PS, QT and RU are medians of triangle PQR. PS lies along the line y = x+3, QT lies along the line y = 2x+4.
The length of PQ is 60 and ∠PRQ = 90^o.

Determine the area of triangle PQR.

I realized pretty quickly that any algebra based of coordinates would be easier if everything were shifted about.  The given lines cross at the median which has coordinates (-1,2)  Translating everything by <1,-2> doesn't change the given lengths or angles but now:
PS lies along y=x, QT lies along y=2x, and the median is (0,0)

Now coordinatize:
P=(2a,2a)
Q=(2b,4b)
U=(a+b,a+2b)
O=(0,0) the median

First idea:
R is on the circle centered at U with radius 900.
R is also on the line through U and O.
Find the intersection and you find R.
The algebra made a mess and I couldn't get both variables to drop out.  (I later realized this doesn't fix the position of R because the other medians are not guaranteed.)

Second idea:
R is on the line through U and O.
This line is y=(a+2b)/(a+b)*x
R is the place on this line that makes PR and QR perpendicular.
Find both slopes and multiply.  Product will be -1.
This turned out to be messier than before because it adds a variable.  None of them dropped out. (Same reason as before.)

Ok lets just go draw the darned thing on Geometer's Sketchpad.
Drawn as in First idea it became clear that Q is based of P which is not fixed.  So R is not fixed.  Move P and R moves but it is clear the other 'medians' are not actually at the midpoints.
Get P to just the right spot and the solution comes into focus and the area is 400.0000

For the record the lengths:
PR = 13.6948
QR = 58.4162
are not square roots of integers.

And the points (using my translation)
P=(-15.2072,-15.2072)
Q=(17.5356,35.0712)
R=(-2.3284,-19.8640)

Posted by Jer on 2015-12-27 21:34:28