Statement
In cartesian coordinates, the distance between the point $ P(x_P, y_P) $ and the line $ l: y = ax + b $ is given by:
$$ d(P, l) = \frac{| ax_P - y_P + b |}{\sqrt{a^2 + 1}} $$
Picture
This picture can be used to better understand this proof. Remember that all points and lines have variable positions.
Line k through P
Let $ k $ be the line where $ k \perp l $ and $ k $ goes through point $ P $.
Since $ k \perp l $, we can find the slope $ m_k $:
$$ m_k * m_l = -1 \implies m_k * a = -1 \implies m_k = -\frac{1}{a} $$
And since $ k $ goes through $ P(x_P, y_P) $, we can also find the $ y $-intercept $ b_k $:
$$ y_P = -\frac{1}{a}x_P + b_k \implies b_k = y_P + \frac{1}{a}x_P $$
Thus, this is line $ k $:
$$ k: y = -\frac{1}{a}x + y_P + \frac{1}{a}x_P $$
Intersection Q between l and k
To find the intersection point $ Q $ for lines $ l $ and $ k $, we can set them equal to eachother:
$$ ax + b = -\frac{1}{a}x + y_P + \frac{1}{a}x_P $$
$$ ax + \frac{1}{a}x = y_P + \frac{1}{a}x_P - b $$
$$ x\left(a + \frac{1}{a}\right) = y_P + \frac{1}{a}x_P - b $$
$$ x = \frac{y_P + \frac{1}{a}x_P - b}{a + \frac{1}{a}} $$
Simplify this, we get:
$$ x = \frac{y_P + \frac{1}{a}x_P - b}{\frac{a^2 + 1}{a}} = \frac{ay_P + x_P - ab}{a^2 + 1} $$
Now to get $ y_Q $, substitute and simplify:
$$ y = a\left(\frac{ay_P + x_P - ab}{a^2 + 1}\right) + b = \frac{a^2y_P + ax_P - a^2b + b(a^2 + 1)}{a^2 + 1} = \frac{a^2y_P + ax_P + b}{a^2 + 1} $$
Thus, the coordinates of point $ Q $ are given by:
$$ Q = \left(\frac{ay_P + x_P - ab}{a^2 + 1}, \frac{a^2y_P + ax_P + b}{a^2 + 1}\right) $$
Distance between Q and P
Since $ k \perp l $ and $ k $ goes through $ Q $ and $ P $, $ d(P, l) = d(Q, P) $. This distance can be found using the Pythagorean theorem.
$$ d(Q, P) = \sqrt{(x_Q - x_P)^2 + (y_Q - y_P)^2} $$
$$ d(Q, P) = \sqrt{\left(\frac{ay_P + x_P - ab}{a^2 + 1} - x_P\right)^2 + \left(\frac{a^2y_P + ax_P + b}{a^2 + 1} - y_P\right)^2} $$
To conclude, the distance between $ P(x_P, y_P) $ and the line $ l: y = ax + b $ is given by:
$$ d(P, l) = \sqrt{\left(\frac{ay_P + x_P - ab}{a^2 + 1} - x_P\right)^2 + \left(\frac{a^2y_P + ax_P + b}{a^2 + 1} - y_P\right)^2} $$
Simplifying the formula
Let's now simplify the formula. First, combine the fractions.
$$ d(P, l) = \sqrt{\left(\frac{ay_P + x_P - ab - x_P(a^2 + 1)}{a^2 + 1}\right)^2 + \left(\frac{a^2y_P + ax_P + b - y_P(a^2 + 1)}{a^2 + 1}\right)^2} $$
Next, expand $ x_P(a^2 + 1) $ and $ y_P(a^2 + 1) $ and combine like terms.
$$ d(P, l) = \sqrt{\left(\frac{ay_P + x_P - ab - a^2x_P - x_P}{a^2 + 1}\right)^2 + \left(\frac{a^2y_P + ax_P + b - a^2y_P - y_P}{a^2 + 1}\right)^2} $$
$$ d(P, l) = \sqrt{\left(\frac{ay_P - ab - a^2x_P}{a^2 + 1}\right)^2 + \left(\frac{ax_P + b - y_P}{a^2 + 1}\right)^2} $$
Append the powers to the numerator and denominator, instead of the whole fraction. Then combine the fractions
$$ d(P, l) = \sqrt{\frac{(ay_P - ab - a^2x_P)^2}{(a^2 + 1)^2} + \frac{(ax_P + b - y_P)^2}{(a^2 + 1)^2}} $$
$$ d(P, l) = \sqrt{\frac{(ay_P - ab - a^2x_P)^2 + (ax_P + b - y_P)^2}{(a^2 + 1)^2}} $$
Factor out $ -a $ from the left square on the numerator, and reorder the terms.
$$ d(P, l) = \sqrt{\frac{\bigg((-a)(-y_P + b + ax_P)\bigg)^2 + (ax_P + b - y_P)^2}{(a^2 + 1)^2}} $$
$$ d(P, l) = \sqrt{\frac{\bigg((-a)(ax_P + b - y_P)\bigg)^2 + (ax_P + b - y_P)^2}{(a^2 + 1)^2}} $$
Append the square on both factors.
$$ d(P, l) = \sqrt{\frac{(-a)^2 * (ax_P + b - y_P)^2 + (ax_P + b - y_P)^2}{(a^2 + 1)^2}} $$
$$ d(P, l) = \sqrt{\frac{a^2 * (ax_P + b - y_P)^2 + 1 * (ax_P + b - y_P)^2}{(a^2 + 1)^2}} $$
Factor out $ (ax_P + b - y_P)^2 $.
$$ d(P, l) = \sqrt{\frac{(a^2 + 1)(ax_P + b - y_P)^2}{(a^2 + 1)^2}} $$
Append the square root on each factor.
$$ d(P, l) = \frac{\sqrt{(a^2 + 1)} * \sqrt{(ax_P + b - y_P)^2}}{\sqrt{(a^2 + 1)^2}} $$
Note that $ \sqrt{x^2} = | x | $. But since $ a^2 + 1 $ is always positive, no absolute value signs are required there.
$$ d(P, l) = \frac{\sqrt{(a^2 + 1)} * | ax_P + b - y_P |}{a^2 + 1} $$
Finally, note that $ \frac{\sqrt{x}}{x} = \frac{1}{\sqrt{x}} $.
$$ d(P, l) = \frac{| ax_P + b - y_P |}{\sqrt{a^2 + 1}} $$