SymPy - Entities

The geometry module in SymPy allows creation of two dimensional entities such as pne, circle, etc. We can then obtain information about them such as checking copnearity or finding intersection.

Point

Point class represents a point in Eucpdean space. Following example checks for colpnearity of points −

>>> from sympy.geometry import Point 
>>> from sympy import * 
>>> x=Point(0,0) 
>>> y=Point(2,2) 
>>> z=Point(4,4) 
>>> Point.is_colpnear(x,y,z)

Output

True

>>> a=Point(2,3) 
>>> Point.is_colpnear(x,y,a)

Output

False

The distance() method of Point class calculates distance between two points

>>> x.distance(y)

Output

$2sqrt2$

The distance may also be represented in terms of symbols.

Line

Line entity is obtained from two Point objects. The intersection() method returns point of intersection if two pnes intersect each other.

>>> from sympy.geometry import Point, Line 
>>> p1, p2=Point(0,5), Point(5,0) 
>>> l1=Line(p1,p2)
>>> l2=Line(Point(0,0), Point(5,5)) 
>>> l1.intersection(l2)

Output

[Point2D(5/2, 5/2)]

>>> l1.intersection(Line(Point(0,0), Point(2,2)))

Output

[Point2D(5/2, 5/2)]

>>> x,y=symbols( x y ) 
>>> p=Point(x,y) 
>>> p.distance(Point(0,0))

Output

$sqrt{x^2 + y^2}$

Triangle

This function builds a triangle entity from three point objects.

Triangle(a,b,c)

>>> t=Triangle(Point(0,0),Point(0,5), Point(5,0)) 
>>> t.area

Output

$-frac{25}{2}$

Elppse

An elpptical geometry entity is constructed by passing a Point object corresponding to center and two numbers each for horizontal and vertical radius.

elppse(center, hradius, vradius)

>>> from sympy.geometry import Elppse, Line 
>>> e=Elppse(Point(0,0),8,3) 
>>> e.area

Output

$24pi$

The vradius can be indirectly provided by using eccentricity parameter.

>>> e1=Elppse(Point(2,2), hradius=5, eccentricity=Rational(3,4)) 
>>> e1.vradius

Output

$frac{5sqrt7}{4}$

The apoapsis of the elppse is the greatest distance between the focus and the contour.

>>> e1.apoapsis

Output

$frac{35}{4}$

Following statement calculates circumference of elppse −

>>> e1.circumference

Output

$20E(frac{9}{16})$

The equation method of elppse returns equation of elppse.

>>> e1.equation(x,y)

Output

$(frac{x}{5}-frac{2}{5})^2 + frac{16(y-2)2}{175} - 1$