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avatar+2765 

Given  \(R(-6, 19)\) and \(S(15, -5)\) , what are the coordinates of the point on segment \(\overline{RS}\) two-thirds of the distance from \(S\)\(R\) to ?

tertre  Dec 29, 2017
 #1
avatar+92805 
+2

Divide the interval  S(15,-5)  R(-619)  in the ratio  2:1

 

\((\frac{1*15+2*-6}{3},\frac{1*-5+2*19}{3})\\ =(\frac{15-12}{3},\frac{-5+38}{3})\\ =(\frac{3}{3},\frac{33}{3})\\ =(1,11)\)

 

 

Melody  Dec 29, 2017
 #2
avatar+2765 
0

is the answer (1,11)? If so, it's wrong...

tertre  Dec 29, 2017
 #3
avatar+87301 
0

It's easier to find 1/3 of the distance from R to S.....so we have

 

[ -6  +  (15 - - 6)/3   , 19  +  ( -5 - 19)/ 3  ]  =

 

[ -6  + 21/3,  19  +  -24/3 ]  =

 

[ -6   + 7,  19  - 8  ]  =

 

(1, 11)

 

Same answer as Melody's  ???

 

Did you leave something out, tertre ???

 

 

cool cool cool

CPhill  Dec 29, 2017
 #4
avatar+2765 
0

Yep, I did. Sorry: Given \(R(-6,19)\) and \(S(15,-5)\), what are the coordinates of the point on segment \(\overline{RS}\)  two-thirds of the distance from \(R\) to \(S\) ?

tertre  Dec 29, 2017
 #5
avatar+87301 
+2

OK....we have

 

[  - 6  +  (15 - - 6)(2/3),  19  + (-5 - 19)(2/3)  ]

 

[ - 6  +  (21)(2/3), 19 + (-24)(2/3) ]

 

[ - 6  + 14,  19  -  16 ]

 

(8, 3)

 

 

cool cool cool

CPhill  Dec 29, 2017
 #6
avatar+2765 
+1

Oh, ok, I get it! Thanks so much!

tertre  Dec 29, 2017
 #7
avatar+122 
+3

We need to find two-thirds of the difference in the x values of points R and S and add this to the X value of point R . Then we will find two-thirds of the difference in the y values and add this to the y value of point R . When we solve for the x value we find that \(\frac{2}{3}\times (15-(-6))=\frac{2}{3} \times 21 = 14\) , and -6+14=8 , so x=8 . When we solve for y we find that \(\frac{2}{3} \times (-5 -19) = \frac{2}{3} \times (-24) = -16 \) , and \(19 + (-16) = 3\) , so y=3 . The coordinates are \(\boxed{(8,3)}\) .

azsun  Dec 29, 2017

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