Asked by Bob
At what point does the graph of the parametric equations y=3-2t, y=-2+5t intersect the graph of the parametric equations x = -1+3s,y=5-9s. The answer is the point of intersection (x,y).
Please help! I am really confused!
Please help! I am really confused!
Answers
Answered by
Reiny
I am a bit confused myself, unless you have a typo.
I will assume you meant your first set to be
x = 3-2t, y = -2 + 5t
so -1+ 3s = 3 - 2t ---> 2t + 3s = 4 **
and
-2 + 5t = 5 - 9s ---> 5t + 9s = 7 ***
3 times ** ---> 6t + 9s = 12
subtract ***
t = 5
back in **
10 + 3s = 4
3s = -6
s = -2
then x = 3 -2t = 3 -10 = -7
y = -2+5t = -2 + 25 = 23
(x,y) = (-7, 23)
OR
If you meant it the way you typed it ....
from the first, since they are both equal to y
-2+5t = 3-2t
7t = 5
t = 5/7
-2 + 5t = 5 - 9s ---> 5t + 9s = 7
sub in t = 5/7
25/7 + 9s = 7
9s = 24/7
s = 8/21
since x = -1+3s
x = -1 + 8/7 = 1/7
y = 3-2t = 3-10/7 = 11/7
point would be ( 1/7, 11/7)
I will assume you meant your first set to be
x = 3-2t, y = -2 + 5t
so -1+ 3s = 3 - 2t ---> 2t + 3s = 4 **
and
-2 + 5t = 5 - 9s ---> 5t + 9s = 7 ***
3 times ** ---> 6t + 9s = 12
subtract ***
t = 5
back in **
10 + 3s = 4
3s = -6
s = -2
then x = 3 -2t = 3 -10 = -7
y = -2+5t = -2 + 25 = 23
(x,y) = (-7, 23)
OR
If you meant it the way you typed it ....
from the first, since they are both equal to y
-2+5t = 3-2t
7t = 5
t = 5/7
-2 + 5t = 5 - 9s ---> 5t + 9s = 7
sub in t = 5/7
25/7 + 9s = 7
9s = 24/7
s = 8/21
since x = -1+3s
x = -1 + 8/7 = 1/7
y = 3-2t = 3-10/7 = 11/7
point would be ( 1/7, 11/7)
Answered by
Bob
Sorry it was a typo it was the first one thanks a lot!!!
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