# Math Help - trisect lines

1. ## trisect lines

- points x and y lie on cordinate axises
- the line CD is disected into 3 parts by x and y equally (trisected is the term?)
- we are given the value of C to find the possible cordinates of D

Can someone lead me in the right direction, I'm not exactly looking for answers here.

(this is for saturday school math, no idea how to, ty, im in grade 9 btw)

2. Originally Posted by girlsrox32
- points x and y lie on cordinate axises
- the line CD is disected into 3 parts by x and y equally (trisected is the term?)
- we are given the value of x to find the possible cordinates of y

Can someone lead me in the right direction, I'm not exactly looking for answers here.

(this is for saturday school math, no idea how to, ty, im in grade 9 btw)
what do we know about CD? in particular, it's end points? how about using the distance formula? the distance between x and the end point close to x would be equal to the distance between y and the end point close to y. you would be able to set up an equation with y as the only unknown

3. Sorry, edit of the first post! didnt see something
its suppose to be
- we are given the value of C to find the possible cordinates of D

T_T can u rehelp me?

4. Originally Posted by girlsrox32
uh we don't know one of CD's endpoints. but we know that x and y lie in coordinate axises.
Alright then, say you know the location of C and the location of X. Then either Y is the midpoint of CX or X is the midpoint of CY.

For example, say C = (1,3) and X = (7, 9).

If Y is the midpoint of CX, then Y = (4, 6). and D = X + <CY> = (7, 9) + <3, 3> = (10, 12).

If X is the midpoint of CY, then Y = (13, 15) and D = Y + <CX> = (13, 15) + <6, 6> = (19, 21).

5. Originally Posted by icemanfan
Alright then, say you know the location of C and the location of X. Then either Y is the midpoint of CX or X is the midpoint of CY.

For example, say C = (1,3) and X = (7, 9).

If Y is the midpoint of CX, then Y = (4, 6). and D = X + <CY> = (7, 9) + <3, 3> = (10, 12).

If X is the midpoint of CY, then Y = (13, 15) and D = Y + <CX> = (13, 15) + <6, 6> = (19, 21).
But if x and y lie on the coordinate axises, wouldn't we have a value and variable as a ordered pair for each one? and there's two axises, so wouldn't there be two different pairs?

6. Ok, I have a better understanding of the question now. You are given C only, and you have to find D, but you know that X and Y lie on the coordinate axes. Assuming that C does not lie on an axis, then you will have two possibilities for the location of D. Let C = (a, b) and define X to be the midpoint of CY. Then, because Y lies on an axis, it is either (0, y) for some y or (x, 0) for some x. If Y is (0, y), then X = $\left(\frac{1}{2}a, 0\right)$, which means Y = (0, -b). If Y is (x, 0), then X = $\left(0, \frac{1}{2}b\right)$, which means Y = (-a, 0). Can you determine the possible locations of D from this information?

7. ^thanks!

could you explain how you got to[ with a formula or something?] (1/2a, 0). I don't get why there would be 1/2

8. X is the midpoint of CY. So if Y = (0, y) and C = (a, b), then the x-coordinate of X must be the average of the x-coordinates of C and Y, and since this value is nonzero and X is on an axis, the y-coordinate of X must be 0.

9. omg ty!

and uh, how did you conclude to Y(0,-b) after finding the x-coordinate midpoint. because when I do it, I have two variables to average out. and i also know that once i have the two coordinates of X and Y, I can find the distance between X and Y using that distance to substitute in a distance formula of Y and D, but for D I have two variables, how would I solve for each.

10. Originally Posted by girlsrox32
and uh, how did you conclude to Y(0,-b) after finding the x-coordinate midpoint. because when I do it, I have two variables to average out. and i also know that once i have the two coordinates of X and Y, I can find the distance between X and Y using that distance to substitute in a distance formula of Y and D, but for D I have two variables, how would I solve for each.
The average of the y-coordinates of C and Y must be zero. So you have $\frac{b + y}{2} = 0$ , hence

$b + y = 0$
$y = -b$

Also, if you are given C = (a, b), then a and b are not variables. They are constants.

11. but for D, i have only variables o_O
, right?

12. Nope. The same vector that connects X and Y also connects Y and D, and you know exactly what X and Y are (in each of the two cases), so you can find that vector and then add it to Y to obtain D.

13. uh sry T_T but whats a vector

14. A vector is a quantity that has length and direction. For example, the vector connecting the two points (1,1) and (4,5) is <3, 4> and it has a length of 5. It represents, in this case, the difference between the coordinates of the two points. If I add <3, 4> to (4, 5) I get (7, 9). Because I added the same vector twice, it turns out that (4, 5) is the midpoint of the line segment connecting (1, 1) and (7, 9).