# Thread: distance between a point and a line.

1. ## distance between a point and a line.

Show that the distance from a point (xZERO,yZERO) to a line Ax+By+C=0 is given by

|AxZERO+ByZERO+C| / sqrt(A^2+B^2)

Why absolute value? I get that hes putting the points into the slope of the other one to get maybe a parallel line. but thats all I get. why the sqrt of (a^2+b^2) ?

2. Hello, frankinaround!

$\text}Show that the distance from a point }(x_o,y_o)\text{ to a line }Ax+By+C\:=\:0$

$\text{is given by: }\;d \;=\;\dfrac{|Ax_o +By_o+C|}{\sqrt{A^2+B^2}}$

Why absolute value? . Because distance is a positive measure.

I get that hes putting the points into the slope of the other one . What?
to get maybe a parallel line, but thats all I get.

Why the sqrt of (A^2+B^2) ? . Why not?

You're expected to derive that formula, aren't you?

Well, do it . . . and you'll see why.

3. Originally Posted by Soroban
Hello, frankinaround!

You're expected to derive that formula, aren't you?

Well, do it . . . and you'll see why.

That's a bit harsh, the OP obviously does not know how to...

Hint, the slope of the shortest line segment going through $\displaystyle (x_0, y_0)$ and touching $\displaystyle Ax + By + C = 0$ will be perpendicular to the slope of $\displaystyle Ax + By + C = 0$.

4. um... so im trying to do this. I can see ax/-b + c/-b = y and that the slop is a/-b, so then bx0/a should be the perpendicular slope. then I think I would need to take the point of intersection and use the distance formula to get to the next part of this question. but Im not sure how, because how do I know the point of intersection ? I have almost 2 equations. y=ax/-b+c/-b and bx/a + ? = y. I can TRY to substitute with y, so like bx/a + ? = ax/-b + c/-b. But it seems strange. what do you think?

5. the equation of a line with slope $\frac{B}{A}$ (perpendicular to the given line) through the point $(x_0, y_0)$ is $y= \frac{B}{A}(x- x_0)+ y_0$. Where does that intersect $y= -\frac{A}{B}x- \frac{C}{A}$?