# Thread: Lines in a Plane 10

1. ## Lines in a Plane 10

Prove that the shortest distance from a point to a line is the distance measured along the perpendicular from the point to the line.

2. Originally Posted by skeske1234
Prove that the shortest distance from a point to a line is the distance measured along the perpendicular from the point to the line.
Assume it isn't -- take a point $p$ which is on the line and not perpendicular to the point and creates the shortest distance. Sketch this and try seeing what you can do with the right triangle formed by the original point and $p$

3. Let (p,q) be a point in the plane and y=ax+b be a line in the plane

Then the distance from (p,q) to any point on the line is $\sqrt{(x^2-p^2)+((ax+b)^2-q^2)}$ here we took x to be x and y to be ax+b

So we need to minimize the distance

It should be obvious that to minimize the distance we can minimize the square of the distance, or in other words, we can ignore the radical sign and minimize what is under it

From calculus, to minimize something look at the derivative when it equals zero

Let E=the square of the distance, then

$E'=2x+2(ax+b)*a=0$

$2x+2a^2x+2ab=0$

$x(1+2a)=-ab$

$x=-\frac{ab}{2a+1}$

Now use the second derivative test to confirm this is a minimum and not a maximum

$E''=2+2a^2$ this quantity is always positive, as a^2 is always positive for all a, so we do have a local min

This means that the closest point to our point (p,q) on the line is $(-\frac{ab}{2a+1}, a(-\frac{ab}{2a+1})+b)=(-\frac{ab}{2a+1}, -\frac{a^2b}{2a+1}+b)$

Now all you have to do is come up with the line perpindicular to our line that goes through point (p,q) and show that it contains the point we just found

*there is probably an easier, shorter way that I can't come up with at the moment, but whenever I see prove things about min and max, my brain jumps to calculus... sorry

4. Originally Posted by Defunkt
Assume it isn't -- take a point $p$ which is on the line and not perpendicular to the point and creates the shortest distance. Sketch this and try seeing what you can do with the right triangle formed by the original point and $p$
ok, so this is what I have.. can you or someone check to see if my method of "proving" this is valid?

assuming it isn't true: (like defunkt says)
Let point be (5,4)
A be (0,0)
B be (-1,0)
let theta be 30 degrees
cos(theta)=a/h
h=a/(costheta)
h=sqrt1/cos30
=1.15

now assuming it is true: (what the question asks to prove)

Let point be (5,4)
A be (0,0)
B be (1,0)
let theta be 30 degrees
tan(theta)=o/a
atan(theta)=o
sqrt1 (tan30)=o
0.577=o

so the shortest distance from a point to a line is the distance measured along the perpendicular from the point to the line (which is the second method above)

So let me know if this is a valid method of proving this question.
Thanks!

5. if you haven't taken a proofs class yet then proofs can be very challenging but you cannot prove a general statement with examples... you've only shown it holds for a very specific case and not for all lines ax+b and points (p,q),

if you are going to assume it isn't you still have to be general and use variables not numbers