[java] Write a two-dimensional transformation library by implementing the following API:public class PolygonTransform { // Returns a new array object that is an exact copy of the given array. // The given array is not mutated. public static double[] copy(double[] array) // Scales the polygon by the factor alpha. public static void scale(double[] x, double[] y, double alpha) // Translates the polygon by (dx, dy). public static void translate(double[] x, double[] y, double dx, double dy) // Rotates the polygon theta degrees counterclockwise, about the origin. public static void rotate(double[] x, double[] y, double theta) // Tests each of the API methods by directly calling them. public static void main(String[] args) } 1. Polygon transformpolygon is defiıned by its sequence of vertices (xo, y o), (x 1, y 1), (x 2, Y 2), .... In Java, we will represent a polygon by storing the x- and y-Write a library of static methods that performs various geometric transforms on polygons. Mathematically, acoordinates of the vertices in two parallel arrays x[] and y[].(1, 2)(0, 1)// Represents the polygon with vertices (0, 0), (1, 0), (1, 2), (0, 1).double x[ ]double y[]{0, 1, 1, о };{0, о, 2, 1 };%3D(0, 0)(1, 0)Three useful geometric transforms are scale, translate and rotate.o Scale the coordinates of each vertex (x i, Y i) by a factor a.Xi = a X¡yi = a yo Translate each vertex (x , Y ) by a given offset (dx, dy).X; = X¡ + dxYi = Yi + dyo Rotate each vertex (x ;, y ) by 0 degrees counterclockwise, around the origin.X¡ = Xj Cos 0– y sin 0Yi = Y cos e + x¡ sin 0 Note that the transformation methods scale(), translate() and rotate() mutate the polygons. Here are some example test cases (testsfor copy () are not shown):// Scales polygon by the factor 2.StdDraw.setScale(-5.0, +5.0);double[] x = { 0, 1, 1, 0 };double[] y = { 0, 0, 2, 1 };double alphaStdDraw.setPenColor(StdDraw.RED);StdDraw.polygon(x, y);scale (x, y, alpha);StdDraw.setPenColor(StdDraw.BLUE);StdDraw.polygon(x, y);|// Translates polygon by (2, 1).StdDraw.setScale(-5.0, +5.0);double[] x =double[] y = { 0, 0, 2, 1 };double dx = 2.0, dyStdDraw.setPenColor(StdDraw.RED);StdDraw.polygon(x, y);translate(x, y, dx, dy);StdDraw.setPenColor(StdDraw.BLUE);StdDraw.polygon(x, y);|// Rotates polygon 45 degrees.StdDraw.setScale(-5.0, +5.0);double[] x = { 0, 1, 1, 0 };double[] y = { 0, 0, 2, 1 };double theta = 45. 0;StdDraw.setPenColor(StdDraw.RED);StdDraw.polygon(x, y);rotate(x, y, theta);StdDraw.setPenColor(StdDraw.BLUE);StdDraw.polygon(x, y);{ 0, 1, 1, 0 };= 2.0;= 1.0;(0, 0)(0, 0)(0, 0)

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[java] Write a two-dimensional transformation library by implementing the following API:

public class PolygonTransform { // Returns a new array object that is an exact copy of the given array. // The given array is not mutated. public static double[] copy(double[] array) // Scales the polygon by the factor alpha. public static void scale(double[] x, double[] y, double alpha) // Translates the polygon by (dx, dy). public static void translate(double[] x, double[] y, double dx, double dy) // Rotates the polygon theta degrees counterclockwise, about the origin. public static void rotate(double[] x, double[] y, double theta) // Tests each of the API methods by directly calling them. public static void main(String[] args) }

1. Polygon transform
polygon is defiıned by its sequence of vertices (xo, y o), (x 1, y 1), (x 2, Y 2), .... In Java, we will represent a polygon by storing the x- and y-
Write a library of static methods that performs various geometric transforms on polygons. Mathematically, a
coordinates of the vertices in two parallel arrays x[] and y[].
(1, 2)
(0, 1)
// Represents the polygon with vertices (0, 0), (1, 0), (1, 2), (0, 1).
double x[ ]
double y[]
{0, 1, 1, о };
{0, о, 2, 1 };
%3D
(0, 0)
(1, 0)
Three useful geometric transforms are scale, translate and rotate.
o Scale the coordinates of each vertex (x i, Y i) by a factor a.
Xi = a X¡
yi = a y
o Translate each vertex (x , Y ) by a given offset (dx, dy).
X; = X¡ + dx
Yi = Yi + dy
o Rotate each vertex (x ;, y ) by 0 degrees counterclockwise, around the origin.
X¡ = Xj Cos 0– y sin 0
Yi = Y cos e + x¡ sin 0

Note that the transformation methods scale(), translate() and rotate() mutate the polygons. Here are some example test cases (tests
for copy () are not shown):
// Scales polygon by the factor 2.
StdDraw.setScale(-5.0, +5.0);
double[] x = { 0, 1, 1, 0 };
double[] y = { 0, 0, 2, 1 };
double alpha
StdDraw.setPenColor(StdDraw.RED);
StdDraw.polygon(x, y);
scale (x, y, alpha);
StdDraw.setPenColor(StdDraw.BLUE);
StdDraw.polygon(x, y);
|// Translates polygon by (2, 1).
StdDraw.setScale(-5.0, +5.0);
double[] x =
double[] y = { 0, 0, 2, 1 };
double dx = 2.0, dy
StdDraw.setPenColor(StdDraw.RED);
StdDraw.polygon(x, y);
translate(x, y, dx, dy);
StdDraw.setPenColor(StdDraw.BLUE);
StdDraw.polygon(x, y);
|// Rotates polygon 45 degrees.
StdDraw.setScale(-5.0, +5.0);
double[] x = { 0, 1, 1, 0 };
double[] y = { 0, 0, 2, 1 };
double theta = 45. 0;
StdDraw.setPenColor(StdDraw.RED);
StdDraw.polygon(x, y);
rotate(x, y, theta);
StdDraw.setPenColor(StdDraw.BLUE);
StdDraw.polygon(x, y);
{ 0, 1, 1, 0 };
= 2.0;
= 1.0;
(0, 0)
(0, 0)
(0, 0)

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