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 20151227 21:34:28 