Question

A, B and C are collinear and B is between a and C. The ratio of AB to AC is 3:2. If A is at (4,8) and B is at (7,2) what are the coordinates of point C

Answer 1

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

 $(x_{1},y_{1})$ and $(x_{2},y_{2})$ at ratio $m_{1}:m_{2}$

 from the first point $(x_{1},y_{1})$, then

 $x=\frac{x_{1}m_{2}+x_{2}m_{1}}{m_{1}+m_{2}}$ and

  $y=\frac{y_{1}m_{2}+y_{2}m_{1}}{m_{1}+m_{2}}$

* Lets solve the problem

- A, B and C are collinear and B is between A and C

∴ Point A is $(x_{1},y_{1})$

∴ Point C is $(x_{2},y_{2})$

∴ Point B is (x , y)

- The ratio of AB to BC is 3 : 2

∴ $m_{1}:m_{2}$ = 3 : 2

* Lets use the rule above to find the coordinates of point C

∵ A = (4 , 8) and B = (7 , 2)

∵ $m_{1}:m_{2}$ = 3 : 2

∴  $7=\frac{(4)(2)+x_{2}(3)}{5}$

- Multiply each side by 5

∴  $35=8+3x_{2}$

- Subtract 8 from both sides

∴ $27=3x_{2}$

- Divide both sides by 3

∴ $9=x_{2}$

∴ The x-coordinate of point C is 9

∴  $2=\frac{(8)(2)+y_{2}(3)}{5}$

- Multiply each side by 5

∴  $10=16+3y_{2}$

- Subtract 16 from both sides

∴ $-6=3y_{2}$

- Divide both sides by 3

∴ $-2=y_{2}$

∴ The y-coordinate of point C is -2

* The coordinates of point C are (9 , -2)