Simpler solution to this geometry/trig problem? i had a geometry/trignometry problem come up at work today, and i've been out of school too long: i've lost my tools.
i'm starting with a rectangle of known width (w) and height (h). For graphical simplification i can convert it into a right-angle triangle:

i'm trying to find the coordinates of that point above which is perpendicular to the origin:

i've labelled the opposite angle t1 (i.e. theta1, but Microsoft Paint cannot easily do greek and subscripts), and i deduce that the two triangles are similar (i.e. they have the same shape):

Now we come to my problem. Given w and h, find x and y.
Now things get very difficult to keep drawing graphically, to explain my attempts so far.
But if i call the length of the line segment common to both triangles M:

then:
M = w∙sin(t1)

Now i can focus on the other triangle, which i'll call O-x-M:

and use trig to break it down, giving:
x = M∙sin(t1)
  = w∙sin(t1)∙sin(t1)

y = M∙cos(t1)
  = w∙sin(t1)∙cos(t1)

with
t1 = atan(h/w)

Now this all works (i think, i've not actually tested it yet), and i'll be giving it to a computer, so speed isn't horribly important.
But my god, there must have been an easier way to get there. i feel like i'm missing something.

By the way, what this will be used for is drawing a linear gradient in along that perpendicular:


Suppose $h = (h_1,h_2)$ and $w = (w_1, w_2)$. Then you know the slope of the line. Let's call this $m_1$. You also know that the line perpendicular to the point $(x,y)$ has slope $-\frac{1}{m_1}$. You get a system of equations. 
