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

Since CD = 6 units is horizontal and is tangent to the circle at point C, we can calculate C's coordinates using the offset in the y-direction. Given the center N(-1, 3), the coordinates of C are: C(-1, 7).

Coordinates of D can be calculated as follows. Since ABCD is a parallelogram, D and E will share the same y-coordinate as C. Thus, D has coordinates D(5, 7), which is determined by moving 6 units to the right on the x-axis from C.

The area of parallelogram ABCD can be calculated using the base and height. With base BC being 6 units (length of CD) and height equal to the vertical distance from A(-1, -1) to D(5, 7), which is 5 units. Then, the area is given by:

$Area_{ABCD} = ext{base} imes ext{height} = 6 imes 5 = 30 ext{ units}^2.$

To find the length NM, use the coordinates N(-1, 3) and M(1, 3) after reflecting N across the line y=x. The distance formula gives:

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

The midpoint of segment AF is given by the average of the coordinates of A(-1, -1) and F(1, 1). Thus:

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