Question

Consider the tetrahedron with the vertices \( A(3,2,4) \), \( B(x_1,y_1,0) \), \( C(x_2,y_2,0) \), and \( D(x_3,y_3,0) \). If the triangle \( BCD \) is formed by the lines \( y = x \), \( x + y = 6 \), and \( y = 1 \), then the centroid of the tetrahedron is:

• A. \( \left(\frac{9}{4}, \frac{7}{4}, 1 \right) \)
• B. \( \left(\frac{11}{4}, \frac{5}{4}, 1 \right) \)
• C. \( \left(3, \frac{7}{4}, 1 \right) \)
• D. \( (3,2,1) \)

Hint:

To find the centroid of a tetrahedron, use the formula: \[ G = \left( \frac{x_1 + x_2 + x_3 + x_4}{4}, \frac{y_1 + y_2 + y_3 + y_4}{4}, \frac{z_1 + z_2 + z_3 + z_4}{4} \right). \]

Answer 1

The Correct Option is C

Step 1: Finding the vertices of \( BCD \) 
The lines given are: 1. \( y = x \) 2. \( x + y = 6 \) 3. \( y = 1 \) Solving for intersections: - Intersection of \( y = x \) and \( x + y = 6 \): \[ x + x = 6 \Rightarrow 2x = 6 \Rightarrow x = 3, y = 3. \] So, \( B(3,3,0) \). - Intersection of \( x + y = 6 \) and \( y = 1 \): \[ x + 1 = 6 \Rightarrow x = 5, y = 1. \] So, \( C(5,1,0) \). - Intersection of \( y = x \) and \( y = 1 \): \[ x = 1, y = 1. \] So, \( D(1,1,0) \). 
Step 2: Finding the centroid of tetrahedron 
The centroid \( G \) of a tetrahedron with vertices \( (x_1, y_1, z_1) \), \( (x_2, y_2, z_2) \), \( (x_3, y_3, z_3) \), and \( (x_4, y_4, z_4) \) is given by: \[ G = \left( \frac{x_1 + x_2 + x_3 + x_4}{4}, \frac{y_1 + y_2 + y_3 + y_4}{4}, \frac{z_1 + z_2 + z_3 + z_4}{4} \right). \] Substituting \( A(3,2,4) \), \( B(3,3,0) \), \( C(5,1,0) \), \( D(1,1,0) \): \[ G_x = \frac{3 + 3 + 5 + 1}{4} = \frac{12}{4} = 3. \] \[ G_y = \frac{2 + 3 + 1 + 1}{4} = \frac{7}{4}. \] \[ G_z = \frac{4 + 0 + 0 + 0}{4} = 1. \] Thus, the centroid is: \[ \left( 3, \frac{7}{4}, 1 \right). \]