1. ## Simple problem X(

Good Morning!!!!

my prof solved this but i didn't get it idk T,T
i must use division of a line segment.

"find the endpoint of the section joining A(2,3) & M(6,1) if it is extended to a distance 2/3 fro its own length."

got midterms coming up , this is the only thing that i didn't understand .please help ^^

2. ## Re: Simple problem X(

Extended in which direction? Does 2/3 mean that it is being extended by 2/3 units? Or is it being stretched by an additional 2/3 of its current length (making it 5/3 its current length)?

3. ## Re: Simple problem X(

Originally Posted by SlipEternal
Extended in which direction? Does 2/3 mean that it is being extended by 2/3 units? Or is it being stretched by an additional 2/3 of its current length (making it 5/3 its current length)?

yes I saw 5/3 on the solution but i dont know how did that happen

4. ## Re: Simple problem X(

correction A is (-2,-3)

5. ## Re: Simple problem X(

Start with its length. That is one times its length. Add 2/3 of its length. That is 1+2/3 of its length. 1+2/3 = 5/3.

6. ## Re: Simple problem X(

Originally Posted by SlipEternal
Start with its length. That is one times its length. Add 2/3 of its length. That is 1+2/3 of its length. 1+2/3 = 5/3.
so line A-M is 1 i think im getting it now

so i must do
for x
a-m/a-p=5/3=-2-6/-2-x

for y
5/3=-3-1/-3-y

is that it??

7. ## Re: Simple problem X(

If $|AM|:|AP| = 3:5$, then you need to flip the fraction. $\dfrac{-2-x}{-2-6} = \dfrac{5}{3}$ and $\dfrac{-3-y}{-3-1} = \dfrac{5}{3}$.

Then $|AM| = \sqrt{(6-(-2))^2+(1-(-3))^2} = \sqrt{80}$ and $|AP| = \sqrt{\left(\dfrac{34}{3}-(-2)\right)^2+\left(\dfrac{11}{3}-(-3)\right)^2} = \sqrt{\dfrac{2000}{9}} = \sqrt{\dfrac{25}{9}\cdot 80} = \dfrac{5}{3}\sqrt{80}$ as desired.

8. ## Re: Simple problem X(

oh! thats IT!! thanks sir!!

9. ## Re: Simple problem X(

Hello, mventurina1!

Find the endpoint of the section joining A(2,3) & M(6,1)
if it is extended to a distance 2/3 of its own length.

Code:
      |   A
|   o(2,3)
|   :   *
|   :       *
| -2:           *
|   :               *
|   :       +4          *   M
|   + . . . . . . . . . . . o(6,1)
|                           :   *
----+---------------------------:-------*--------
|                           :           *   B
|                           + . . . . . . . o
|
Going from A to M, we move:
. . 4 units right,
. . 2 units down.

Going from M to B, we move:
. . $\tfrac{2}{3}(4) = \tfrac{8}{3}$ units right,
. . $\tfrac{2}{3}(2) = \tfrac{4}{3}$ units down.

Hence: . $\begin{Bmatrix} x &=& 6 + \frac{8}{3} &=& \frac{26}{3} \\ y &=& 1 - \frac{4}{3} &=& \text{-}\frac{1}{3} \end{Bmatrix}$

Therefore: . $B\left(\frac{26}{3},\;\text{-}\frac{1}{3}\right)$

10. ## Re: Simple problem X(

Hi,
Soroban's excellent solution works because of similar triangles: