# How to automatically solve the coordinates of a fixed point passing through a parametric linear equation? [closed]

After previous calculations, this equation was obtained like this:

y == 1 - 2 k + k x


How to find the coordinates of the fixed point it passes through?

The principle of manual calculation is as follows:

1. Extract the common factor for parameter k to obtain the equation:
y==1+k(x-2)

1. The formula in parentheses containing parameters is equal to 0, and the corresponding fixed point coordinate value is obtained by solving it
x-2==0     so  x==2


3.Since the parameter containing part is 0, we can naturally obtain the remaining coordinate values of the fixed point

y==1


So this straight line equation passes through a fixed point {2, 1}

Once again, you should put more effort in understanding the answers you've obtained:

Solve[ForAll[{k}, y == 1 - 2 k + k x], {x, y}]
(* {{x -> 2, y -> 1}} *)

SolveAlways[y == 1 - 2 k + k x, k]
(* {{x -> 2, y -> 1}} *)


The method is the same as that used here, and Domen already mentioned the relevant post here.

• Solve[ForAll[{k}, y == k/(k^2 - 1) x + k/(k^2 - 1) - 1], {x, y}]Why does this equation have no solution at a fixed point? Apr 22, 2023 at 11:31
• @csn899 Because k^2 != 1. Similar issue has been mentioned in this answer you received, so, for the third time: please put more effort in understanding the answers you've obtained. Apr 22, 2023 at 11:40
• Solve[ForAll[{k}, k^2 != 1, y == k/(k^2 - 1) x + 2 k/(k^2 - 1) - 1], {x, y}]It's ok! Apr 22, 2023 at 11:51
• SolveAlways[y == k/(k^2 - 1) x + 2 k/(k^2 - 1) - 1, k]the same Apr 22, 2023 at 11:52