+4609 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 ?
+118608 Divide the interval S(15,-5) R(-619) in the ratio 2:1
\((\frac{1*15+2*-6}{3},\frac{1*-5+2*19}{3})\\ =(\frac{15-12}{3},\frac{-5+38}{3})\\ =(\frac{3}{3},\frac{33}{3})\\ =(1,11)\)
+128407 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 ???
+4609 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 \(S\) ?
+128407 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)
+198 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 \(\boxed{(8,3)}\) .
