1. ## shortest distance

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

2. Hello, strwbrry869!

Determine the shortest distance from point $\displaystyle A(\text{-}5,\:\text{-}2\sqrt{7})$ to the x-axis to point $\displaystyle B(4,\:\text{-}\sqrt7)$
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 $\displaystyle B$ over the $\displaystyle x$-axis to point $\displaystyle B'.$
Draw line $\displaystyle AB'$, intersecting the $\displaystyle x$-axis at $\displaystyle P.$

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

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

Using calculus, the problem gets VERY messy!

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

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

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

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

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

And that is the function we must minimize . . .

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

4. Continuing the Calculus way (the best I can):

The distance formula, $\displaystyle D \;=\;\sqrt{x^2+10x + 53} + \sqrt{x^2 - 8x + 23}$, 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:

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

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

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

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

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

$\displaystyle x = 0.34110898362998....$

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.