# Thread: Coordinate geometry

1. ## Coordinate geometry

Given a straight line BD where point E lies on it.

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

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

2. Hello, Punch!

Was that the original wording of the problem?
It's written very badly.

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

Find the coordinates of point D so that $BE : ED = 1:3$
Here's a back-door approach . . .

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

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

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

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

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

3. Originally Posted by Soroban
Hello, Punch!

Was that the original wording of the problem?
It's written very badly.

Here's a back-door approach . . .

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

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

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

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

4. Originally Posted by Soroban
Hello, Punch!

Was that the original wording of the problem?
It's written very badly.

Here's a back-door approach . . .

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

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

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

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

5. Hello, mrmohamed!

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

That point is also correct . . . Good work!

Silly me . . . I assumed directed distances,

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

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

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

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

with my best wishes
mrmohamed

7. Thanks! but in the solution sheet shows only point is: $\left(\tfrac{5}{2},\;\text{-}5\right)$ 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?