Develop Reference

Determining if a Line intersects with a rotated rectangle on a canvas

I have managed to find code that determines if a Line intersects with a rectangle. My problem is when the Rectangle is rotated. I have looked hi and low to find code that will give me the coordinates of the corners of the rectangle after the transform is performed with no luck. My rectangle when it is rotated is rotated around CenterX and CenterY of 0,0.
I'd appreciate any code that you may have that does this.
Thanks!
More information:
I am working on a program that I want to be able to select one or more rectangles on a canvas by drawing a line. The rectangles can be rotated.
I have tried the following code. The second function works properly for non rotated rectangles but not for rotated rectangles.:
public bool AdjustedIntersects(FrameworkElement elem, Line line)
{
double x = Canvas.GetLeft(elem);
double y = Canvas.GetTop(elem);
double X = x * Math.Cos(((RotateTransform)elem.RenderTransform).Angle)
- y * Math.Sin(((RotateTransform)elem.RenderTransform).Angle);
double Y = x * Math.Sin(((RotateTransform)elem.RenderTransform).Angle)
+ y * Math.Cos(((RotateTransform)elem.RenderTransform).Angle);
x = Canvas.GetLeft(elem) + elem.Width;
y = Canvas.GetTop(elem) + elem.Height;
double X2 = x * Math.Cos(((RotateTransform)elem.RenderTransform).Angle)
- y * Math.Sin(((RotateTransform)elem.RenderTransform).Angle);
double Y2 = x * Math.Sin(((RotateTransform)elem.RenderTransform).Angle)
+ y * Math.Cos(((RotateTransform)elem.RenderTransform).Angle);
return SegmentIntersectRectangle(X, Y,X2,Y2,
line.X1, line.Y1, line.X2, line.Y2);
}
public bool SegmentIntersectRectangle(
double rectangleMinX,
double rectangleMinY,
double rectangleMaxX,
double rectangleMaxY,
double p1X,
double p1Y,
double p2X,
double p2Y)
{
// Find min and max X for the segment
double minX = p1X;
double maxX = p2X;
if (p1X > p2X)
{
minX = p2X;
maxX = p1X;
}
// Find the intersection of the segment's and rectangle's x-projections
if (maxX > rectangleMaxX)
{
maxX = rectangleMaxX;
}
if (minX < rectangleMinX)
{
minX = rectangleMinX;
}
if (minX > maxX) // If their projections do not intersect return false
{
return false;
}
// Find corresponding min and max Y for min and max X we found before
double minY = p1Y;
double maxY = p2Y;
double dx = p2X - p1X;
if (Math.Abs(dx) > 0.0000001)
{
double a = (p2Y - p1Y) / dx;
double b = p1Y - a * p1X;
minY = a * minX + b;
maxY = a * maxX + b;
}
if (minY > maxY)
{
double tmp = maxY;
maxY = minY;
minY = tmp;
}
// Find the intersection of the segment's and rectangle's y-projections
if (maxY > rectangleMaxY)
{
maxY = rectangleMaxY;
}
if (minY < rectangleMinY)
{
minY = rectangleMinY;
}
if (minY > maxY) // If Y-projections do not intersect return false
{
return false;
}
return true;
}

A polygon intersects with a line if one of lines of the poligon overlap with the line. Then try this:
Find four conners of the rectangle by the above way.
Check if one of lines make from sequence of conners overlap with the line. I have a ideal for it:
private bool IsStraightLineOverlap(System.Windows.Shapes.Line line1, Line line2)
{
var line1Min_X = Math.Min(line1.X1, line1.X2);
var line1Max_X = Math.Max(line1.X1, line1.X2);
var line1Min_Y = Math.Min(line1.Y1, line1.Y2);
var line1Max_Y = Math.Max(line1.Y1, line1.Y2);
var line2Min_X = Math.Min(line2.X1, line2.X2);
var line2Max_X = Math.Max(line2.X1, line2.X2);
var line2Min_Y = Math.Min(line2.Y1, line2.Y2);
var line2Max_Y = Math.Max(line2.Y1, line2.Y2);
var isOverlap_X = (line1Min_X <= line2Max_X && line1Max_X >= line2Min_X);
var isOverlap_Y = (line1Min_Y <= line2Max_Y && line1Max_Y >= line2Min_Y);
return isOverlap_X && isOverlap_Y;
}

In our game suite I have to find rotated points for various purposes.
Here's an extension method:
public static class PointExtension
{
public static Point GetRotatedPoint(this Point point, Point centerPoint, double angleInDegrees)
{
double angleInRadians = angleInDegrees * (Math.PI / 180.0);
double cosTheta = Math.Cos(angleInRadians);
double sinTheta = Math.Sin(angleInRadians);
return new Point
{
X =
(int)
(cosTheta * (point.X - centerPoint.X) -
sinTheta * (point.Y - centerPoint.Y) + centerPoint.X),
Y =
(int)
(sinTheta * (point.X - centerPoint.X) +
cosTheta * (point.Y - centerPoint.Y) + centerPoint.Y)
};
}
}
I also have to work out the cells on a grid that a line passes through.
To do this I use a bresenham line algorithm.
You can google this and find various implementations.
Here's mine:
public static IEnumerable<Point> GetOrderedPointsOnLine(int x0, int y0, int x1, int y1)
{
bool steep = Math.Abs(y1 - y0) > Math.Abs(x1 - x0);
if (steep)
{
Swap<int>(ref x0, ref y0);
Swap<int>(ref x1, ref y1);
}
int dx = Math.Abs(x1 - x0);
int dy = Math.Abs(y1 - y0);
int error = (dx / 2);
int ystep = (y0 < y1 ? 1 : -1);
int xstep = (x0 < x1 ? 1 : -1);
int y = y0;
for (int x = x0; x != (x1 + xstep); x += xstep)
{
yield return new Point((steep ? y : x), (steep ? x : y));
error = error - dy;
if (error < 0)
{
y += ystep;
error += dx;
}
}
yield break;
}
The outline of a rectangle is of course 4 lines.
There is an edge case where both an edge and the line can be a variation of 4 degrees and intersection exactly picks cells don't match.
To obviate that you could use a bresenham variation which picks both cells the line partially passes through.
Or you can build two geometries and see if you get anything overlaps when you use the wpf library to detect intersection.
I'm not sure what happens if you apply a transform to a geometry and then use geometry.fillcontainswithdetail
https://learn.microsoft.com/en-us/dotnet/api/system.windows.media.geometry.fillcontainswithdetail?redirectedfrom=MSDN&view=net-5.0#overloads
There's also combinedgeometry to consider.
Pick the right options and it'll only give you what overlaps between 2 geometries.
I use neither because in game has to be very fast.
For pre game calculations I render offscreen and examine the bytes of the image I get. This is the fastest simplest way I've found to see which cells (px) an irregular shape occupies.
And another way you could approach this.

Solution 1: break the problem down into something that is easier to solve:
Pick any edge on the rectangle, represented by points P1 and P2.
Translate both the points in your rectangle and the points that define your line by -P1, so that the P1 point is now at the origin.
Calculate the angle of your edge i.e. Math.Atan2(deltaY, deltaX).
Rotate the points for both the rectangle and line in the opposite direction, so that you now have an axis-aligned rectangle.
Do your line/rectangle hit test between these new primitives, using the algorithm you already have for axis-aligned rectangles.
If you need the actual point of intersection in proper world space coordinates then rotate it forward by the angle and translate by +P1.
Solution 2: test the line against each line that forms the rectangle, your problem is now 4 line-to-line intersection tests.

Disclaimer: this is not c#. I only know javascript but the math should be the same.
The way I attack this is by creating an array for my shape that I use to store each vertex. I then use those points to determine where the objects boundaries are. I am assuming that translate and rotate work the same in c# and the objects coordinate are always axis-aligned.
When I draw my rectangle I draw it with the x (left) as -width/2 and y (top) as -height/2. This is because I am going to use translate to position it where I want it and also allow it to rotate from the center.
ctx.save();
ctx.translate(this.x, this.y)
ctx.rotate(this.rotation);
ctx.fillStyle = this.color;
ctx.fillRect(-this.width/2, -this.height/2, this.width, this.height)
ctx.restore();
I don't know if c# uses save and restore but it would be the same as just translating and rotating it back after i.e.
ctx.translate(this.x, this.y);
ctx.rotate(this.rotation);
ctx.fillStyle = this.color;
ctx.fillRect(-this.width/2, -this.height/2, this.width, this.height);
ctx.rotate(-this.rotation);
ctx.translate(-this.x, -this.y)
Also you'll notice in the snippet code that I created an array to store my x and y coordinates of each vertex
this.vertices = [];
for (let i = 0; i < 4; i++) {
//initially set to (0, 0) and updated in updateVertices()
this.vertices.push({ x: 0, y: 0 });
}
This is part of my rectangle object by the array could be global also.
The part that really matters is the math associated to updating the position of the vertices. In this snippet I use a function called updateVertices(). In this function I need to first calculate the sin and cos based on the current rotation.
let cos = Math.cos(this.rotation); //passing in radians in JS
let sin = Math.sin(this.rotation); //passing in radians in JS
Once I have that I just update the array of vertices that I created
//update Top Left Corner
this.vertices[0].x =
(this.x - this.centerX) * cos -
(this.y - this.centerY) * sin +
(this.centerX - this.width / 2);
this.vertices[0].y =
(this.x - this.centerX) * sin +
(this.y - this.centerY) * cos +
(this.centerY - this.height / 2);
Do that with all four corners. The math is slightly different for each vertex.
That's it. You have an array with all 4 vertices and can use them how you want. In this example I iterate over them passing two (adjacent) at time to my intersectLines() function to see if my vector lines intersects. Since I am testing 4 edges I use a loop to test all four sides against my vector, I do this in my passToIntersectFunction() function.
In this snippet you can use the mouse to move the vector around and see how the intersect points move.
let canvas = document.getElementById("canvas");
let ctx = canvas.getContext("2d");
canvas.width = 400;
canvas.height = 400;
let ptX, ptY;
let intersectPoints = [];
let mouse = {
x: null,
y: null
};
canvas.addEventListener("mousemove", (e) => {
mouse.x = e.x - canvas.getBoundingClientRect().x;
mouse.y = e.y - canvas.getBoundingClientRect().y;
});
class Square {
constructor() {
this.x = 100;
this.y = 100;
this.width = 50;
this.height = 50;
this.centerX = this.x + this.width / 2;
this.centerY = this.y + this.height / 2;
this.color = "red";
this.angle = 0;
this.rotation = (this.angle * Math.PI) / 180; //convert to rads
//used to store all four vertices
this.vertices = [];
for (let i = 0; i < 4; i++) {
//initially set to (0, 0) and updated in updateVertices()
this.vertices.push({ x: 0, y: 0 });
}
}
draw() {
this.angle += 0.5;
this.rotation = (this.angle * Math.PI) / 180;
ctx.translate(this.x, this.y);
ctx.rotate(this.rotation);
ctx.fillStyle = this.color;
ctx.fillRect(-this.width / 2, -this.height / 2, this.width, this.height);
ctx.rotate(-this.rotation);
ctx.translate(-this.x, -this.y);
}
drawIntersectPoint() {
ctx.fillStyle = "black";
ctx.beginPath();
ctx.arc(ptX, ptY, 3, 0, Math.PI * 2);
ctx.fill();
}
drawVertices() {
ctx.fillStyle = "blue";
ctx.beginPath();
for (let i = 0; i < this.vertices.length; i++) {
ctx.arc(this.vertices[i].x, this.vertices[i].y, 3, 0, Math.PI * 2);
ctx.fill();
ctx.closePath();
}
}
updateVertices() {
let cos = Math.cos(this.rotation);
let sin = Math.sin(this.rotation);
//update Top Left Corner
this.vertices[0].x =
(this.x - this.centerX) * cos -
(this.y - this.centerY) * sin +
(this.centerX - this.width / 2);
this.vertices[0].y =
(this.x - this.centerX) * sin +
(this.y - this.centerY) * cos +
(this.centerY - this.height / 2);
//updates Top Right Corner
this.vertices[1].x =
(this.x + this.width - this.centerX) * cos -
(this.y - this.centerY) * sin +
(this.centerX - this.width / 2);
this.vertices[1].y =
(this.x + this.width - this.centerX) * sin +
(this.y - this.centerY) * cos +
(this.centerY - this.height / 2);
//updates Bottom Right Corner
this.vertices[2].x =
(this.x + this.width - this.centerX) * cos -
(this.y + this.height - this.centerY) * sin +
(this.centerX - this.width / 2);
this.vertices[2].y =
(this.x + this.width - this.centerX) * sin +
(this.y + this.height - this.centerY) * cos +
(this.centerY - this.height / 2);
//updates Bottom Left Corner
this.vertices[3].x =
(this.x - this.centerX) * cos -
(this.y + this.height - this.centerY) * sin +
(this.centerX - this.width / 2);
this.vertices[3].y =
(this.x - this.centerX) * sin +
(this.y + this.height - this.centerY) * cos +
(this.centerY - this.height / 2);
}
}
let square = new Square();
class Vector {
constructor() {
this.x1 = 200;
this.y1 = 100;
this.x2 = mouse.x;
this.y2 = mouse.y;
}
draw() {
ctx.strokeStyle = "black";
ctx.beginPath();
ctx.moveTo(this.x1, this.y1);
ctx.lineTo(this.x2, this.y2);
ctx.stroke();
}
updateVector() {
this.x2 = mouse.x;
this.y2 = mouse.y;
this.draw();
}
}
let vector = new Vector();
function intersectLines(coord1, coord2, vector) {
//this if statement just keeps the array from constantly growing
if (intersectPoints.length > 2) {
intersectPoints.shift();
}
let x1 = coord1.x;
let x2 = coord2.x;
let y1 = coord1.y;
let y2 = coord2.y;
let x3 = vector.x1;
let x4 = vector.x2;
let y3 = vector.y1;
let y4 = vector.y2;
let d = (x1 - x2) * (y3 - y4) - (y1 - y2) * (x3 - x4);
if (d == 0) {
return;
}
let t = ((x1 - x3) * (y3 - y4) - (y1 - y3) * (x3 - x4)) / d;
let u = ((x2 - x1) * (y1 - y3) - (y2 - y1) * (x1 - x3)) / d;
if (t > 0 && t < 1 && u > 0) {
intersectPoints.push({ x: x1 + t * (x2 - x1), y: y1 + t * (y2 - y1) });
}
return;
}
function passToIntersectFunction() {
for (let i = 0; i < square.vertices.length; i++) {
intersectLines(
square.vertices[i],
square.vertices[i + 1] ?? square.vertices[0],
vector
);
}
}
function drawIntersectPoints() {
for (let i = 0; i < intersectPoints.length; i++) {
ctx.fillStyle = "black";
ctx.beginPath();
ctx.arc(intersectPoints[i].x, intersectPoints[i].y, 3, 0, Math.PI * 2);
ctx.fill();
}
}
function animate() {
ctx.clearRect(0, 0, canvas.width, canvas.height);
square.draw();
square.updateVertices();
square.drawIntersectPoint();
square.drawVertices();
vector.updateVector();
drawIntersectPoints();
passToIntersectFunction();
requestAnimationFrame(animate);
}
animate();
<canvas id="canvas"></canvas>
Sorry its not c# but maybe it can help.

Where is the flaw in my logic for finding all the coordinates whose center is covered by a square

.. a square whose opposite ends are given by (x1,y1), (x3, y3).
For example, see the image
(x1,y1)=(2,6), (x3,y3)=(8,4)
I'm thinking in terms of basic "y = mx + b" Algebra 1 stuff.
public void DrawSquare(double x1, double y1, double x3, double y3)
{
// Increment input coordinates so that they now are
// at the center of the squares
x1 += 0.5;
y1 += 0.5;
x3 += 0.5;
y3 += 0.5;
// Ensure x1 <= x2
if(x1 > x3)
{
SwapDoubles(ref x1, ref x3);
SwapDoubles(ref y1, ref y3);
}
// Get midpoints of x-y pairs
double midx = (x1 + x3) / 2;
double midy = (y1 + y3) / 2;
// Get the two lines f(x) = m1*x + b1 and g(x) = m3*x + b2
// that parallel the sides of the square and run through the center.
double m = (y3 - y1) / (x3 - x1);
if (m == 0)
m = double.Epsilon; // better way to do this?
double m1 = -m;
double m3 = 1 / m;
double b1 = midy - m1 * midx;
double b3 = midy - m3 * midx;
// Get distance from center to lines that run across the sides of the square
double d = Math.Sqrt(Math.Pow((double)x1 - midx, 2) + Math.Pow((double)y1 - midy, 2)) / Math.Sqrt(2);
DrawInRange((Coordinate<int> coord) =>
{
// distance from middle of coordinate to line 1
double d1 = ClosestDistanceToLine(m1, b1, coord.X + 0.5, coord.Y + 0.5);
// distance from middle coordinate to line 2
double d2 = ClosestDistanceToLine(m3, b3, coord.X + 0.5, coord.Y + 0.5);
return d1 <= d && d2 <= d;
});
}
// finds the closest distance beteen the line y = mx + b and the point (x0, y0)
// http://www.intmath.com/plane-analytic-geometry/perpendicular-distance-point-line.php
static double ClosestDistanceToLine(double m, double b, double x0, double y0)
{
double A = m;
double B = -1;
double C = b;
double numerator = Math.Abs(A * x0 + B * y0 + C);
double denominator = Math.Sqrt(Math.Pow(A, 2.0) + Math.Pow(B, 2.0));
return numerator / denominator;
}
Any idea why this isn't getting the proper coordinates for me?

Calculate the X -aixes Limmets of a arch from G-Code

I am trying to determine the limits of an arch for a couple of days now without much successes.I know this is more of a geometry type question than programming but here it goes.Let me first show this image to elaborate on what I am trying to achieve in C#.
In the image above you can see a Green Arc, I am trying to determine where the ‘Left Most’ point is on the X axis (and the 'right most' eventually, but for the sake of simplicity I will focus on the ‘left most’ point)
The answer of this image ‘left most’ in the X axis is 5,note: even if the arch has a greater sweep angle the answer is still 5, but if the sweep angle is less than the answer will be more than 5.
I am reading this information from a text based G-code file and here is a sample of the arc drawn in the G-code file.
G1 X10. Y15.
G3 X5. Y10. I10. J10.
The first line indicates the starting-point of the arch, witch is in this case X 10 and Y 15. The second line indicates the end-point, arch-center-point and rotation.
The G3 indicates a Counter-Clockwise rotation of the arch (G2 indicates a Clockwise Rotation).
The X & Y indicates the End-Point and the I & J is the (X & Y respectively) arch-center-point values.The first thing that I have tried is to see if I can detect the arch crosses the 180* red line without much successes mainly because I cannot seem to get angle calculations working correctly when I try to process different types of coordinates.Here is a sample of code I’m unable to finish:
else if (Gval == 3)
{
//Values read from G-Code File
double startX = Convert.ToDouble(oldXval);
double startY = Convert.ToDouble(oldYval);
double endX = Convert.ToDouble(Xval);
double endY = Convert.ToDouble(Yval);
double midX = Convert.ToDouble(Ival);
double midY = Convert.ToDouble(Jval);
//Get Start angle of Line
double startAngle = Math.Atan2(startY - midY, startX - midX);
startAngle = Math.Round(startAngle * (180.0 / Math.PI), 5);//Radiants to Degrees
startAngle = ((startAngle % 360) + 360) % 360;//Normalise
//Get End angle of line
double EndAngle = Math.Atan2(endY - midY, endX - midX);
EndAngle = Math.Round(EndAngle * (180.0 / Math.PI), 5);//Radiants to Degrees
EndAngle = ((EndAngle % 360) + 360) % 360;//Normalise
if (EndAngle == 0) EndAngle = 360;
//Get Raiduis
double raduis = Math.Round((Math.Sqrt(Math.Pow(Math.Abs(startX - midX), 2) + Math.Pow(Math.Abs(startY - midY), 2))),5);
double deltaY = (midY) - (startX);
double deltaX = (midX) - (startY);
double angleInDegrees = Math.Atan(deltaY / deltaX)*180/Math.PI;
if (startAngle <= 180 && EndAngle >= 180)
{
double LeftValue = midX - raduis;
}
}
The code above only works for some specific arcs.
I have Googled around and found mainly topics on Lines intersections and line and circle intersections but only one specifically about lines intersecting arcs, the answer is too vague to make out what is supposed to happen here is the Link
In addition to that there is also a type of arch that I think will have to be processed differently probably by some sort of Pathegorem, but I have not started on how I can do that one. But here is an image of it:
Here is the G-code :
G1 X30. Y15.
G3 X25.4 Y11.96 I30. J10.
And I think the ‘Left most’ X value is 25.4
I would like to know, if you know of a Method or a Library that can help out with this one .Thanks

This should calculate the X and Y bounds of an arc. There are many optimizations that could be made for better performance, but I think this is easier to understand.
class Bounds
{
public double MinX, MinY, MaxX, MaxY;
}
Bounds GetArcBounds(float cx, float cy, float x1, float y1, float x2, float y2)
{
var a1 = GetAngle(y1 - cy, x1 - cx);
var a2 = GetAngle(y2 - cy, x2 - cx);
var r = (float)Math.Sqrt(Math.Pow(y1 - cy, 2) + Math.Pow(x1 - cx, 2));
var bounds = new Bounds();
bounds.MinX = double.MaxValue;
bounds.MinY = double.MaxValue;
bounds.MaxX = double.MinValue;
bounds.MaxY = double.MinValue;
ExpandBounds(bounds, x1, y1);
ExpandBounds(bounds, x2, y2);
if (IsAngleInArc(a1, a2, 0))
ExpandBounds(bounds, cx + r, cy);
if (IsAngleInArc(a1, a2, (float)Math.PI * 0.5f))
ExpandBounds(bounds, cx, cy + r);
if (IsAngleInArc(a1, a2, (float)Math.PI))
ExpandBounds(bounds, cx - r, cy);
if (IsAngleInArc(a1, a2, (float)Math.PI * 1.5f))
ExpandBounds(bounds, cx, cy - r);
return bounds;
}
float GetAngle(float dy, float dx)
{
var a = (float)Math.Atan2(dy, dx);
if (a < 0) a += (float)Math.PI * 2.0f;
return a;
}
void ExpandBounds(Bounds bounds, float x, float y)
{
if (x < bounds.MinX) bounds.MinX = x;
if (y < bounds.MinY) bounds.MinY = y;
if (x > bounds.MaxX) bounds.MaxX = x;
if (y > bounds.MaxY) bounds.MaxY = y;
}
bool IsAngleInArc(float a1, float a2, float test)
{
if (a2 < a1)
{
a2 += (float)Math.PI * 2.0f;
}
if (test < a1)
{
test += (float)Math.PI * 2.0f;
}
return a1 <= test && test <= a2;
}
If you wanted to be able to do both counter-clockwise and clockwise, I believe you could change the logic to the following, although I have not tested this yet.
if (IsAngleInArc(a1, a2, 0) ^ clockwise)
ExpandBounds(bounds, cx + r, cy);
if (IsAngleInArc(a1, a2, (float)Math.PI * 0.5f) ^ clockwise)
ExpandBounds(bounds, cx, cy + r);
if (IsAngleInArc(a1, a2, (float)Math.PI) ^ clockwise)
ExpandBounds(bounds, cx - r, cy);
if (IsAngleInArc(a1, a2, (float)Math.PI * 1.5f) ^ clockwise)
ExpandBounds(bounds, cx, cy - r);

Rotate bunch of points thourgh an origin [duplicate]

This question already has answers here:
Rotating a point about another point (2D)
(6 answers)
Closed 9 years ago.
I have list of points containing x and y locations of a page. I want to apply rotation on all these points relative to any pivot point of page (currently lets assume its center).
var points = new List<Point>();
points.Add(1,1);
points.Add(15,18);
points.Add(25,2);
points.Add(160,175);
points.Add(150,97);
const int pageHeight = 300;
const int pageWidth = 400;
var pivotPoint = new Point(200, 150); //Center
var angle = 45; // its in degree.
// Apply rotation.
Do I need some formula here?

public static Point Rotate(Point point, Point pivot, double angleDegree)
{
double angle = angleDegree * Math.PI / 180;
double cos = Math.Cos(angle);
double sin = Math.Sin(angle);
int dx = point.X - pivot.X;
int dy = point.Y - pivot.Y;
double x = cos * dx - sin * dy + pivot.X;
double y = sin * dx + cos * dy + pivot.X;
Point rotated = new Point((int)Math.Round(x), (int)Math.Round(y));
return rotated;
}
static void Main(string[] args)
{
Console.WriteLine(Rotate(new Point(1, 1), new Point(0, 0), 45));
}

If you have a large number of points to rotate, you might want to precompute the rotation matrix…
[C -S U]
[S C V]
[0 0 1]
…where…
C = cos(θ)
S = sin(θ)
U = (1 - C) * pivot.x + S * pivot.y
V = (1 - C) * pivot.y - S * pivot.x
You then rotate each point as follows:
rotated.x = C * original.x - S * original.y + U;
rotated.x = S * original.x + C * original.y + V;
The above formula is the result of combining three transforms…
rotated = translate(pivot) * rotate(θ) * translate(-pivot) * original
…where…
translate([x y]) = [1 0 x]
[0 1 y]
[0 0 1]
rotate(θ) = [cos(θ) -sin(θ) 0]
[sin(θ) cos(θ) 0]
[ 0 0 1]

if u rotate a point(x,y) around point (x1,y1) by an angle some a then you need a formula...
x2 = cos(a) * (x-x1) - sin(a) * (y-y1) + x1
y2 = sin(a) * (x-x1) + cos(a) * (y-y1) + y1
Point newRotatedPoint = new Point(x2,y2)

How to find a point where a line intersects an ellipse in 2D (C#)

I need to find a point where a line (its origin is ellipse' center) intersects an ellipse in 2D... I can easily find a point on a circle, because I know an angle F and the circle' radius (R):
x = x0 + R * cosF
y = y0 + R * sinF
However I just can't figure how am I supposed to deal with an ellipse... I know it's dimensions (A & B), but what is the way of finding parameter T?!
x = x0 + A * cosT
y = y0 + B * sinT
From what I understand the parameter T (T angle) is not far from the F angle (approximately +-15 degrees in some cases), but I just can't figure how to calculate it!!!
If there is a kind hearted soul, please help me with this problem...

The standard equation of an ellipse, stationed at 0,0, is:
1 = (x)^2 / (a) + (y)^2 / (b)
Where a is 1/2 the diameter on the horizontal axis, and b is 1/2 the diameter on the vertical axis.
you have a line, assuming an equation:
y = (m)(x - x0) + y0
So, let us plug-and-play!
1 = (x)^2 / (a) + (m(x - x0) + y0)^2 / (b)
1 = x^2 / a + (mx + (y0 - mx0))^2 / b
1 = x^2 / a + (m^2 * x^2 + 2mx*(y0 - mx0) + (y0 - mx0)^2) / b
1 = x^2 / a + (m^2 x^2) / b + (2mx*(y0 - mx0) + (y0^2 - 2y0mx0 + m^2*x0^2)) / b
1 = ((x^2 * b) / (a * b)) + ((m^2 * x^2 * a) / (a * b)) + (2mxy0 - 2m^2xx0)/b + (y0^2 - 2y0mx0 + m^2*x0^2)/b
1 = ((bx^2 + am^2x^2)/(ab)) + (x*(2my0 - 2m^2x0))/b + (y0^2 - 2y0mx0 + m^2*x0^2)/b
0 = x^2*((b + a*m^2)/(ab)) + x*((2my0 - 2m^2x0)/b) + (((y0^2 - 2y0mx0 + m^2*x0^2)/b) - 1)
That last equation follows the form of a standard quadratic equation.
So just use the quadratic formula, with:
((b + a*m^2)/(ab))
((2my0 - 2m^2x0)/b)
and
(((y0^2 - 2y0mx0 + m^2*x0^2)/b) - 1)
to get the X values at the intersections; Then, plug in those values into your original line equation to get the Y values.
Good luck!

Don't do it this way. Instead check the equation that forms an ellipse and that forming a line and solve the set:
The ellipse: (x/a)^2 + (y/b)^2 = 1
Your line: y = cx
You know a, b and c, so finding a solution is going to be easy. You'll find two solutions, because the line crosses the ellipse twice.
EDIT: Note I moved your ellipse's center to (0,0). It makes everything easier. Just add (x0,y0) to the solution.

public Hits<float2> EllipseLineIntersection ( float rx , float ry , float2 p1 , float2 p2 )
{
Hits<float2> hits = default(Hits<float2>);
float2 p3, p4;
Rect rect = default(Rect);
{
rect.xMin = math.min(p1.x,p2.x);
rect.xMax = math.max(p1.x,p2.x);
rect.yMin = math.min(p1.y,p2.y);
rect.yMax = math.max(p1.y,p2.y);
}
float s = ( p2.y - p1.y )/( p2.x - p1.x );
float si = p2.y - ( s * p2.x );
float a = ( ry*ry )+( rx*rx * s*s );
float b = 2f * rx*rx * si * s;
float c = rx*rx * si*si - rx*rx * ry*ry;
float radicand_sqrt = math.sqrt( ( b*b )-( 4f * a * c) );
p3.x = ( -b - radicand_sqrt )/( 2f*a );
p4.x = ( -b + radicand_sqrt )/( 2f*a );
p3.y = s*p3.x + si;
p4.y = s*p4.x + si;
if( rect.Contains(p3) ) hits.Push( p3 );
if( rect.Contains(p4) ) hits.Push( p4 );
return hits;
}
public struct Hits<T>
{
public byte count;
public T point0, point1;
public void Push ( T val )
{
if( count==0 ) { point0 = val; count ++; }
else if( count==1 ) { point1 = val; count ++; }
else print("This structure can only fit 2 values");
}
}

I wrote a C# code for your problem and I hope you can find it helpful. the distance function inside this code calculates euclidean distance between two points in space.
wX denotes horizontal radios of ellipse and wY denotes vertical radios.
private PointF LineIntersectEllipse(PointF A, PointF B, float wX, float wY)
{
double dx = B.X - A.X;
double dy = B.Y - A.Y;
double theta = Math.Atan2(dy, dx);
double r = distance(A, B) - ((wX * wY) / Math.Sqrt(Math.Pow(wY * Math.Cos(theta), 2) + Math.Pow(wX * Math.Sin(theta), 2)));
return PointF((float)(A.X + r * Math.Cos(theta)), (float)(A.Y + r * Math.Sin(theta)));
}

Andrew Łukasik posted a good and useful answer, however it is not using regular C# types. As I wrote in the comments, I converted the code using System.Drawing objects PointF and RectangleF. I found out that if the points given as parameters are aligned as a vertical or horizontal line, then "rect" will have a width or a height equal to 0. Then, rect.Contains(point) will return false even if the point is on this line.
I also modified the "Hits" structure to check if the point pushed is not already existing, which is the case if the line is perfectly tangent, then p3 and p4 will have same coordinates, as the exact tangent point is the only crossing point.
Here is the new code taking care of all the cases :
public static Hits<PointF> EllipseLineIntersection0(float rx, float ry, PointF p1, PointF p2)
{
Hits<PointF> hits = default(Hits<PointF>);
PointF p3 = new PointF();
PointF p4 = new PointF();
var rect = default(RectangleF);
rect.X = Math.Min(p1.X, p2.X);
rect.Width = Math.Max(p1.X, p2.X) - rect.X;
rect.Y = Math.Min(p1.Y, p2.Y);
rect.Height = Math.Max(p1.Y, p2.Y) - rect.Y;
float s = (p2.Y - p1.Y) / (p2.X - p1.X);
float si = p2.Y - (s * p2.X);
float a = (ry * ry) + (rx * rx * s * s);
float b = 2f * rx * rx * si * s;
float c = rx * rx * si * si - rx * rx * ry * ry;
float radicand_sqrt = (float)Math.Sqrt((b * b) - (4f * a * c));
p3.X = (-b - radicand_sqrt) / (2f * a);
p4.X = (-b + radicand_sqrt) / (2f * a);
p3.Y = s * p3.X + si;
p4.Y = s * p4.X + si;
if (rect.Width == 0)
{
if (p3.Y >= rect.Y && p3.Y <= rect.Y + rect.Height) hits.Push(p3);
if (p4.Y >= rect.Y && p4.Y <= rect.Y + rect.Height) hits.Push(p4);
}
else if (rect.Height == 0)
{
if (p3.X >= rect.X && p3.X <= rect.X + rect.Width) hits.Push(p3);
if (p4.X >= rect.X && p4.X <= rect.X + rect.Width) hits.Push(p4);
}
else
{
if (rect.Contains(p3)) hits.Push(p3);
if (rect.Contains(p4)) hits.Push(p4);
}
return hits;
}
public struct Hits<T>
{
public byte Count;
public T P0, P1;
public void Push(T val)
{
if (Count == 0) { P0 = val; Count++; }
else if (Count == 1) { if (!P0.Equals(val)) { P1 = val; Count++; } }
else throw new OverflowException("Structure Hits can only fit 2 values.");
}
}