I have doubt in derivation. Pls clarify
Let points be $(a, f(a)), (c,0), (b,f(b))$
$y-f(a) = m (x-a) = \frac{f(b)-f(a)}{b-a}(x-a)$
Substituting y=0, x=c,
$0-f(a) = \frac{f(b)-f(a)}{b-a}(c-a) \implies \fbox {c=a-f(a) ( $\frac{b-a}{f(b)-f(a)}$)}$ ---->(1)
But if i use geometry in uploaded figure
$c = a+ f(a) cot \theta$
where $\cot \theta = (1/slope ) = (1/tan \theta) = \frac{(f(b)-f(a))}{(b-a)}$
here $\theta = \angle ACB =$ slope of line AC = $ \tan^{-1}\frac{(f(b)-f(a))}{(b-a)}$
this means $c=a+f(a)/\tan \theta = \fbox {$a+f(a) \frac{b-a}{f(b)-f(a)}$}$ ---> (2)
If we compare (1) and (2) equations, i am getting sign change i..e, a+() and a-()
But in textbook regular falsi method final equation is given by (1) and not (2).

Geometry is independent of coordinate axis. Even if i remvoe coordinate axis it is magnitude that we need to consider right... Kindly enlighten me ...
– Magneto Jun 10 '18 at 18:27