# Thread: work out the co ordinates

1. ## work out the co ordinates

hi,

i need to work out the co ordinates of two points that the graph of y=2x-1 passes through.

i don't really know how to do this so could somone explain what i have to do please?!?

thanks!

2. Pick any old values of x and put the corresponding y value in.

For example (0,-1) is a point on the line

3. oh so would (1, -1) be one aswell?

4. Originally Posted by andyboy179
oh so would (1, -1) be one aswell?
No. (x, y) = (1, -1) is NOT on the line since using these values in
y = 2x - 1 gives us
-1 = 2(1) -1, which is FALSE.

5. No, what makes you say that?

The equation is a straight line. If you want to use x=1 then it would be $(1, [2 \cdot 1 - 1])$

6. would (-1,0) work?

7. Originally Posted by andyboy179
would (-1,0) work?
Sub it into the equation and check. Does $(2 \times -1) -1$ equal 0?

8. When x = -1,
y = 2x - 1 becomes
y = 2(-1) - 1 =
-2 -1 = -3
S0 (-1, -3) is the point

9. oh i see now, thanks alot!

10. no its -1

11. Uhm, you don't simply search numbers that verify the equation, but take the intersection with the axes (make x=0 and find y, then make y=0 and find x).

12. That's only necessary to find the x- and y-intercepts. For any two points on the line, any two values of x suffice.

13. Originally Posted by andyboy179
oh so would (1, -1) be one aswell?
Quote what you are responding to otherwise the thread will become confusing.

CB

14. Originally Posted by e^(i*pi)
No, what makes you say that?

The equation is a straight line. If you want to use x=1 then it would be $(1, [2 \cdot 1 - 1])$
Quote what you are responding to, otherwise the thread will become incomprehensible.

CB

15. Originally Posted by andyboy179
would (-1,0) work?
Quote, or don't bother replying thank you.

CB

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