Write the coordinates of A, B and C - Junior Cycle Mathematics - Question Question 1 - 2013

To find D, the mid-point of segment [AB], use the formula: $D = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$ Substituting the coordinates of A and B: $D = \left( \frac{3 + (-6)}{2}, \frac{6 + 0}{2} \right) = \left( -\frac{3}{2}, 3 \right)$

To find the equation of the line AB, we first calculate the slope: $\text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 6}{-6 - 3} = \frac{-6}{-9} = \frac{2}{3}$ Using point-slope form, we find the equation of the line: $y - y_1 = m(x - x_1)$ So, $y - 6 = \frac{2}{3}(x - 3)$ This simplifies to: $y = \frac{2}{3}x + 4$

The slope of line AB is $\frac{2}{3}$, so the perpendicular slope is the negative reciprocal: $\text{perpendicular slope} = -\frac{3}{2}$ Using point-slope form at point C (4, -2): $y - (-2) = -\frac{3}{2}(x - 4)$ This can be rewritten and simplified: $2y + 4x - 8 = 0$

To find E, solve the equations of the lines obtained earlier: Line AB: $y = \frac{2}{3}x + 4$ Perpendicular line through C: $2y + 4x - 8 = 0$ Substituting from AB into the perpendicular line gives: $2\left( \frac{2}{3}x + 4 \right) + 4x - 8 = 0$ Solving this leads to: $x = 0, y = 4$ Thus, E is (0, 4).

First, calculate the distances:

1. For |CD|: $|CD| = \sqrt{(4 - (-\frac{3}{2}))^2 + (-2 - 3)^2} = \sqrt{(\frac{8}{2})^2 + (-5)^2} = \sqrt{16 + 25} = \sqrt{41}$
2. For |CE|: $|CE| = \sqrt{(0 - 4)^2 + (4 - (-2))^2} = \sqrt{(-4)^2 + (6)^2} = \sqrt{16 + 36} = \sqrt{52}$ Since $\sqrt{41} < \sqrt{52}$, |CD| is the shorter distance.