How does changing the constant parameter in a linear equation affect the line? Imagine we have the equation: $$11x - 9y = D$$
What affect does changing D have on the line? Here's what I have so far. Changing D amounts to moving the line along its orthogonal complement by some distance. How might we find an expression for this distance? Thanks
 A: Changing $D$ will increase the y-intercept of the line by D/9.
Rewriting into slope intercept form we have $y = -\frac{11}{9}x+\frac{D}{9}$. So increasing D by 1 will increase the y intercept of the line by $\frac{1}{9}$. All lines you get from changing D are parallel because they are just vertical shifts of each other.
A: Eric Jin helped me see this. But in general, if we have the line: $$Ax + By = D$$
the amount D we shift by does indeed change the x-intercept by: $$\frac{D}{A}$$ and the y-intercept by: $$\frac{D}{B}$$
So we can then say that since changing $D$ does move our line along it's orthogonal compliment (remains to be proven), the distance it will move is given by: $$(\frac{D}{2A})^2 + (\frac{D}{2B})^2$$ The extra factor of 1/2 comes in because it's the midpoint between the x and y intercepts that the orthogonal compliment bisects.
A: The perpendicular through the origin to the line $Ax+By=D$ has the equation $Bx-Ay=0$. You can find that the intersection of these two lines is $\left({AD\over A^2+B^2},{BD\over A^2+B^2}\right)$. The distance of this point from the origin is equal to ${\lvert D\rvert\over\sqrt{A^2+B^2}}$. So, the absolute value of $D$ is equal to the distance of the line from the origin times the length of the line’s normal vector $\mathbf n=(A,B)$. When $D=0$, the line of course passes through the origin.  
On the other hand, when $D\ne0$, we can also look at which multiple of $\mathbf n$ lies on the line. Substituting the coordinates of $\lambda\mathbf n$ into its equation produces the equation $\lambda(A^2+B^2)=D$, from which we can see that the sign of $D$ tells you if $\mathbf n$ points toward or away from the line.  
So, if you rescale the equation of the line by dividing both sides by $\lVert\mathbf n\rVert=\sqrt{A^2+B^2}$ to get the equation $A'x+B'y=D'$, changing $D'$ changes the distance of the line from the origin by the same amount; whether that movement is toward or away from the origin depends on the sign of $D$. So changing $D$ in the original equation makes a proportional change to the distance from the origin.
