Please help! I really need the answer.

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?

7 answers

X problem
delta x = (15 - -6) = 21
delta x * 2/3 = 14
14 + -6 = 8
Y problem
delta y = -5 -19 = -24
-24 *2/3 = -16
19 - 16 = 3
(8,3)
How about just typing:
Given R(-6, 19) and S(15, -5), what are the coordinates of the point on segment RS two-thirds of the distance from R to S?

Let P(x,y) be that point. You want RP : PS = 2 : 3

for the x:
(x+6)/(15-x) = 2/3
3x + 18 = 30 - 2x
5x = 12
x = 12/5 = 2.4

for the y:
(y-19)/(-5-y) = 2/3
3y - 57 = -10 - 2y
5y = 47
t = 47/5 = 9.4
small typo, last line should say
y = 47/5 = 9.4

So the point is (2.4 , 9.4)
2.4 is 2/3 of the way from -6 to + 15 ?
for the x:
I THINK doing it your way 15 - -6 = 15+6 = 21 = the whole distance that you want 2/3 of
(x+6)/(15--6) = 2/3
or
(x+6) / 21 = 2/3
42 = 3 x + 18
24 = 3 x
x = 8
You are right, Anonymous
my ratio should have been 2 : 1 , not 2:3
I split the line into 5 parts, not 3

A simple way would have been to use vectors for the partition of a line segment:
<x,y> = (1/3)<-6,19> + (2/3)<15,-5>
= <-2,19/3> + <10,-10/3>
= <8,3>

my bad!
Whew ! Thanks :)