Perpendicular distance of a point from the line.
The length of perpendicular distance of point $P(x_1,y_1)$ from line $ax+by+c=0$.
Let foot of perpendicular from point $P$ on line $L\,:\,ax+by+c=0$ be $Q(h,k)$
$$\therefore ah+bk+c=0$$ $$m_L.m_{PQ}=-1$$
$$\Rightarrow -(\frac{a}{b})(\frac{k-y_1}{h-x_1})=-1$$
$$\Rightarrow \frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{a(h-x_1)+b(k-y_1)}{a^2+b^2}$$
$$\Rightarrow \frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{(ah+bk+c)-(ax_1+by_1+c)}{a^2+b^2}$$
$$\Rightarrow \frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{-(ax_1+by_1+c)}{a^2+b^2}$$
$$\Rightarrow (h-x_1)=-a( \frac{(ax_1+by_1+c)}{a^2+b^2})$$
$$ And \;\; (k-y_1)=-b( \frac{(ax_1+by_1+c)}{a^2+b^2})$$
$$\Rightarrow PQ=\sqrt{(h-x_1)^2+(k-y_1)^2}=\cfrac{\vert ax_1+by_1+c \vert}{\sqrt{a^2+b^2}}$$
$\therefore$ Distance of point $P(x_1,y_1)$ from the line $ax+by+c=0$ is $PQ= \cfrac{\vert ax_1+by_1+c \vert}{\sqrt{a^2+b^2}}$
NOTE : – (i) Distance of origin $O(0,0)$ from the line $L:ax+by+c=0$ is $OQ=\cfrac{\vert c \vert}{\sqrt{a^2+b^2}}$
(ii) Distance between two parallel lines .
(a) If origin is on side with respect to the two lines.
Distance between line $L_1$ and $L_2$ is $AB=OB-OA$
$$=\cfrac{\vert c_2 \vert}{\sqrt{a^2+b^2}}- \cfrac{\vert c_2 \vert}{\sqrt{a^2+b^2}} $$
$$= \frac{\vert c_2-c_1 \vert}{\sqrt{a^2+b^2}}$$
Since $c_1.c_2\gt 0$ origin lie on same side with respect to the lines.
(b) If origin lie between the two lines.
If origin lies between the lines then $c_1.c_2 \lt 0 $.
And Distance between lines $L_1\;,\,L_2$ is $AB=OA+OB$.
$$=\cfrac{\vert c_1 \vert}{\sqrt{a^2+b^2}}+ \cfrac{\vert c_2 \vert}{\sqrt{a^2+b^2}} $$ $$= \cfrac{\vert c_1-c_2 \vert}{\sqrt{a^2+b^2}} \;,Since \;\;c_1.c_2 \lt 0$$
Foot of Perpendicular.
Foot of perpendicular from the given point to fixed line.
Let $Q(h,k)$ be the foot of perpendicular from $P(x_1,y_1)$ on the line $L\;:\;ax+by+c=0$
$$\Rightarrow ah+bk+c=0 \;—–(1)$$ $$Since\;\;\;m_L.m_{PQ}=-1$$
$$\therefore -\Bigl(\frac{a}{b}\Bigr)\Bigl( \frac{k-y_1}{h-x_1} \Bigr)=-1$$
$$\Rightarrow \cfrac{h-x_1}{a}=\cfrac{k-y_1}{b}=(\frac{(ah+bk+c)-(ax_1+by_1+c)}{a^2+b^2}$$
Using (i) we get $\cfrac{h-x_1}{a}=\cfrac{k-y_1}{b}=-\Bigl(\cfrac{ax_1+by_1+c}{a^2+b^2}\Bigr)$
Where (h,k) is foot of perpendicular from $(x_1,y_1)$ on the line $L\;: \;ax+by+c=0$
NOTE :- let $R(\alpha,\beta)$ be image of point $P(x_1,y_1)$ with respect to line $L\;:\;ax+by+c=0$
$\therefore Q(h,k)=Q\Bigl(\cfrac{\alpha +x_1}{2},\cfrac{\beta + y_1}{2}\Bigr)$ Lies on line $L=0$
$$\therefore a(\frac{\alpha+x_1}{2})+b(\frac{\beta+y_1}{2})+c=0$$
$$\therefore a\alpha +b\beta+c+(ax_1+by_1+c)=0$$
$$\Rightarrow \;a\alpha+b\beta+c=-(ax_1+by_1+c)\,——(2)$$
$$m_{PR}.m_L=-1\;\;\Rightarrow (\cfrac{\beta-y_1}{\alpha-x_1})(-\cfrac{a}{b})=-1$$
$$\Rightarrow \cfrac{\alpha – x_1}{a}=\cfrac{\beta-y_1}{b}=\cfrac{a(\alpha-x_1)+b(\beta-y_1)}{a^2+b^2}$$
$$\Rightarrow \cfrac{\alpha – x_1}{a}=\cfrac{\beta-y_1}{b}=\cfrac{(a\alpha+b\beta+c)-(ax_1+by_1+c)}{a^2+b^2}$$
Using (2) we get $ \Rightarrow \cfrac{\alpha – x_1}{a}=\cfrac{\beta-y_1}{b}=\cfrac{-2(ax_1+by_1+c)}{a^2+b^2}$
Where $(\alpha,\beta)$ is image of $(x_1,y_1)$ with respect to line $ax+by+c=0$