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helppppp

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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 ?

Dec 29, 2017

#1
+2

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

$$(\frac{1*15+2*-6}{3},\frac{1*-5+2*19}{3})\\ =(\frac{15-12}{3},\frac{-5+38}{3})\\ =(\frac{3}{3},\frac{33}{3})\\ =(1,11)$$ Dec 29, 2017
#2
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is the answer (1,11)? If so, it's wrong...

Dec 29, 2017
#3
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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)

Did you leave something out, tertre ???   Dec 29, 2017
#4
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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 $$S$$ ?

Dec 29, 2017
#5
+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)   Dec 29, 2017
#6
+1

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

Dec 29, 2017
#7
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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 $$\boxed{(8,3)}$$ .

Dec 29, 2017