Find the lengths of the medians AD and BE of Triangle ABC whose vertices areA (7, − 3), B (5, 3) and C (3, −1).
The coordinates of the triangle ABC are A(7,-3), B(5,3) and C(3,-1)
Now, as AD is a median on BC.
Therefore, D is the mid-pt. of BC.
D= (5+3/2, 3-1/2)
D=( 4,1)
Now, as BE is a median on AC.
Therefore, E is the mid-pt. of AC.
E= (7+3/2, -3-1/2)
E= (5, -2)
Now,
AD= under root (7-4)2 + (-3-1)2 = under root 9+16= under root 25= 5 square units
BE= under root (5-5)2 + (3+2)2= under root 25= 5 square units
Therefore, AD=BE= 5 square units.
Now, as AD is a median on BC.
Therefore, D is the mid-pt. of BC.
D= (5+3/2, 3-1/2)
D=( 4,1)
Now, as BE is a median on AC.
Therefore, E is the mid-pt. of AC.
E= (7+3/2, -3-1/2)
E= (5, -2)
Now,
AD= under root (7-4)2 + (-3-1)2 = under root 9+16= under root 25= 5 square units
BE= under root (5-5)2 + (3+2)2= under root 25= 5 square units
Therefore, AD=BE= 5 square units.