
> indicating that there won't be any problems with vertical lines.
>
> > yi = (xi-x11)*(y12-y11)/(x12-x11)+y11
>
> Better use the term symmetric to the one used for xi, i.e.:
>
> yi =
>
[(y21-y22)*(y11*x12-y12*x11)+(y11-y12)*(y22*x21-y21*x22)]/[(y21-y22)*(x12-x11)+(y11-y12)(x21-x22)]
>
> (Note that the denominator term is the same as for xi except for the
> sign, so you can save a few computations there.)
>
No need to even do that. You can leave the denominator the same and simply swap
only the (x21-x22) & (x11-x12) in the numerator to the equivalent ys:
[(y21-y22)*(x11*y12-x12*y11)+(y11-y12)*(x22*y21-x21*y22)]/[(x21-x22)*(y12-y11)+(x11-x12)(y21-y22)]
In this case both denominators are the same. (If you notice, in your version in
the numerator, the second terms are the same as well except for the swapped
sign)
You do also need to watch for non-intersecting lines (i.e. parallel). Parallel
lines can be recognized by the above formulation as when the denominator
[(x21-x22)*(y12-y11)+(x11-x12)(y21-y22)] is equal to zero (divide by zero).
If the lines are finite in length (only exist between the defined points), then
once the intersecton is solved, it is simply a matter of checking that the x
intersect lies between the x endpoints of both lines (or y).
For the general "y=mx +b" formulation, I did forget about the vertical line
situation, howver the solution would be quite simple as you would already know
the x intercept and would only need to solve the other, non-vertical line for
the 'y' intercept.
-tgq
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