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I have two different line on a coordinate plane and I'm looking for the intersection point of those lines which is shown on third quadrant of the plane.
Here are the values for each two lines:
d1: (0, 2) , (-1, 0)
d2: (2, 0) , (0, -1)

How can I find the intersection point which is showed with "P(x, y)" now?

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:D Thank you... :D Yes, I'm asking for "intersection" between those two line on the plane. :) Sorry about my Math(s) terms by the way... :D Thank you, again... :) – Kerim Atasoy Mar 2 '12 at 20:53
:) Humm... Yes, David, you two are right and...I'm still looking for a better, proper way to ask the right question now... :D :) Thanks anyway, your replies might still work here, just let me try it now... :) – Kerim Atasoy Mar 2 '12 at 20:59
up vote 2 down vote accepted

One way to find the intersection points of the two lines is to first find the equations of the lines (but see the remark at the end of this answer). Recall that the equation of a line with slope $m$ and $y$-intercept $b$ is $y=mx+b$.

The line $d_1$ (assuming the two points you gave define it) has slope ${2-0\over 0-(-1)}=2$ and $y$-intercept $2$ (looking at the point $(0,2)$), so its equation is $$\tag{1}y=2x+2.$$

The line $d_2$ has slope ${0-(-1)\over 2-0}={1\over2}$ and $y$-intercept $-1$ (looking at the point $(0,-1)$), so its equation is $$\tag{2}y={1\over2}x-1.$$

To find the intersection point of the lines, set the right hand sides of $(1)$ and $(2)$ equal to each other $$ {1\over2}x-1 = 2x+2 $$ and solve for $x$. The solution to the above equation is $x=-2$. This gives the $x$-coordinate of the point of intersection. the $y$-coordinate can be found by substitutng this $x$-value into either equation $(1)$ or $(2)$. Using equation $(1)$, we get $y=2\cdot(-2)+2=-2$.

So the point of intersection is $(-2,-2)$.

Remark: Note if you draw the lines, you can take advantage of the symmetry displayed to find the point of intersection. (It lies on the line $y=x$; note the slopes of the lines are negative reciprocals of each other.)

enter image description here

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A very interesting draw!... I would like to be able to code and draw like this soon... Thank you, David... – Kerim Atasoy Mar 2 '12 at 21:33

Assuming that you mean that the four pairs of numbers are the coordinates of two points on each line, and you want to find their intersection:

Just grab a piece of graph paper and draw a diagram!

  +   +   A   +   +   +     A(0,2)
  +   + / +   +   +   +
       /  |         _,-     B(-1,0)
--+---B---+---+--_C'--+--   C(2,0)
     /    | _,--'
  + / +  _D'  +   +   +     D(0,-1)
   /_,--' |
 _P'  +   +   +   +   +     P(-2,-2)
 /        |

(Or, as I did before typing in the ascii art, grab a piece of blank paper and start by drawing a half-dozen parallel lines and another half-dozen parallel lines across them).

Of course, you can also throw some algebra at it, but that will be overkill unless the problem is homework where you need to demonstrate that you can do it with algebra instead of the diagram.

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I assume you name is Henning, sir. Well, you answer that shows P(-2 , -2) is different from the answer key actually, but, you might be right because I'didn't revise the solution yet... – Kerim Atasoy Mar 2 '12 at 21:04
The answer key shows P(-2, 2) by the way. Thank you, Henning... – Kerim Atasoy Mar 2 '12 at 21:05
Now, I'm sure that the answer key for this problem is wrong, yes, I finally got the current solution approaching for the question. Thank you all... :) – Kerim Atasoy Mar 4 '12 at 3:28

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