Question

The mid points of the sides of triangle are (1, 5, -1) (0, 4, -2) and (2, 3, 4) then centroid of the triangle

• A. (1, 4, 3)
• B. \((1, 4,\frac{1}{3})\)
• C. (-1, 4, 3)
• D. \((\frac{1}{3},2,4)\)

Answer 1

The Correct Option is B

The centroid of a triangle is the point where the three medians intersect. It is also the average of the coordinates of the three vertices of the triangle. Since the given points are midpoints of the sides of the triangle, we can use these points to find the centroid. Let the three midpoints of the triangle be: \[ A = (1, 5, -1), \quad B = (0, 4, -2), \quad C = (2, 3, 4) \] The centroid \((G_x, G_y, G_z)\) is the average of the coordinates of the points: \[ G_x = \frac{1 + 0 + 2}{3} = 1, \quad G_y = \frac{5 + 4 + 3}{3} = 4, \quad G_z = \frac{-1 + (-2) + 4}{3} = \frac{1}{3} \] Therefore, the centroid of the triangle is \( (1, 4, \frac{1}{3}) \).

Answer 2

Let the vertices of the triangle be A, B, and C. Let the midpoints of the sides be D(1, 5, -1), E(0, 4, -2), and F(2, 3, 4), where D is the midpoint of BC, E is the midpoint of AC, and F is the midpoint of AB.

Let the coordinates of A, B, and C be (x₁, y₁, z₁), (x₂, y₂, z₂), and (x₃, y₃, z₃) respectively.

Then, we have:

D: \(\frac{x_2 + x_3}{2} = 1\), \(\frac{y_2 + y_3}{2} = 5\), \(\frac{z_2 + z_3}{2} = -1\)

E: \(\frac{x_1 + x_3}{2} = 0\), \(\frac{y_1 + y_3}{2} = 4\), \(\frac{z_1 + z_3}{2} = -2\)

F: \(\frac{x_1 + x_2}{2} = 2\), \(\frac{y_1 + y_2}{2} = 3\), \(\frac{z_1 + z_2}{2} = 4\)

Thus,

x₂ + x₃ = 2, y₂ + y₃ = 10, z₂ + z₃ = -2

x₁ + x₃ = 0, y₁ + y₃ = 8, z₁ + z₃ = -4

x₁ + x₂ = 4, y₁ + y₂ = 6, z₁ + z₂ = 8

The centroid G of the triangle ABC is given by \((\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3})\).

Adding the x-coordinate equations: (x₂ + x₃) + (x₁ + x₃) + (x₁ + x₂) = 2 + 0 + 4

2(x₁ + x₂ + x₃) = 6, so x₁ + x₂ + x₃ = 3

Adding the y-coordinate equations: (y₂ + y₃) + (y₁ + y₃) + (y₁ + y₂) = 10 + 8 + 6

2(y₁ + y₂ + y₃) = 24, so y₁ + y₂ + y₃ = 12

Adding the z-coordinate equations: (z₂ + z₃) + (z₁ + z₃) + (z₁ + z₂) = -2 - 4 + 8

2(z₁ + z₂ + z₃) = 2, so z₁ + z₂ + z₃ = 1

Therefore, the centroid is \((\frac{3}{3}, \frac{12}{3}, \frac{1}{3}) = (1, 4, \frac{1}{3})\).