Question

In a triangle ABC, co-ordinate of orthocentre, centroid and vertex A are (3, 2), (3, 1) and (1, 2) respectively. Then x-coordinate of vertex B, is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (D)

4

Answer 2

1. Centroid Formula: The centroid G of a triangle with vertices A$(x_A, y_A)$, B$(x_B, y_B)$, and C$(x_C, y_C)$ is given by $G = \left(\frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3}\right)$. Given A(1, 2) and G(3, 1), we use the x-coordinate: $3 = \frac{1 + x_B + x_C}{3}$ $9 = 1 + x_B + x_C$ $x_B + x_C = 8$ (Equation 1)

2. Orthocentre Property: The orthocentre H is the intersection of the altitudes of a triangle. An altitude from a vertex is perpendicular to the opposite side. The line segment AH lies on the altitude from vertex A. Given A(1, 2) and H(3, 2). The slope of the line segment AH is $m_{AH} = \frac{y_H - y_A}{x_H - x_A} = \frac{2 - 2}{3 - 1} = \frac{0}{2} = 0$. Since the slope of AH is 0, the line segment AH is horizontal. This means the altitude from vertex A is a horizontal line. A horizontal line is perpendicular to a vertical line. Therefore, the side BC must be a vertical line. For BC to be a vertical line, the x-coordinates of vertices B and C must be equal: $x_B = x_C$ (Equation 2)

3. Solving for $x_B$: Substitute Equation 2 into Equation 1: $x_B + x_B = 8$ $2x_B = 8$ $x_B = 4$

The x-coordinate of vertex B is 4.