You have a point with coordinates $(w,z)$.
Part I:
Suppose there were another point on the plane. Any point on the plane. We aren't worry about how to get the point you want; just any point with coordinate $(a,b)$.
So what is the distance between $(w,z)$ and $(a,b)$? Well, in the horizontal direction $w$ and $a$ are $|w - a|$ apart. In the vertical direction $b$ and $z$ are $|z-b|$ apart. The vertical and the horizontal distances form two legs of a right triangle. The hypotenuse of that triangle will be the distance between the points.
The pythagorian theorems says this hypotenuse squared is the sum of the squares of the legs. So $distance^2 = |w-a|^2 + |z-b|^2$ and so $distance = \sqrt {(w-a)^2 + (z-b)^2}$.
This is the famous distance formula. $d((x_0,y_0) =$ distance between point $(x_0, y_0)$ and $(x_1,y_1)$ $= \sqrt {(x_1-x_0)^2 + (y_1-y_0)^2}$.
Put a pin in that.
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Part II
You have a slope that is $m$. That means if you move $k$ units in the horizontal direction you will move $km$ units in the vertical direction. This work forward and backwards. If you move $-k$ units in the horizontal direction you will move $-km$ units in the vertical direct.
So if you move $k$ units forward from point $(w,z)$ your $x$- coordinate will be $w + k$, and your $y$- coordinate will be $y +km$ or the point $(w + k, z + km)$.
We can use this do find an equation for the line. A point of the line will be any point $(x,y)$ where $x = w +k$ and $y = z + km$. Manipulating this algebraically we get $k = x-w$ so $y=z +(x-w)m = mx + (z-wm)$. (Note: $z-wm = b$ equals the $y$ intercept because we had to go $w$ units back to get to $x=0$ and that means vertically we needed to go back $mw$ units.)
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Part III
So we have a point $(w,z)$ and the line. We want to move it forward (or back) $k$ units to get the point $(w + k, z + km)$ and we want $d((w,z), (w+k,z+km)) = $ to a specific distance $D$.
So we want $d((w+k, z+km)) = \sqrt{[(w+k)-w]^2 + [(z+km) - z)]^2} = D$.
Well.... let's do it. Let's solve for $k$ (which is how for we must move forward horizontally).
$\sqrt{[(w+k)-w]^2 + [(z+km) - z)]^2} = D$
$\sqrt {k^2 + (km)^2 } = D$
$k^2 + k^2m^2 = D^2$
$k^2(1 + m^2) = D^2$
$k^2= \frac {D^2}{1+m^2}$
$k = \frac {D}{\sqrt{1+m^2}}$.
So the point we want is that we go $\frac {D}{\sqrt{1+m^2}}$ horizontally, and $m*\frac {D}{\sqrt{1+m^2}}$ vertically.
So the point is $(w + \frac {D}{\sqrt{1+m^2}}, z + m\frac {D}{\sqrt{1+m^2}})$.
To go in the opposite direct you get $(w - \frac {D}{\sqrt{1+m^2}}, z - m\frac {D}{\sqrt{1+m^2}})$
So $(w \pm \frac {D}{\sqrt{1+m^2}}, z \pm m\frac {D}{\sqrt{1+m^2}})$
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Example:
Suppose you have a point $(8, 9)$ and we have a slope of $\frac 34$ and we want to find the two points that are $5$ units away.
Our formula is $(x \pm \frac {D}{\sqrt{1+m^2}}, y \pm m\frac {D}{\sqrt{1+m^2}})$
So $(8\pm \frac 5{\sqrt{1+ \frac 9{16}}}, 9\pm \frac 34\frac 5{1 + \frac 9{16}}) = $
$(8\pm \frac 5{\sqrt{\frac {25}{16}}}, 9\pm \frac 34\frac 5{\sqrt{\frac {25}{16}}})=$
$(8 \pm \frac 5{\frac 54}, 9\pm \frac 34*\frac 5{\frac 54})=$
$(8\pm 4, 9\pm 3)$
so the points are $(4, 6)$ and $(12, 12)$.
So $(
