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[SOLVED] Shortest distance between line and point

karush

Well-known member
Jan 31, 2012
2,775
Show that the shortest distance from the point $\left(x_1,y_1\right)$ to a straight line
$$Ax_1+By_1+C=0$$ is
$$\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$$

ok, well a line from a point to a line is shortest if it is perpendicular to that line

obviously we are trying to find out a min value to this but taking a derivative of this without numbers is rather daunting and the question is asking for a proof

anyway not much of a start, but caught in the bushes already...

I did read the commentary on
"Finding the distance between a point and a line"

but this problem is under applications of differentiation which seem hard to set up

the book didn't give an answer to this...:cool:
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: shortest distance between line and point

Given that this is a calculus exercise, I would say you should not assume the shortest distance is perpendicular, but rather you should define the distance function $D(x)$ and then optimize it.

I would let \(\displaystyle (x,y)=\left(x,-\frac{Ax+C}{B} \right)\) be an arbitrary point on the line and then let$f$ represent the square of the distance function, since minimizing it will also minimize the distance function itself.

\(\displaystyle f(x)=D^2(x)=\left(x-x_1 \right)^2+\left(\frac{Ax+C}{B}+y_1 \right)^2\)

Now, differentiate with respect to $x$, and equate to zero to find the critical value.
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,502
Re: shortest distance between line and point

I added another proof to the commentary thread, but it relies on the properties of dot product rather than calculus.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Another way to prove the distance formula by calculus is to use Lagrange multipliers. We want to minimise the square of the distance, $d^2 = (x-x_1)^2 + (y-y_1)^2$ subject to the constraint $Ax+By+C = 0.$ So we take the partial derivatives of $(x-x_1)^2 + (y-y_1)^2 - \lambda(Ax+By+C)$ with respect to $x$ and $y$, and put them equal to $0$: $$2(x-x_1) - \lambda A = 0,\qquad 2(y-y_1) - \lambda B = 0.$$ Thus $x = x_1 + \frac12\lambda A,\ y = y_1 + \frac12\lambda B$. Substitute those into the constraint equation to get $A\bigl(x_1 + \frac12\lambda A\bigr) + B\bigl(y_1 + \frac12\lambda B\bigr) + C = 0$, from which $\lambda = - \dfrac{2(Ax_1+By_1+C)}{A^2+B^2}.$ Then $$d^2 = (x-x_1)^2 + (y-y_1)^2 = \bigl(\tfrac12\lambda A\bigr)^2 + \bigl(\tfrac12\lambda B\bigr)^2 = \bigl(\tfrac12\lambda\bigr)^2(A^2+B^2) = \frac{(Ax_1+By_1+C)^2}{A^2 + B^2}.$$ Now take the square root to get $d = \dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}.$
 

karush

Well-known member
Jan 31, 2012
2,775
Re: shortest distance between line and point

\(\displaystyle f(x)=D^2(x)=\left(x-x_1 \right)^2+\left(\frac{Ax+C}{B}+y_1 \right)^2\)

Now, differentiate with respect to $x$, and equate to zero to find the critical value.
I got lost trying $d/dx$ this
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: shortest distance between line and point

I got lost trying $d/dx$ this
Okay, let's have a look see...

Using the power and chain rules, we obtain:

\(\displaystyle \frac{df}{dx}=2\left(x-x_1 \right)(1)+2\left(\frac{Ax+C}{B}+y_1 \right)\left(\frac{A}{B} \right)=0\)

Multiplying through by \(\displaystyle \frac{B^2}{2}\), we may write:

\(\displaystyle B^2\left(x-x_1 \right)+A\left(Ax+C+By_1 \right)=0\)

Solving for $x$, there results:

\(\displaystyle x=\frac{B^2x_1-ABy_1-AC}{A^2+B^2}\)

Because $f$ is the sum of two squares, we know this critical value must be at the global minimum, but we could also use the argument that:

\(\displaystyle \frac{d^2f}{dx^2}=2+2\frac{A^2}{B^2}>0\,\forall (A,B)\in\mathbb{R}\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
A much simpler approach involves trigonometry.

Let $p$ be the perpendicular distance between the point and the line. we may then state:

\(\displaystyle d(\theta)=p\csc(\theta)\)

Now, minimize this function. :D