Foot of perpendicular proof. I know there is few answer on Foot of perpendicular, but my doubt is different, so please don't mark this as the duplicate.
My book says:

Foot of the perpendicular from a point $(x_1,y_1)$ on the line $ax+by+c=0$ is $$\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=-\dfrac{ax_1+by_1+c}{a^2+b^2}$$

My proof:
Slope of line $\perp$ to $ax+by+c=0$ is $\frac{b}{a}\implies\tan\theta=\frac{b}{a}\implies\begin{cases}\cos\theta &=\pm\frac{a}{\sqrt{a^2+b^2}}\\ \sin\theta &=\frac{b}{\sqrt{a^2+b^2}}\end{cases}$
Let $L$ passes through point $(x_ 1,y_1)$ perpendicular to $ax+by+c=0$.
Let $r$ be the algebraic distance of $(x_ 1,y_1)$ from $ax+by+c=0$ $\implies r=\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}$.
Now co-ordinate of any point on $L$ distance $r$ from point $(x_1,y_1)$ can be given as:$$\bigg(x_1+r\cos \theta,\ y_1+r\sin\theta\bigg)$$, where $\theta$ is angle $L$ makes with positive direction of $x$-axis.
Substituting the values:$$\bigg(x_1+\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{a}{\sqrt{a^2+b^2}},\ y_1+\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{b}{\sqrt{a^2+b^2}}\bigg)$$ Now I didn't wrote $\pm$ with $\cos$, whose sign can be calculated as of $\tan$.
Now camparing gives :
$x=x_1+\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{a}{\sqrt{a^2+b^2}}$,   $\ \ \ y=y_1+\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{b}{\sqrt{a^2+b^2}}$
Note, I didn't mixed the denominators $\bigg(\sqrt{a^2+b^2}\bigg)$ to $\bigg(a^2+b^2\bigg)$, as I've calculated $\frac{a}{\sqrt{a^2+b^2}}$ from $\tan\theta$ with sign.
When I applied this on some problems gives me the correct answers but is not in the bookish form please help me to do this.

First the algebraic distance $r$ should be $\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$ depending on the position of $(x_1;y_1)$ for a given system of coordinates.
The correct formula is $$\dfrac{x-x_1}{a}=-\dfrac{ax_1+by_1+c}{a^2+b^2}.$$
To see this take the scalar product to be zero for $(x,-\dfrac{ax}{b}-\frac{c}{b})$ is on the line of direction $\overrightarrow{v}(1;-\frac{a}{b})$:
$$(x_1-x)\times 1+(y_1-(-\dfrac{ax}{b}-\frac{c}{b}))(\dfrac{-a}{b})=0.$$
