Question

The vertices of triangle \( \triangle ABC \) are \( A(1, 1, 0), B(1, 2, 1) \), and \( C(-2, 2, -1) \). Find the equations of the medians through A and B. Use the equations so obtained to find the coordinates of the centroid.

Hint:

To find the centroid of a triangle, first find the mid-points of two sides and then find the intersection of the medians. The centroid divides each median in a 2:1 ratio.

Answer 1

Step 1: {Find the mid-point of \( BC \)}
The coordinates of \( B \) are \( (1, 2, 1) \) and the coordinates of \( C \) are \( (-2, 2, -1) \). The mid-point of \( BC \) is: \[ \left( \frac{1 + (-2)}{2}, \frac{2 + 2}{2}, \frac{1 + (-1)}{2} \right) = \left( -\frac{1}{2}, 2, 0 \right). \] Step 2: {Find the equation of the median through \( A \)}
The equation of the median through \( A(1, 1, 0) \) is the line joining \( A \) to the mid-point of \( BC \left( -\frac{1}{2}, 2, 0 \right) \). Using the parametric form of the equation of a line: \[ x - 1 = \lambda \left( -\frac{1}{2} - 1 \right), \quad y - 1 = \lambda \left( 2 - 1 \right), \quad z - 0 = \lambda (0 - 0). \] Simplifying: \[ x - 1 = \lambda \left( -\frac{3}{2} \right), \quad y - 1 = \lambda, \quad z = 0. \] Thus, the equation of the median through \( A \) is: \[ x = 1 - \frac{3\lambda}{2}, \quad y = 1 + \lambda, \quad z = 0. \quad \cdots (1) \] Step 3: {Find the mid-point of \( AC \)}
The coordinates of \( A \) are \( (1, 1, 0) \) and the coordinates of \( C \) are \( (-2, 2, -1) \). The mid-point of \( AC \) is: \[ \left( \frac{1 + (-2)}{2}, \frac{1 + 2}{2}, \frac{0 + (-1)}{2} \right) = \left( -\frac{1}{2}, \frac{3}{2}, -\frac{1}{2} \right). \] Step 4: {Find the equation of the median through \( B \)}
The equation of the median through \( B(1, 2, 1) \) is the line joining \( B \) to the mid-point of \( AC \left( -\frac{1}{2}, \frac{3}{2}, -\frac{1}{2} \right) \). Using the parametric form of the equation of a line: \[ x - 1 = \mu \left( -\frac{1}{2} - 1 \right), \quad y - 2 = \mu \left( \frac{3}{2} - 2 \right), \quad z - 1 = \mu \left( -\frac{1}{2} - 1 \right). \] Simplifying: \[ x - 1 = \mu \left( -\frac{3}{2} \right), \quad y - 2 = \mu \left( -\frac{1}{2} \right), \quad z - 1 = \mu \left( -\frac{3}{2} \right). \] Thus, the equation of the median through \( B \) is: \[ x = 1 - \frac{3\mu}{2}, \quad y = 2 - \frac{\mu}{2}, \quad z = 1 - \frac{3\mu}{2}. \quad \cdots (2) \] Step 5: {Find the point of intersection of the two lines (the centroid)}
Any point on the line (1) is \( (-3\lambda + 1, \lambda + 1, 0) \).
Any point on the line (2) is \( (-3\mu + 1, -\mu + 2, -3\mu + 1) \).
For the point of intersection: \[ -3\lambda + 1 = -3\mu + 1, \quad \lambda + 1 = -\mu + 2, \quad 0 = -3\mu + 1. \] From the third equation, \( 3\mu = 1 \), so \( \mu = \frac{1}{3} \). Substitute \( \mu = \frac{1}{3} \) into the second equation: \[ \lambda + 1 = -\frac{1}{3} + 2 = \frac{5}{3} \quad \Rightarrow \quad \lambda = \frac{5}{3} - 1 = \frac{2}{3}. \] Step 6: {Find the coordinates of the centroid}
The coordinates of the centroid are: \[ x = -3\left( \frac{2}{3} \right) + 1 = 0, \quad y = \frac{5}{3} + 1 = \frac{5}{3}, \quad z = 0. \] Conclusion: The coordinates of the centroid are \( (0, \frac{5}{3}, 0) \).