Answer:
The coordinates of point C are (9 , -2)
Step-by-step explanation:
* Lets explain how to solve the problem
- If point (x , y) divides a line segment whose end points are
[tex](x_{1},y_{1})[/tex] and [tex](x_{2},y_{2})[/tex] at ratio [tex]m_{1}:m_{2}[/tex]
from the first point [tex](x_{1},y_{1})[/tex], then
[tex]x=\frac{x_{1}m_{2}+x_{2}m_{1}}{m_{1}+m_{2}}[/tex] and
[tex]y=\frac{y_{1}m_{2}+y_{2}m_{1}}{m_{1}+m_{2}}[/tex]
* Lets solve the problem
- A, B and C are collinear and B is between A and C
∴ Point A is [tex](x_{1},y_{1})[/tex]
∴ Point C is [tex](x_{2},y_{2})[/tex]
∴ Point B is (x , y)
- The ratio of AB to BC is 3 : 2
∴ [tex]m_{1}:m_{2}[/tex] = 3 : 2
* Lets use the rule above to find the coordinates of point C
∵ A = (4 , 8) and B = (7 , 2)
∵ [tex]m_{1}:m_{2}[/tex] = 3 : 2
∴ [tex]7=\frac{(4)(2)+x_{2}(3)}{5}[/tex]
- Multiply each side by 5
∴ [tex]35=8+3x_{2}[/tex]
- Subtract 8 from both sides
∴ [tex]27=3x_{2}[/tex]
- Divide both sides by 3
∴ [tex]9=x_{2}[/tex]
∴ The x-coordinate of point C is 9
∴ [tex]2=\frac{(8)(2)+y_{2}(3)}{5}[/tex]
- Multiply each side by 5
∴ [tex]10=16+3y_{2}[/tex]
- Subtract 16 from both sides
∴ [tex]-6=3y_{2}[/tex]
- Divide both sides by 3
∴ [tex]-2=y_{2}[/tex]
∴ The y-coordinate of point C is -2
* The coordinates of point C are (9 , -2)
