Path of intersection point between two lines

[apple2357]

Been playing with a problem and trying to work out the path of the point of intersection of two lines as 'a' changes for the following two equations:

y= 2x+a
y= ax+3

On software, I get this picture but i am a little stuck determining what the equation of the dotted traced line is ?

I am thinking 2x+a = ax+3 and a rational function of some kind, but i get a little stuck? solving for x, i get x= (3-a)/(2-a), something a bit weird is going on.

 

[tkhunny]

I'm not seeing the problem, except that you are only half done.

You solved for x in terms of a. How about y in terms of a?

 

[apple2357]

So do i get a parametric equation?

x= (3-a)/(2-a), y= 2(3-a)/(2-a) + a

And directly finding a cartesian is not that simple i guess?

 

[Mr. Bland]

There is an asymptote at [MATH]x = 1[/MATH] as [MATH]a[/MATH] approaches either [MATH]\infty[/MATH] or [MATH]-\infty[/MATH], where no point of intersection will ever fall. It should be possible to graph the path of these points as a function of [MATH]x[/MATH].

 

[Steven G]

You want to find the point of intersection of the two lines, which depends on a. The point of intersection will still be an (x,y) point on the plane, in terms of a. Yes, you need to involve y in this experiment!

 

[Dr.Peterson]

So do i get a parametric equation?

x= (3-a)/(2-a), y= 2(3-a)/(2-a) + a

And directly finding a cartesian is not that simple i guess?

Solve the first equation for a in terms of x, then put that in place of a in the second equation. That will give you an equation for y as a function of x. No, it isn't easy, but it's straightforward. (You may want to simplify your second equation first.)

 

[apple2357]

Solve the first equation for a in terms of x, then put that in place of a in the second equation. That will give you an equation for y as a function of x. No, it isn't easy, but it's straightforward. (You may want to simplify your second equation first.)

Yes after doing the algebra, i get y= (2x^2-3)/(x-1) which appears correct. Its interesting that the path of an intersection point between two lines can produce something like this! Before i explored it on software, i wasn't expecting anything so different.

Does anyone know of other surprising paths ( well surprising to you!) that are produced by looking at intersection points?