1. The problem asks why the gradient of the perpendicular bisector of segment AB is -2. 2. First, recall that the gradient (slope) of a line perpendicular to another line is the negative reciprocal of the original line's gradient. 3. To find the gradient of AB, we need the coordinates of points A and B. We know A is (-2, 2) and B lies on the y-axis, so B's x-coordinate is 0. Let B be (0, y_B). 4. The midpoint of AC is given as (4, 5), but this is not directly related to AB's slope. Instead, let's find the slope of AB using A(-2, 2) and B(0, y_B): $$m_{AB} = \frac{y_B - 2}{0 - (-2)} = \frac{y_B - 2}{2}$$ 5. The problem states the gradient of BD is -2, and BD is perpendicular to AC. Since BD is perpendicular to AC, and the gradient of AC is $\frac{1}{2}$, the gradient of BD is the negative reciprocal: $$m_{BD} = -\frac{1}{m_{AC}} = -\frac{1}{\frac{1}{2}} = -2$$ 6. The perpendicular bisector of AB has the same gradient as BD because it is perpendicular to AB, and BD is perpendicular to AC. Since AC's gradient is $\frac{1}{2}$, the perpendicular bisector's gradient is $-2$. 7. Therefore, the gradient of the perpendicular bisector of AB is $-2$ because it is perpendicular to AB, whose gradient is $\frac{1}{2}$, and the negative reciprocal of $\frac{1}{2}$ is $-2$. 8. In summary, perpendicular lines have gradients that are negative reciprocals, so if AB has gradient $\frac{1}{2}$, the perpendicular bisector must have gradient $-2$.