First, find the midpoint M of BD. The formula for the midpoint is:
\(M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)
Substitute the coordinates of B (2, 10) and D (6, 2):
\(M = \left( \frac{2 + 6}{2}, \frac{10 + 2}{2} \right) = (4, 6)\)
Next, find the gradient of BD:
\(m_{BD} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 10}{6 - 2} = -2\)
Since ABCD is a rhombus, the diagonals are perpendicular. Therefore, the gradient of AC is the negative reciprocal of \(m_{BD}\):
\(m_{AC} = \frac{1}{2}\)
The equation of line AC passing through M (4, 6) is:
\(y - 6 = \frac{1}{2}(x - 4)\)
Since A lies on the x-axis, \(y = 0\):
\(0 - 6 = \frac{1}{2}(x - 4)\)
\(-6 = \frac{1}{2}x - 2\)
\(-4 = \frac{1}{2}x\)
\(x = -8\)
Thus, A is (-8, 0).
To find C, use the vector method. The vector from A to M is:
\(\overrightarrow{AM} = (4 - (-8), 6 - 0) = (12, 6)\)
Since M is the midpoint, C is:
\(C = M + \overrightarrow{AM} = (4, 6) + (12, 6) = (16, 12)\)
