# Coordinate geometry

• May 2nd 2010, 10:41 PM
Punch
Coordinate geometry
Given a straight line BD where point E lies on it.

Coordinates $\displaystyle B(0.5,3)$ and $\displaystyle E(1,1)$

Find the coordinates of D given that $\displaystyle BE : ED = 1:3$ and the equation of the line BD is $\displaystyle y=-4x+5$
• May 3rd 2010, 02:18 AM
Soroban
Hello, Punch!

Was that the original wording of the problem?

Quote:

Given points $\displaystyle B\left(\tfrac{1}{2},\:3\right)$ and $\displaystyle E(1,\:1)$

Find the coordinates of point D so that $\displaystyle BE : ED = 1:3$

Here's a back-door approach . . .

Code:

              ½       B(½,3)o - +             \  :             d \ : 2               \:           E(1,1)o - - - +                 \      :                   \    :                   \    :                 3d \  :                     \  :                       \ :                       \:                       D o

Going from $\displaystyle B$ to $\displaystyle E$, we moved $\displaystyle \tfrac{1}{2}$ unit right and 2 units down.
. . And we moved a diagonal distance $\displaystyle d.$

Going from $\displaystyle E$ to $\displaystyle D$, we want to move a diagonal distance $\displaystyle 3d.$
. . Hence, we must triple our previous moves.

From $\displaystyle E$, we move $\displaystyle {\color{blue}\tfrac{3}{2}}$ units right and 6 units down.

Therefore, point $\displaystyle D$ is: .$\displaystyle \left(\tfrac{5}{2},\;\text{-}5\right)$

• May 3rd 2010, 04:42 AM
mrmohamed
Quote:

Originally Posted by Soroban
Hello, Punch!

Was that the original wording of the problem?

Here's a back-door approach . . .

Code:

              ½       B(½,3)o - +             \  :             d \ : 2               \:           E(1,1)o - - - +                 \      :                   \    :                   \    :                 3d \  :                     \  :                       \ :                       \:                       D o
Going from $\displaystyle B$ to $\displaystyle E$, we moved $\displaystyle \tfrac{1}{2}$ unit right and 2 units down.
. . And we moved a diagonal distance $\displaystyle d.$

Going from $\displaystyle E$ to $\displaystyle D$, we want to move a diagonal distance $\displaystyle 3d.$
. . Hence, we must triple our previous moves.

From $\displaystyle E$, we move $\displaystyle {\color{blue}\tfrac{3}{2}}$ units right and 6 units down.

Therefore, point $\displaystyle D$ is: .$\displaystyle \left(\tfrac{5}{2},\;\text{-}5\right)$

HI all
Excuse me Mr soroban
what about Point D (-1/2 , 7 )????
in my solution I found D(5/2 , -5 ) and D(-1/2 , 7 )
mrmohamed
• May 3rd 2010, 05:49 AM
Punch
Quote:

Originally Posted by Soroban
Hello, Punch!

Was that the original wording of the problem?

Here's a back-door approach . . .

Code:

              ½       B(½,3)o - +             \  :             d \ : 2               \:           E(1,1)o - - - +                 \      :                   \    :                   \    :                 3d \  :                     \  :                       \ :                       \:                       D o
Going from $\displaystyle B$ to $\displaystyle E$, we moved $\displaystyle \tfrac{1}{2}$ unit right and 2 units down.
. . And we moved a diagonal distance $\displaystyle d.$

Going from $\displaystyle E$ to $\displaystyle D$, we want to move a diagonal distance $\displaystyle 3d.$
. . Hence, we must triple our previous moves.

From $\displaystyle E$, we move $\displaystyle {\color{blue}\tfrac{3}{2}}$ units right and 6 units down.

Therefore, point $\displaystyle D$ is: .$\displaystyle \left(\tfrac{5}{2},\;\text{-}5\right)$

Hi! its you again :) I have understood your workings and I knew this before I asked this question on MHF. However, I am still a little puzzled on how to present my workings...
• May 3rd 2010, 06:13 AM
Soroban
Hello, mrmohamed!

Quote:

What about Point D (-1/2 , 7 )?

That point is also correct . . . Good work!

Silly me . . . I assumed directed distances,

. . as in .$\displaystyle \overrightarrow{BE} : \overrightarrow{ED} \:=\:1:3$

• May 3rd 2010, 06:47 AM
mrmohamed
Quote:

Originally Posted by Punch
Given a straight line BD where point E lies on it.

Coordinates $\displaystyle B(0.5,3)$ and $\displaystyle E(1,1)$

Find the coordinates of D given that $\displaystyle BE : ED = 1:3$ and the equation of the line BD is $\displaystyle y=-4x+5$

HI all
http://i69.servimg.com/u/f69/12/89/99/77/almonk12.jpg
with my best wishes
mrmohamed
• May 4th 2010, 03:31 AM
Punch
Thanks! but in the solution sheet shows only point is: $\displaystyle \left(\tfrac{5}{2},\;\text{-}5\right)$ and the marks allocated to this question is probably only 2-3... judging from the long workings, perhaps your working is a more elaborated one compared to what is required of me?