# Thread: Co-ordinate Geometry: Colinear points

1. ## Co-ordinate Geometry: Colinear points

A, B and C are three colinear points. A and B are respectively (3,4) and (7,7) and AC is equal to 10 units. Find the coordinates of C.

2. Originally Posted by Ilsa
A, B and C are three colinear points. A and B are respectively (3,4) and (7,7) and AC is equal to 10 units. Find the coordinates of C.
Let $\displaystyle Ch,k)$. From the given we know:
$\displaystyle \displaystyle\frac{7-4}{7-3}=\displaystyle\frac{7-k}{7-h}$ and $\displaystyle (h-3)^2+(k-4)^2=100$.

Solve for $\displaystyle h~\&~k$.

3. could u pls explain which formula u have used?

4. Originally Posted by Ilsa
could u pls explain which formula u have used?
Slope: collinear points determine the same slope.
Distance.

5. Hello, Ilsa!

Did you make a sketch?

$\displaystyle A, B, C$ are three collinear points.
$\displaystyle A$ and $\displaystyle B$ are (3,4) and (7,7), respectively,
and $\displaystyle AC$ is equal to 10 units.
Find the coordinates of $\displaystyle C.$
Code:
      |                                      C
|                                      o
|                                  *   :
|                              *       :3
|                      B   *           :
|                 (7,7)o - - - - - - - +
|                  *   :       4
|              *       :3
|      A   *           :
| (3,4)o - - - - - - - +
|             4
|
|
- + - - - - - - - - - - - - - - - - - - - - -
|

To go from point $\displaystyle A$ to point $\displaystyle B$, we move: right 4, up 3.

The length of $\displaystyle AB$ is: .$\displaystyle \sqrt{4^2+3^2} \:=\:\sqrt{25} \:=\:5$

Since the length of $\displaystyle AC$ is 10, then: .$\displaystyle BC = 5.$

To go from point $\displaystyle B$ to point $\displaystyle C$, we move (again): right 4, up 3.

Therefore, point C is at: .$\displaystyle (11,10)$

6. pls sir i wud be grateful of u if u pls help me doing the sum in intermediate level or ordinary level or according to the ratio theorem or distance formula