一点到直线的距离

Let us say we have a point and a line, as shown.

How would we find the distance between the point and the line? Hint: Think about what this means. The distance between a point and a line is defined as the shortest possible distance between the point and a point on the line.

So we, of course drop a perpendicular from the point to the line, because the Pythagorean theorem tells us that this gives the minimum distance. So we do so.

The distance $AD$ is our answer! So how would you do that with coordinates? If we are given the equation of the line, we can find the slope of any line perpendicular to it. To find the one that crosses $A$, we just substitute the coordinates for $A$. Then, we can find the coordinates of $D$ and we'd just have to use the distance formula to find $AD$.

We leave it as an exercise for the reader to prove that the distance between the line represented $ax+by+c=0$ and $(x_0,y_0)$ is $$\frac {|ax_0+by_0 + c|}{\sqrt{a^2+b^2}}.$$

Review of the Cartesian Plane

We all know the coordinate system: $(x,y)$ refers to the point that lies $x$ along the x-axis and $y$ along the y-axis, lines have certain common forms like $y=mx+b$, even weirder shapes like all the conics (circles, hyperbolas, parabolas, etc.) have nice quadratic representations. So if we can make all these nice shapes, why not try to do some geometry? Common Formulae

Distance from point to a line

Think about the logic used to answer that in general. If you were given a random point, let's call it $(x_0, y_0)$, and a random line, let's call it $ax + by + c = 0$, then how would we find the required distance? Well, first we'd have to find the foot of the projection from our point to our line; this would require us to find a new line through our point perpendicular to our original line. But we know how to do this! And once we do, we can solve for the projection and then use the distance formula! After all this work, we actually end up with the nice looking formula of: $\frac{|ax_0 + by_0 + c|}{\sqrt{a^2+b^2}}$. Try to work out all the details for yourself!

Here is the solution to the above problem:

$$\begin{align} \frac{|ax_0 + by_0 + c|}{\sqrt{a^2+b^2}} &= \frac{|1(0)+ 2(0) + (-2)|}{\sqrt{1^2+2^2}}\\ &= \frac{|-2|}{\sqrt{1+4}}\\ &= \frac{2}{\sqrt{1+4}}\\ &= \frac{2}{\sqrt{5}}\\ \end{align}$$