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Home » The midpoints of the sides of a triangle are (1, 5, -1), (0, 4, -2) and (2, 3, 4). Find its vertices.
The midpoints of the sides of a triangle are (1, 5, -1), (0, 4, -2) and (2, 3, 4). Find its vertices.
CBSE | Class 11 | Exercise 26C | Maths | RS Aggarwal | Three Dimensional Geometry
The midpoints of the sides of a triangle are (1, 5, -1), (0, 4, -2) and (2, 3, 4). Find its vertices.

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(x1,y1,z1), B(x2,y2,z2), C(x3,y3,z3) .

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

\frac{{{x_2} + {x_1}}}{2} = 1

x1 + x2 = 2 …..(1)

\frac{{{y_2} + {y_1}}}{2} = 5

y1 + y2 = 10 …..(2)

\frac{{{z_2} + {z_1}}}{-1} = 5

z1 + z2 = -2 …..(3)

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

\frac{{{x_3} + {x_1}}}{2} = 2

x1 + x3 = 4 …..(4)

\frac{{{y_3} + {y_1}}}{2} = 3

y1 + y3 = 6 …..(5)

\frac{{{z_3} + {z_1}}}{2} = 4

z1 + z3 = 8 …..(6)

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

\frac{{{x_2} + {x_3}}}{2} = 0

x2 + x3 = 0 …..(7)

\frac{{{y_2} + {y_3}}}{2} = 4

y3 + y2 = 8 …..(8)

\frac{{{z_2} + {z_3}}}{2} = -2

z3 + z2 = -4 …..(9)

Adding the equations 1,4 and 7, and divide it by 2,

x1 + x2 + x3 = 3

Subtracting 1, 4, 7 individually,

x1 = 3

x2 = -1

x3 = 1

Adding the equations 2,5 and 8, and divide it by 2,

y1 + y2 + y3 = 12

Subtracting 1, 4, 7 individually,

y1 = 4

y2 = 6

y3 = 2

Adding the equations 3,6 and 9, and divide it by 2,

z1 + z2 + z3 = 1

Subtracting 1, 4, 7 individually,

z1 = 5

z2 = -7

z3 = 3

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

 

 

 

 

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