The building shown in the picture below, has sides in the form of quadrilaterals - NSC Technical Mathematics - Question 1 - 2019 - Paper 2

The length of AF can be calculated using the distance formula, which is:

[ AF = \sqrt{(-2 - (-8))^2 + (8 - (-4))^2} ] [ = \sqrt{(6)^2 + (12)^2} ] [ = \sqrt{36 + 144} = \sqrt{180} = 6\sqrt{5} ]

To find the gradient (m) of the line BD, we use the tangent of the angle θ. Since ( \theta = 76° ):

[ m = \tan(76°) \approx 4 ]
(Rounded to the nearest integer)

The coordinates of the midpoint of AF can be calculated as follows:

[ M_{AF} = \left( \frac{-2 + (-8)}{2}, \frac{8 + (-4)}{2} \right) = \left( -5, 2 \right) ]

To find the perpendicular bisector, we first need the slope of line AF. Since the midpoint is ( M_{AF} ) and the slope is opposite to the slope of AF:

Let ( m_{AF} ) be calculated: [ m_{AF} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-4 - 8}{-8 - (-2)} = \frac{-12}{-6} = 2 ]

The perpendicular gradient will be: [ m_{perpendicular} = -\frac{1}{m_{AF}} = -\frac{1}{2} ]

Using the point-slope form of the equation: [ y - 2 = -\frac{1}{2}(x + 5) ]

Rearranging gives us: [ y = -\frac{1}{2}x + \frac{1}{2} ]