# Thread: [SOLVED] Linear substitution

1. ## [SOLVED] Linear substitution

D is the triangle with corners in (0,0), (1,-1) and (3,0).

If we make the substitution: $u=x+y$ and $v=x-2y$ we end up with a triangle E with corners in (0,0), (0,3) and (3,3).

The triangle D is formed by the three lines $x+y=0$, $x-2y=3$ and $y=0$, but I'm not quite sure how you're supposed to think when you transform this to uv coordinates (the E triangle).

2. Originally Posted by Spec
D is the triangle with corners in (0,0), (1,-1) and (3,0).

If we make the substitution: $u=x+y$ and $v=x-2y$ we end up with a triangle E with corners in (0,0), (0,3) and (3,3).

The triangle D is formed by the three lines $x+y=0$, $x-2y=3$ and $y=0$, but I'm not quite sure how you're supposed to think when you transform this to uv coordinates (the E triangle).
just transform each sides of the triangle of D: the side $x+y=0, \ 0 \leq x \leq 1, \ -1 \leq y \leq 0$ goes to $u=x+y=0, \ v=x-2y=-3y.$ since $-1 \leq y \leq 0,$ we'll get: $0 \leq v \leq 3.$

so that side of D is transformed to $u=0, \ 0 \leq v \leq 3.$ now let's see what happens to the second side of D, i.e. $x-2y=3, \ 1 \leq x \leq 3, \ -1 \leq y \leq 0.$ we'll have: $v=x-2y=3$

and $u=x+y=3y+3.$ since $-1 \leq y \leq 0,$ we'll have $0 \leq 3y+3 \leq 3.$ so the second side of D goes to $0 \leq u \leq 3, \ v = 3$ in uv plane. the third side of D is $y=0, \ 0 \leq x \leq 3.$ thus

$u=x+y=x, \ v=x-2y=x.$ so $u=v=x.$ hence the image of the third side of D in uv plane is the line segment $v=u, \ 0 \leq u \leq 3.$