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
[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, 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.
∵ RS is a line segment whose end points are R (-4 , 3) and S (8 , -5)
∵ Point (-1 , 1) divides it at ratio [tex]m_{1}:m_{2}[/tex] from R
- By using the rule above
∴ [tex]-1=\frac{-4m_{2}+8m_{1}}{m_{1}+m_{2}}[/tex]
- By using cross multiplication
∴ [tex](-1)m_{1}+(-1)m_{2}[/tex] = [tex]-4m_{2}+8m_{1}[/tex]
- By collecting [tex]m_{1}[/tex] in one side and [tex]m_{2}[/tex] in the
other side
∴ [tex]3m_{2}=9m_{1}[/tex]
- Divide both sides by 3
∴ [tex]m_{2}=3m_{1}[/tex]
∴ [tex]m_{1}=\frac{1}{3}m_{2}[/tex]
∴ (-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 [tex]m_{1}:m_{2}[/tex] from R
- By using the rule above
∴ [tex]5=\frac{-4m_{2}+8m_{1}}{m_{1}+m_{2}}[/tex]
- By using cross multiplication
∴ [tex]5m_{1}+5m_{2}[/tex] = [tex]-4m_{2}+8m_{1}[/tex]
- By collecting [tex]m_{1}[/tex] in one side and [tex]m_{2}[/tex] in the
other side
∴ [tex]9m_{2}=3m_{1}[/tex]
- Divide both sides by 3
∴ [tex]3m_{2}=m_{1}[/tex]
∴ [tex]\frac{m_{1}}{m_{2}}=3[/tex]
∴ (5 , -3) is the point at 3 : 1 of the way from R to S
