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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}

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1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – csn899
    Apr 22, 2023 at 11:31
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    $\begingroup$ @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. $\endgroup$
    – xzczd
    Apr 22, 2023 at 11:40
  • $\begingroup$ Solve[ForAll[{k}, k^2 != 1, y == k/(k^2 - 1) x + 2 k/(k^2 - 1) - 1], {x, y}]It's ok! $\endgroup$
    – csn899
    Apr 22, 2023 at 11:51
  • $\begingroup$ SolveAlways[y == k/(k^2 - 1) x + 2 k/(k^2 - 1) - 1, k]the same $\endgroup$
    – csn899
    Apr 22, 2023 at 11:52

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