# Coordinate Geometry

• Jul 11th 2010, 07:40 AM
Punch
Coordinate Geometry
http://i952.photobucket.com/albums/a...g/P7090025.jpg

In the diagram above, triangle $\displaystyle ACD$ is such that the coordinates of $\displaystyle A$ and $\displaystyle B$ are $\displaystyle (2, 12)$ and $\displaystyle (6, 9)$ respectively. $\displaystyle BE$ is parallel to $\displaystyle CD$ and $\displaystyle BC = \frac{2}{3}AC$. The gradient of $\displaystyle BE$ is $\displaystyle \frac{2}{7}$. The equation of $\displaystyle AD$ is $\displaystyle 3y=5x+26$.

Calculate the coordinates of C.
• Jul 11th 2010, 08:39 AM
red_dog
Let $\displaystyle M_1(x_1,y_1), \ M_2(x_2,y_2)$ be two points and $\displaystyle M(x_M,y_M)$ a point on the segment $\displaystyle M_1M_2$ such as $\displaystyle \displaystyle\frac{M_1M}{MM_2}=k$.

Then $\displaystyle x_M=\displaystyle\frac{x_1+kx_2}{1+k}, \ y_M=\displaystyle\frac{y_1+ky_2}{1+k}$

In this case $\displaystyle \displaystyle\frac{AB}{AC}=\frac{1}{2}$

Then $\displaystyle x_B=\displaystyle\frac{x_A+\frac{1}{2}x_C}{1+\frac {1}{2}}$

$\displaystyle y_B=\displaystyle\frac{y_A+\frac{1}{2}y_C}{1+\frac {1}{2}}$

You know the coordinates of A and B, so you can find the coordinates of C.
• Jul 11th 2010, 08:43 AM
Soroban
Hello, Punch!

I assume there are more parts to this problem.
There is way too much information . . .

Quote:

In the diagram, $\displaystyle \Delta ACD$ has points: $\displaystyle A(2,12)$ and $\displaystyle B(6,9).$

$\displaystyle BE$ is parallel to $\displaystyle CD$ and $\displaystyle BC = \frac{2}{3}AC$.
The gradient of $\displaystyle BE$ is $\displaystyle \frac{2}{7}$.
The equation of $\displaystyle AD$ is: $\displaystyle 3y\:=\:5x+26$.

Calculate the coordinates of $\displaystyle C.$

Here is a simplified diagram:

Code:

|
|      A
| (2,12)o - - +
|        \    :
|        \  :
|          \  :
|          \ :
|            \:
|      B(6,9)o - - - - +
|              \        :
|              \      :
|                \      :
|                \    :
|                  \    :
|                  \  :
|                    \  :
|                    \ :
|                      \:
|                    C o
|
- - + - - - - - - - - - - - - - -
|

$\displaystyle \text{Since }BC = \frac{2}{3}AC,\; BC\text{ is twice }AB$

Going from $\displaystyle A$ to $\displaystyle B$, we move: 4 units right and 3 units down.

So, going from $\displaystyle B$ to $\displaystyle C$, we move: 8 units right and 6 units down.

Therefore, point $\displaystyle C$ is $\displaystyle (14,3).$

• Jul 11th 2010, 04:26 PM
Punch
you are right soroban