(i) To find the equation of CD, first calculate the slope of AB. The slope \(m\) of a line through points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\(m = \frac{y_2 - y_1}{x_2 - x_1}\)
For AB, \(m = \frac{11 - 3}{13 - 1} = \frac{8}{12} = \frac{2}{3}\).
The slope of the perpendicular line CD is the negative reciprocal of \(\frac{2}{3}\), which is \(-\frac{3}{2}\).
Using point C (6, 15) and the slope \(-\frac{3}{2}\), the equation of CD is:
\(y - 15 = -\frac{3}{2}(x - 6)\)
(ii) To find the coordinates of D, first find the equation of AB using point A (1, 3):
\(y - 3 = \frac{2}{3}(x - 1)\)
Rearrange to get:
\(3y - 9 = 2x - 2\)
\(3y = 2x + 7\)
\(y = \frac{2}{3}x + \frac{7}{3}\)
Now solve the system of equations:
1. \(y - 15 = -\frac{3}{2}(x - 6)\)
2. \(y = \frac{2}{3}x + \frac{7}{3}\)
Substitute equation 2 into equation 1:
\(\frac{2}{3}x + \frac{7}{3} - 15 = -\frac{3}{2}(x - 6)\)
\(\frac{2}{3}x + \frac{7}{3} - 15 = -\frac{3}{2}x + 9\)
Clear fractions by multiplying through by 6:
\(4x + 14 - 90 = -9x + 54\)
\(4x - 76 = -9x + 54\)
\(13x = 130\)
\(x = 10\)
Substitute \(x = 10\) back into equation 2:
\(y = \frac{2}{3}(10) + \frac{7}{3} = \frac{20}{3} + \frac{7}{3} = 9\)
Thus, the coordinates of D are (10, 9).
