Find the lengths of the medians AD and BE of triangle ABC whose vertices are A(7,-3) B(5,3) and C(3,-1).
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Hii friend,
Since D and E are the midpoints of BC and AC respectively.
So their coordinates are,
D(X1+X2/2) , (Y1+Y2/2)
D[(5+3/2) , (3+(-1)/2]
and,
E(X1+X2/2 ) , (Y1+Y2/2)
E(7+3/2), ( -3-1/2)]
Therefore,
D(4,1) and E(5,-2)
Therefore,
Using distance formula,
AD = under root (7-4)² + (-3-1)² = ✓3² + (-4)²
= ✓9+16 = ✓25 = 5 units.
And,
BE = under root (5-5)² + (-2-3)² = ✓0+(-5)² = ✓25 = 5 units..
HOPE IT WILL HELP U.. :-)
Since D and E are the midpoints of BC and AC respectively.
So their coordinates are,
D(X1+X2/2) , (Y1+Y2/2)
D[(5+3/2) , (3+(-1)/2]
and,
E(X1+X2/2 ) , (Y1+Y2/2)
E(7+3/2), ( -3-1/2)]
Therefore,
D(4,1) and E(5,-2)
Therefore,
Using distance formula,
AD = under root (7-4)² + (-3-1)² = ✓3² + (-4)²
= ✓9+16 = ✓25 = 5 units.
And,
BE = under root (5-5)² + (-2-3)² = ✓0+(-5)² = ✓25 = 5 units..
HOPE IT WILL HELP U.. :-)
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