shortest distance

**strwbrry869** · May 3rd 2009, 05:49 AM

Determine the shortest distance from point A (-5, -2sqrt7) to the X-axis to point B ( 4, -sqrt7).

**Soroban** · May 3rd 2009, 08:01 AM

Hello, strwbrry869!

Determine the shortest distance from point $A(\text{-}5,\:\text{-}2\sqrt{7})$ to the x-axis to point $B(4,\:\text{-}\sqrt7)$

There is a back-door approach to this problem.

Code:

                    |             _
                    |         (4,√7)
                    |           o B'
                    |        *  :
    -5              |  P  *     :
  ---+--------------+--o--------+----
     :              *     *     :
     :           *  |        *  :
     :        *     |           o B_
     :     *        |         (4,-√7)
     :  *           |
   A o   _          |
  (-5,-2√7)         |

Reflect point $B$ over the $x$ -axis to point $B'.$
Draw line $AB'$ , intersecting the $x$ -axis at $P.$

$P$ gives the shortest distance from $A$ to $P$ plus from $P$ to $B.$
. . (Do you see why?)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Using calculus, the problem gets VERY messy!

The distance from $A(-5,-2\sqrt{7})$ to $P(x,y)$ is:

. $d_1 \:=\:\sqrt{(x+5)^2 + (2\sqrt{7})^2} \:=\:\sqrt{x^2 + 10x + 53}$

The distance from $B(4,-\sqrt{7})$ to $P(x,y)$ is:

. . $d_2 \;=\;\sqrt{(x-4)^2 + (\sqrt{7})^2} \:=\:\sqrt{x^2+10x + 23}$

The total distance is: . $D \;=\;\sqrt{x^2+10x + 53} + \sqrt{x^2 - 8x + 23}$

And that is the function we must minimize . . .

**strwbrry869** · May 3rd 2009, 09:54 AM

thanks but im not good at math at all... do you know what the answer to that would be?

**333qaz333** · May 3rd 2009, 04:41 PM

Continuing the Calculus way (the best I can):

The distance formula, $D \;=\;\sqrt{x^2+10x + 53} + \sqrt{x^2 - 8x + 23}$ , looks right. I think from there you would have to take the derivative of that and set it equal to 0 to find the minimum value.

So:

$ \frac{dD}{dx} = \frac{1}{2}(x^2+10x+53)^\frac{-1}{2}*(2x+10) + \frac{1}{2}(x^2+10x+23)^\frac{-1}{2}*(2x-8) $

$ \frac{dD}{dx} = \frac{2x+10}{2\sqrt{x^2+10x+53}} + \frac{2x-8}{2\sqrt{x^2+10x+23}} $

To find the minimum, we substitute 0 for $\frac{dD}{dx}$ .

$ 0 = \frac{2x+10}{2\sqrt{x^2+10x+53}} + \frac{2x-8}{2\sqrt{x^2+10x+23}} $

That is an extremely difficult equation to factor, so I just put it in my calculator and used the Equation Solver.
I found that:

$ x = 0.34110898362998.... $

So, I guess that that (0.3411, 0), would be your point of intersection on the x-axis.
So you can now just find the distance between A and the point of intersection, and then the distance between B and the point of intersection, and add them together.

Hope that helps a little. Seems like an extremely hard problem for an Algebra student. I'm in Calculus now, and that seems pretty difficult to me.

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