
Distances (in general)
I've asked several of my friends about this question, but nobody knows where to start with it, and unfortunately, neither do I. Here it is:
Given point A(x1,y1) and the line ax+by+c=0, show the distance from Point A to the line is d=[ax1+by1+c/root(a^2 + b^2)]. Note that ax+by+c=0 corresponds to a vertical line if b=0 and to a horizontal line if a=0.
Any help would be appreciated. Thank you

Hello,
It depends on what you can use as tools. (For example, what is the definition of a "distance"? Can we use vectors or the complex plane? etc.)
The easiest way would be to use the CauchySchwarz inequality: $\displaystyle \sqrt{a^2+b^2}\sqrt{p^2+q^2}\geq ap+bq$.
We have to minimize (xx_1)^2+(yy_1)^2 where ax+by+c=0.
By the above inequality, $\displaystyle \sqrt{a^2+b^2}\sqrt{(xx_1)^2+(yy_1)^2}\geq a(xx_1)+b(yy_1)$. Substitute ax+by=c into this and you get the formula.
Let B(x_0, y_0) be the point on the line realizing the distance: d=AB. If you know that AB is perpendicular to the line, there are many more proofs.For example, the line AB is b(xx_1)+a(yy_1)=0, so you can determine B with the condition ax_0+by_0+c=0. (You can rephrase this by using vectors and saying that (xx_1, yy_1) is parallel to (a, b).)
If you know trigonometrics, represent any point on the line by $\displaystyle (x_1+r\cos t, y_1+r\sin t).$ As it is on the line, $\displaystyle a(x_1+r\cos t)+b(y_1+r\sin t)+c=0$ which is $\displaystyle ax_1+by_1+c+r\sqrt{a^2+b^2}\sin (t+\theta)=0$ for some $\displaystyle \theta$. The minimum of r is attained when $\displaystyle \sin(t+\theta)$ is either 1 or 1.
Bye.