Given the problem of finding the intersection of 2 parameterised lines
L1: $x=2-t ; y=1+t$ and L2: $x=2+t ; y=4+t$.
Recovering original eqns $y=3-x$ and $y=2x$ yields the correct answer of $(1,2)$ for the point of intersection. However, when I set each $x$'s $2-t=2+t$ and $y$'s equal $1+t=4+t$ and solve I get $(2,-2)$. Why? If $x=2-t$ specifies how $x$ varies (with $t$) for L1 and likewise $x=2+t$ for L2 why does setting the $x$'s equal not yield the $x$ common to both, ie their point of intersection? What is the significance of setting $x$'s equal, and the significance of $(2,-2)$? (And similarly for $y$?)
Furthermore, if I change the paramters eg $x=\ln A; y=2\ln A$ for L2 and $x=t ; y=3-t$ for L1, and then equate the $x$'s and $y$'s I get the correct answer of $(1,2)$. Why is this so? Thanks
The mistake you’ve made is in using the same parameter $t$ when equating $x$ and $y$ values of the two lines. There’s no reason to believe that the point at which the lines intersect will correspond to the same value of $t$ for both lines. Instead, you need to rename the parameter for one of the lines before equating them, e.g., $$ x = 2-s = 2+t \\ y = 1+s = 4+t. $$ Solving this system results in $s=\frac32$, $t=-\frac32$, which when substituted back into the parametric equations of either line gives $x=\frac12$, $y=\frac52$ for the intersection point. Notice that the two parameters are indeed unequal at the intersection.
This disagrees with your first solution, but that’s because you’ve also made an error in rewriting the equation for $L2$. It’s not $y=2x$, but $y=x+2$. Using the correct equation results in the solution $x=\frac12$, $y=\frac52$ as expected.
There’s no particular significance to the pair $(2,-2)$. Indeed, you’re kind of lucky that you found a “solution” at all. Using your second parametrization of $L1$, $x=t$, $y=3-t$, together with the original one for $L2$, your method results in the equation $x = t = 2+t$, which has no solution.