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The building shown in the picture below, has sides in the form of quadrilaterals - NSC Technical Mathematics - Question 1 - 2019 - Paper 2
Question 1
The building shown in the picture below, has sides in the form of quadrilaterals. Quadrilateral ABDF models a side view of the building, in the Cartesian plane, wi... show full transcript
Worked Solution & Example Answer:The building shown in the picture below, has sides in the form of quadrilaterals - NSC Technical Mathematics - Question 1 - 2019 - Paper 2
Step 1
1.1 The size of \(\alpha\)
Answer
Given that the angle formed by the x-axis and line BD is 76°, we can find (\alpha) as follows:
Since lines BD and the x-axis form a complementary angle, we have:
[ \alpha = 90° - 76° = 14° ]
Thus, the size of (\alpha) is (104°) (this represents the angle at D, which is supplementary to the angle with the x-axis).
Step 2
1.2 The length of AF (leave answer in simplified surd form)
Answer
To find the length of AF, we use the distance formula:
[ AF = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ]
Substituting the coordinates of points A and F:
[ AF = \sqrt{(-8 - (-2))^2 + (-4 - 8)^2} ] [ = \sqrt{(-6)^2 + (-12)^2} ] [ = \sqrt{36 + 144} ] [ = \sqrt{180} ] [ = 6\sqrt{5} ]
Hence, the length of AF is (6\sqrt{5}).
Step 3
1.3 The gradient of BD (rounded off to the nearest integer)
Answer
The gradient (slope) of a line is given by the formula:
[ m = \tan \theta ]
where (\theta) is the angle of the line with the horizontal. We already know (\theta = 76°):
[ m = \tan(76°) \approx 4 ] (rounded to the nearest integer)
Therefore, the gradient of BD is approximately 4.
Step 4
1.4 The coordinates of the midpoint of AF
Answer
The midpoint can be calculated using the midpoint formula:
[ M_{AF} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) ]
Substituting the coordinates of points A and F:
[ M_{AF} = \left(\frac{-2 + (-8)}{2}, \frac{8 + (-4)}{2}\right) ] [ = \left(\frac{-10}{2}, \frac{4}{2}\right) ] [ = (-5, 2) ]
So, the coordinates of the midpoint of AF are (-5, 2).
Step 5
1.5 Hence, determine the perpendicular bisector of AF in the form \(y = ...\)
Answer
To find the equation of the perpendicular bisector, we first need the gradient of AF:
The gradient of AF is given by:
[ m_{AF} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-4 - 8}{-8 - (-2)} = \frac{-12}{-6} = 2 ]
The gradient of the perpendicular bisector is the negative reciprocal of (m_{AF}):
[ m_{perpendicular} = -\frac{1}{2} ]
Using the midpoint (-5, 2) in the point-slope form:
[ y - y_1 = m(x - x_1) ]
[ y - 2 = -\frac{1}{2}(x + 5) ]
Rearranging this to find the y-intercept:
[ y = -\frac{1}{2}x - \frac{5}{2} + 2 ] [ y = -\frac{1}{2}x + \frac{3}{2} ]
Thus, the equation of the perpendicular bisector of AF is (y = -\frac{1}{2}x + \frac{3}{2}).