The correct option is
A 400Given line
y=x+3−−−−−(1)
y=2x+4−−−−−(2)
By shiftiing centroid to origin
Thus now equation of median are y=x and y=2x
Now the coordinates of A and B can be taken as (a,a) and B(b,2b)
Using centroif formula C(−a−b,−a−2b)
Length of hypotenuse AB=60
AB2=(a−b)2+(a−2b)2
3600=a2+b2−2ab+a2+4b2−4ab
2a2+5b2−6ab=3600−−−(3)
Slope of line AC from point A,C is mAC=−a−2b−a−a−b−a
mAC=2a+2b2a+b
Slope of line BC from point B,C is mBC=−a−2b−2b−a−b−b
mBC=a+3ba+2b
Angle C=90 means line AC and BC are perpendicular
Hence mACmBC=−1
2a+2b2a+b×a+3ba+2b=−1
2a2+6ab+2ab+6b2=−2a2−4ab−ab−2b2
4a2+8b2+13ab=0−−−−−−−−(4)
Solving both (3) and (4) equations we get
ab=−8003
Areaoftraingle=32(ab)=400