The
acute angle, q, between the two lines with gradients 1 and 2
is given by tan(q) = (2 - 1)/(1 + 2x1) = 1/3.

The point of intersection of the lines, G, is the centroid of triangle
ABC. Let a, b and c be the position
vectors of A, B and C,
relative to G, and a, b and c be their magnitudes.

Since /ACB = 90^{o}, C lies on a circle with diameter AB. So the

median from C is a radius of length 30, and GC = c = 20.

Since G is the centroid:a
+ b + c= 0(1)

Since /ACB = 90^{o}:(a – c).(b – c) = 0 (2)

From (2):a.b – c.(a + b) + c^{2} = 0

and using (1):a.b = -2c^{2}

which gives:a*b*cos(180^{o}
- q) = -800

a*b*cos(q)
= 800

Since the medians of a triangle divide it into 6 triangles with
equal areas, it follows that