Hi Ruchi12,

Thank you for posting here.

As I know, if you want to get the quadrilateral diagonal intersection, you could use the points with formula which you used to draw the lines.

A(x1,y1) B(x2,y2) C(x3,y3) D(x4,y4)
DiagonalAC：y-y1=(y3-y1)/(x3-x1)*(x-x1)
DiagonalBD：y-y2=(y4-y2)/(x4-x2)*(x-x2)
(y3-y1)/(x3-x1)*(x-x1)+y1=(y4-y2)/(x4-x2)*(x-x2)+y2
(y3-y1)(x4-x2)(x-x1)+y1(x3-x1)(x4-x2)=(y4-y2)(x3-x1)(x-x2)+y2(x3-x1)(x4-x2)
(y3-y1)(x4-x2)(x-x1)-(y4-y2)(x3-x1)(x-x2)=(x3-x1)(x4-x2)(y2-y1)
x[(y3-y1)(x4-x2)-(y4-y2)(x3-x1)]=(x3-x1)(x4-x2)(y2-y1)+x1(y3-y1)(x4-x2)-x2(y4-y2)(x3-x1)
x=[(x3-x1)(x4-x2)(y2-y1)+x1(y3-y1)(x4-x2)-x2(y4-y2)(x3-x1)]/[(y3-y1)(x4-x2)-(y4-y2)(x3-x1)]
y=(y3-y1)/(x3-x1)*(x-x1)+y1
=(y3-y1)[(x4-x2)(y2-y1)+(x1-x2)(y4-y2)]/[(y3-y1)(x4-x2)-(y4-y2)(x3-x1)]+y1
Hence point is
([(x3-x1)(x4-x2)(y2-y1)+x1(y3-y1)(x4-x2)-x2(y4-y2)(x3-x1)]/[(y3-y1)(x4-x2)-(y4-y2)(x3-x1)],(y3-y1)[(x4-x2)(y2-y1)+(x1-x2)(y4-y2)]/[(y3-y1)(x4-x2)-(y4-y2)(x3-x1)]+y1)

You could use the formula to get the points in your code.

But, for curves, I am nor sure what curves you used to get the point. Please provide more information about that.

Best Regards,

Wendy

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