Q. 134.1( 8 Votes )

# The midpoints of

Answer :

The midpoints of the sides of a triangle are (1, 5, -1), (0, 4, -2) and (2, 3, 4).

Let its vertices be A(x_{1,}y_{1},z_{1}), B(x_{2},y_{2},z_{2}), C(x_{3},y_{3},z_{3}) .

The mid point of AB is (1,5,-1), therefore

x_{1} +x_{2} = 2………..eq.1

y_{1} + y_{2} = 10……eq.2

z_{1} + z_{2} = -2……..eq.3

Mid point of AC is (2,3,4), therefore

x_{1} +x_{3} = 4………..eq.4

y_{1} + y_{3} = 6……eq.5

z_{1} + z_{3} = 8……..eq.6

Mid point of BC is (0,4,-2), therefore

x_{2} +x_{3} = 0………..eq.7

y_{3} + y_{2} = 8……eq.8

z_{3} + z_{2} = -4……..eq.9

now, adding the equations 1,4 and 7, and divide it by two we get,

x_{1} + x_{2} + x_{3} = 3

now subtracting 1, 4, 7 individually, we get

x_{1} = 3, x_{2} = -1 and x_{3} = 1

now, adding the equations 2,5 and 8, and divide it by two we get,

y_{1} + y_{2} + y_{3} = 12

now subtracting 1, 4, 7 individually, we get

y_{1} = 4, y_{2} = 6 and y_{3} = 2

now, adding the equations 3,6 and 9, and divide it by two we get,

z_{1} + z_{2} + z_{3} = 1

now subtracting 1, 4, 7 individually, we get

z_{1} = 5, z_{2} = -7 and z_{3} = 3

therefore, the coordinates are A(3,4,5), B(-1,6,-7) and C(1,2,3).

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