Geometry question

[Frizzeness]

Hi, I am a sophomore in highschool, and I was wondering if anyone knew what a minimum was? The chapter I am on is transformations, and the section is on reflections.

an example of one of the problems I do not understand is:

Minimum Distance - Find point C on the x-axis so AC + BC is a minimum.

37. A (1,5), B(6,3)

I am totally lost please help me?

 

[Deleted member 4993]

Hi, I am a sophomore in highschool, and I was wondering if anyone knew what a minimum was? The chapter I am on is transformations, and the section is on reflections.

an example of one of the problems I do not understand is:

Minimum Distance - Find point C on the x-axis so AC + BC is a minimum.

37. A (1,5), B(6,3)

I am totally lost please help me?

Have you studied quadratic equations and parabola?

 

[Denis]

Did you take some ye olde graph paper and plot A and B?
Put a point C somewhere on x-axis, join AC and BC?
If so, drop perpendiculars AD and BE (D and E on x-axis)?
Then you see 2 mouth watering right triangles ADC and BEC, right?
If you let k = DC, then EC = 5 - k, right?

Soooo....let's see you go!

 

[soroban]

Hello, Frizzeness!

37. Given: .$\displaystyle A(1,5),\;B(6,3)$

Find point $\displaystyle C$ on the x-axis so distance $\displaystyle AC + BC$ is a minimum.

There is a "back door" approach to this problem . . . rather sneaky.
. . but since your section is on reflections, this method may be appropriate.

Look at the diagram of the situation . . .

        |   A
        |   o (1,5)
        |    *            B
        |     *           o (6,3)
        |      *        *
        |       *     *
        |        *  *
    - - + - - - - o - - - - - -
        |       C(x,0)
        |

$\displaystyle \text{We want to locate point }C(x,0)\text{ so that the distance }AC + BC\text{ is a minimum.}$
We could apply the Distance Formula and some calculus to get the answer.
. . But there is a very sneaky solution . . .


$\displaystyle \text{Reflect point }B\text{ across the }x\text{-axis to point }B'\!:\,(6,-3).$
$\displaystyle \text{Then draw line }AB'\text{, intersecting the }x\text{-axis at }C.$

        |   A
        |   o(1,5)
        |     *             B
        |       *           o (6,3)
        |         *       * :
        |           *   *   :
    - - + - - - - - - o - - + - - - -
        |             C *   :
        |                 * :
        |                   * (6,-3)
        |                   B'

$\displaystyle \text{Then }C\text{ is the desired point!}$

$\displaystyle \text{Since }CB' = CB\text{, the distance }AC + CB'\text{ is a minimum because }AB'\text{ is a straight line.}$

$\displaystyle \text{Therefore, }AC + CB\text{ is a minimum.}$


Determine the equation of the line through $\displaystyle A$ and $\displaystyle B$, and find its x-intercept.

$\displaystyle \text{I found that point }C\text{ is: }\:\left(\tfrac{33}{8},\,0\right)$