<s>[INST] Equation for Distance of the Straight line from the Origin. By reduction of the equation $ax + by + c = 0 $ of a straight line to the normal form , we get 
$$\left(\frac{-a}{\sqrt{a^2 + b^2}}\right)x + \left(\frac{-b}{\sqrt{a^2 + b^2}}\right)y = \frac{c}{\sqrt{a^2 + b^2}}$$
And,
$$p= \frac{∣c∣}{\sqrt{a^2 + b^2}}$$
And my textbook says that $p$ is the distance of the straight line from the origin. I don't know why we are getting it as a distance from origin?
I know $p= x\cos\theta + y\sin\theta$ ( where $p$ is distance of line from origin).
Also I want to know how can we relate both equations?
 [/INST] For this derivation, we won’t be needing any trigonometric equation nor you have to prior study trigonometry.
For this derivation we just need the following things:

*

*Pythagoras Theorem

*Algebra

*Area of triangle $$\frac{1}{2} \times \text{height} \times \text{base}$$
Let us take the line equation $Ax+By+C = 0$.
So by putting first $y = 0$ solving for $x$ and then putting $x = 0$ and solving for $y$.

*

*By putting $y = 0$ and solve for $x$:
\begin{align}
Ax+By+C &= 0, \quad y = 0 \\
Ax+B(0)+C &= 0 \\
Ax+C &= 0 \\
Ax &= -C \\
x &= -\frac C A
\end{align}


*By putting $x = 0$ and solve for $y$:
\begin{align}
Ax+By+C &= 0, \quad x = 0 \\
A(0)+By+C &= 0 \\
By+C &= 0 \\
By &= -C \\
y &= -\frac C B
\end{align}
By doing the above we get the base distance and perpendicular distance of the triangle formed by the line equation $Ax+By+C = 0$.
Since the triangle formed is a right-angled triangle, we can get its area and hypotenuse with no problem.

*

*So first, deriving the hypotenuse:
$$\text{Rise} = \left|\frac{-C}{A}\right|,\ \text{Run} = \left|\frac{-C}{B}\right|$$
Pythagoras Thm.:  Hyp.${}^2$ = Rise${}^2$ + Run${}^2$:
\begin{align}
\text{Hyp.}^2 &= \left|\frac{-C}{A}\right|^2 + \left|\frac{-C}{B}\right|^2 \\
&= \frac{C^2}{A^2} + \frac{C^2}{B^2} \\
&= \frac{C^2(A^2+B^2)}{(AB)^2}; \\
\text{Hyp.} &= \sqrt\frac{C^2(A^2+B^2)}{(AB)^2} \\
&= \frac{|C|\sqrt{A^2+B^2}}{|AB|}.
\end{align}


*Deriving the Area:
\begin{align}
\text{Base} &= \left|\frac{-C}{A}\right|, \\
\text{Height} &= \left|\frac{-C}{B}\right|; \\
\text{Area} &= \frac 1 2 \times \left|\frac{-C}{A}\right| \times \left|\frac{-C}{B}\right| \\
&= \frac{C^2}{2|AB|}.
\end{align}
We can also write the area as half the hypotenuse times the height that is perpendicular to the hypotenuse (Note: the height perpendicular to the hypotenuse is the shortest distance to the origin, so let us call this height $w$).
\begin{align}
\text{Area} &= \frac 1 2 \times \text{hypotenuse} \times \text{perp. height} \\
\frac{C^2}{2|AB|} &= \frac 1 2 \times \frac{|C| \sqrt{A^2+B^2}}{|AB|} \times w \\
\frac{C^2}{|AB|} &= \frac{|C|\sqrt{A^2+B^2}}{|AB|} \times w \\
|C| &= \sqrt{A^2+B^2} \times w \\
\frac{|C|}{\sqrt{A^2+B^2}} &= w.
\end{align}
This is the answer.
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