如何使用 Sympy 根据指定变量重新排列方程？

Question

I can write, for example,

Line(Point(3,-4), Point(-2,2)).equation()

to generate an equation of a line that passes through those points, but the output is given as

-6x - 5y - 2

presumably being equivalent to -6x - 5y - 2 = 0 . How can I instead set the output to be

y = (-6/5)x - (2/5)

I thought it might have to do with some formatting settings in the equation() method, so I checked the documention , but it didn't say anything about it.

EDIT 1: It appears that if I input

solve(Eq(Line(Point(3,-4), Point(-2,2)).equation()))

I will get the (albeit ugly) output

⎡⎧     5⋅y   1⎫⎤
⎢⎨x: - ─── - ─⎬⎥
⎣⎩      6    3⎭⎦

This does give a rearranged solution, but the issue is that this is only in the form of x=f(y) . I'm not sure how I would get it to instead be in terms of y=f(x) .

EDIT 2: I think this might actually be a bug with solve() , or Eq() . If I instead manually type

solve(-6*x-5*y-2,y)

or

solve(Eq(-6*x-5*y-2),y)

I will get the (somewhat ugly, but correct) output of

⎡  6⋅x   2⎤
⎢- ─── - ─⎥
⎣   5    5⎦

Now if I were to instead type

solve(Eq(Line(Point(3,-4), Point(-2,2)).equation()),y)

or

solve(Eq(Line(Point(3,-4), Point(-2,2)).equation()),x)

I get the output of

[]

This is rather strange, though, because

Eq(-6*x-5*y-2)

and

Eq(Line(Point(3,-4),Point(-2,2)).equation())

both output

-6⋅x - 5⋅y - 2 = 0

so I'm really not sure what Is going on here.

Answer 1

Since you need to pass the symbol you want to solve for, you can do something like this (Although the equation function expects strings )

x, y = symbols('x, y')
eq = (Line(Point(3,-4), Point(-2,2)).equation(x,y))

print(solve(eq, y))# prints [-6*x/5 - 2/5]

or you can get the symbol from the expression and pass it to solve like

eq = (Line(Point(3,-4), Point(-2,2)).equation())

print(solve((eq), list(eq.atoms(Symbol))[1])) # prints [-6*x/5 - 2/5]

Answer 2

The symbols used in the equation have assumptions on them and that makes them different from the plain symbols you might create with Symbol('x') . So that's why it has the option to pass in the symbols you want to use for x and y .

>>> var('x y')
(x, y)
>>> Line(Point(3,-4), Point(-2,2)).equation(x,y)
-6*x - 5*y - 2
>>> Eq(y, solve(_, y)[0])
Eq(y, -6*x/5 - 2/5)

That's also why you didn't get a solution in one of your examples -- the variable you were solving for wasn't in the equation. It looked the same but it wasn't since it had different assumptions on it.