In the diagram, the circle centred at N(-1; 3) passes through A(-1; -1) and C B(-4; 2) - NSC Mathematics - Question 4 - 2021 - Paper 2

The point C is directly above point A due to the vertical nature of the circle. As A is at (-1, -1) and the radius is 4 units, the coordinates of C are:

$C = (-1, -1 + 4) = (-1, 3)$

Now, we find the coordinates of D. The line CD is horizontal (parallel to the x-axis) and has a distance of CD = 6 units. Therefore, D can be located at:

$D = (-1 + 6, 3) = (5, 3)$

The area of parallelogram ABCD can be obtained using the formula:

$ext{Area} = ext{base} imes ext{height}$

Here, the base AB is given as the distance between A and B, which is 6 units, and the height (distance from line AB to line CD) is 5 units. Therefore:

$ext{Area}_{ABCD} = 6 imes 5 = 30 ext{ square units}$

To find the length of NM, we determine the coordinates of M, the reflection of N(-1, 3) across the line y = x. The reflection results in:

$M = (3, -1)$

Using the distance formula, we get:

$NM = ext{sqrt}((-1 - 3)^2 + (3 - (-1))^2) = ext{sqrt}(16 + 16) = ext{sqrt}(32) = 4 ext{ units}$

The coordinates of A are (-1, -1) and we need to find coordinates of F, the reflection of A through the line y = x. Thus:

$F = (-1, -1) \Rightarrow F = (-1, -1) ext{ reflected becomes } F(1, -1)$

Now, the midpoint AF is calculated as:

$ext{Midpoint} = \left( \frac{-1 + 1}{2}, \frac{-1 + (-1)}{2} \right) = (0, -1)$