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I have been trying to find the intersections between a rectangle and a line, following the example given in the Solve function:

Solve[{x, y} ∈ Line[{{-1, 0}, {2, 1}}] && {x, y} ∈ 
   Circle[], {x, y}]
Graphics[{{Blue, Line[{{-1, 0}, {2, 1}}], 
   Circle[]}, {PointSize[Large], Red, Point[{x, y}] /. %}}]

this code works well of course, finding both intersections between the circle and the line. However, if I try to substitute the circle with a rectangle the code stops working:

lin := Line[{{-2, -3}, {2, 1}}]
rec := Rectangle[{-1, -1}, {1, 1}]
Solve[{x, y} ∈ lin && {x, y} ∈ rec, {x, y}]
Graphics[{{Red, lin, Blue, rec}, {PointSize[Large], Yellow, 
   Point[{x, y}] /. %}}]

Any idea of how to make it work?

Thanks in advance

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The problem is that Rectangle has dimension 2, so the intersection of lin and recis a line and not 2 points. You can fix this by taking the boundary of the Rectangle instead:

lin = Line[{{-2, -3}, {2, 1}}]
rec = RegionBoundary @ Rectangle[{-1, -1}, {1, 1}]
Solve[{x, y} ∈ lin && {x, y} ∈ rec, {x, y}]
Graphics[{
    {Red, lin, Blue, rec},
    {PointSize[Large], Yellow, Point[{x, y}] /. %}
}]

Line[{{-2, -3}, {2, 1}}]

Line[{{-1, -1}, {1, -1}, {1, 1}, {-1, 1}, {-1, -1}}]

{{x -> 0, y -> -1}, {x -> 1, y -> 0}}

enter image description here

You could also just use RegionIntersection instead:

Graphics[{Red, lin, Blue, rec, PointSize[Large], Yellow, RegionIntersection[lin, rec]}]

enter image description here

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  • $\begingroup$ thanks, I was about to write that I solved the issue using a polyline instead of a rectangle. Your solution seems much simpler in practice though. I have been trying to implement your code, but for some reason the RegionBoundary function seems not to work (it remains blue in the code). It looks like the built-in function is read as a variable with no value assigned... I am perplexed $\endgroup$ – saimon Apr 16 at 14:40
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    $\begingroup$ @saimon RegionBoundary was introduced in M10, so you must be using an earlier version. In that case, you need to convert the Rectangle to a polyline yourself $\endgroup$ – Carl Woll Apr 16 at 14:51

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