Using The Pearson Complete Guide For Aieee 2/e
By Khattar as a point of reference for the items below.
You want to convert
$$\tag 1 x-2y-3=0$$
into normal (perpendicular) form.
We have:
$$Ax + By = -C \rightarrow (1)x + (-2)y = -(-3)$$
This means $C \lt 0$, so we divide both sides of $(1)$ by:
$$\sqrt{A^2 + B^2} = \sqrt{(1)^2 + (-2)^2} = \sqrt{5}$$
This yields:
$$\dfrac{1}{\sqrt{5}} x - \dfrac{2}{\sqrt{5}} y = -\dfrac{-3}{\sqrt{5}}$$
In normal form, this is:
$$\cos(\alpha)~x + \sin(\alpha)y = p \rightarrow \dfrac{1}{\sqrt{5}} x - \dfrac{2}{\sqrt{5}} y = -\dfrac{-3}{\sqrt{5}}$$
Note that this has a positive $x$ coordinate, and a negative $y$ coordinate, which puts us in the $4^{th}$ quadrant.
So, we have:
$$\cos(\alpha) = \dfrac{1}{\sqrt{5}}, \sin(-\alpha) = -\dfrac{2}{\sqrt{5}} , p = -\dfrac{-3}{\sqrt{5}} = \dfrac{3}{\sqrt{5}}$$
This gives us:
$$\alpha \approx 1.10714871779409 ~\mbox{radians}~ \approx 63.434948822922 {}^{\circ}$$
Since we are in the $4^{th}$ quadrant, we can write this angle as:
$$\alpha \approx 2 \pi - 1.10714871779409 ~\mbox{radians}~ \approx 5.176036589385497 ~\mbox{radians}~ \approx 296.565 {}^{\circ}$$
As another reference point, see $10.3.4$ (including examples) at Straight Lines.
There is a simple derivation of what you want to know. Without going into details, let me introduce the result:
The perpendicular distance of a line $(Ax+By+c=0)$from a point is equal to $|\frac{Ax'+By'+c}{\sqrt{A^2+B^2}}|$. Where $(x',y')$ are coordinates of the point.
Since you want distance of line from origin, the coordinates become $(0,0)$ and hence the perpendicular distance of a line from origin is $|\frac{Ax'+By'+c}{\sqrt{A^2+B^2}}|=|\frac{0+0+c}{\sqrt{A^2+B^2}}|=|\frac{c}{\sqrt{A^2+B^2}}|$.
Best Answer
The two points $A$ and $B$ lie on the circle of radius $c$ centred on the origin $O$, having arguments $\alpha$ and $\beta$ respectively. Now consider $\triangle AOB$; because $AO=OB$ the triangle is isosceles and the perpendicular to $AB$ through $O$ bisects $AB$. This completes the proof.