|
Question 1119600: Find the trisection points of the line segment joining (-6,2) and (3,8)
Found 2 solutions by greenestamps, josmiceli:
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
Solve for the x and y components separately.
The difference between the two x coordinates is 3-(-6) = 3+6 = 9; the difference between the two y coordinates is 8-2 = 6.
Divide each of those differences by 3: 9/3 = 3; 6/3 = 2.
From (-6,2) to (3,8) you move 9 in the x direction and 6 in the y direction. For each "step" you take, you want to move 1/3 of those distances: 3 in the x direction (right) and 2 in the y direction (up).
3 right and 2 up from the starting point (-6,2) is (-3,4); 3 right and 2 up from (-3,4) is (0,6). And to check your calculations, 3 right and 2 up from (0,6) is (3,8), which is where it is supposed to be.
Answer: (-3,4) and (0,6) are the trisection points.
Answer by josmiceli(19441)  (Show Source):
You can put this solution on YOUR website! A( -6,2 )
D( 3,8 )
Let trisection points be B and C. so in order
the points are A, B, C, and D
-----------------------------
AB = BC
BC = CD
AB = CD
using x-values
(1) 
(2) 
so
(3) 
(3) 
Plug this into (2)
(2) 
(2) 
(2) 
(2) 
and
(3) 
(3) 
--------------------------------
Using y-values
(1) 
(2) 
so
(3) 
(3) 
Plug into (2)
(2) 
(2) 
(2) 
(2) 
and
(3) 
(3) 
-----------------------
( x[B], y[B] ) = ( -3,4 )
( x[C], y[C] ) = ( 0, 6 )
( -3,4 ) and ( 0,6 ) are trisection points
|
|
|
| |
