Question

The endpoints of segment RS are R(-4, 3) and S(8, -5). Complete each statement using a fraction.

Answer 1

Answer:

a. (-1 , 1) is the point at 1 : 3 of the way from R to S

b. (5 , -3) is the point at 3 : 1 of the way from R to S

Step-by-step explanation:

* Lets explain how to solve the problem

- If point (x , y) divide 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, 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.

∵ RS is a line segment whose end points are R (-4 , 3) and S (8 , -5)

∵ Point (-1 , 1) divides it at ratio $m_{1}:m_{2}$ from R

- By using the rule above

∴ $-1=\frac{-4m_{2}+8m_{1}}{m_{1}+m_{2}}$

- By using cross multiplication

∴ $(-1)m_{1}+(-1)m_{2}$ = $-4m_{2}+8m_{1}$

- By collecting $m_{1}$ in one side and $m_{2}$ in the

 other side

∴ $3m_{2}=9m_{1}$

- Divide both sides by 3

∴ $m_{2}=3m_{1}$

∴ $m_{1}=\frac{1}{3}m_{2}$

∴ (-1 , 1) is the point at 1 : 3 of the way from R to S

b.

∵ RS is a line segment whose end points are R (-4 , 3) and S (8 , -5)

∵ Point (5 , -3) divides it at ratio $m_{1}:m_{2}$ from R

- By using the rule above

∴ $5=\frac{-4m_{2}+8m_{1}}{m_{1}+m_{2}}$

- By using cross multiplication

∴ $5m_{1}+5m_{2}$ = $-4m_{2}+8m_{1}$

- By collecting $m_{1}$ in one side and $m_{2}$ in the

 other side

∴ $9m_{2}=3m_{1}$

- Divide both sides by 3

∴ $3m_{2}=m_{1}$

∴ $\frac{m_{1}}{m_{2}}=3$

∴ (5 , -3) is the point at 3 : 1 of the way from R to S