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

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

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

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

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

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

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

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

Calculate the coordinates of $C.$

Here is a simplified diagram:

Code:

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

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

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

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

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

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