\(\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 0}}\quad ,&{{ 32}})\quad
% (c,d)
&({{ 100}}\quad ,&{{ 212}})
\end{array}
\\\quad \\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{9}{5}
\\ \quad \\\\
% point-slope intercept
y-{{ \underline{32}}}={{ \cfrac{9}{5}}}(x-{{ 0}})\implies y=\cfrac{9}{5}x+\underline{32}\)
now... notice, your values are 0, 32 and 100, 212 for x, y
that means "x" is the Celsius and "y" is the Farenheit
--------------------------------------------------------------------------
\(\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 32}}\quad ,&{{ 0}})\quad
% (c,d)
&({{ 212}}\quad ,&{{ 100}})
\end{array}
\\\quad \\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{5}{9}
\\ \quad \\\\
% point-slope intercept
y-0={{ \cfrac{5}{9}}}(x-{{ 32}})\implies y=\cfrac{5}{9}x-\cfrac{5\cdot 32}{9}+0
\\\\\\
y=\cfrac{5}{9}x-\cfrac{160}{9}\)
--------------------------------------------------------------------------
now, let us solve the first equation for "x"
\(\bf y=\cfrac{9}{5}x+32\implies y-32=\cfrac{9}{5}x\implies 5y-160=9x
\\\\
\cfrac{5y-160}{9}=x\implies \cfrac{5}{9}y-\cfrac{160}{9}=x\)
so, you do end up with the same value, if you make"y" or "x" for the Farenheit
