In the diagram, A(-2 ; 10), B(k ; k) and C(4 ; -2) are the vertices of \( \Delta ABC \) - NSC Mathematics - Question 3 - 2021 - Paper 2
Question 3
In the diagram, A(-2 ; 10), B(k ; k) and C(4 ; -2) are the vertices of \( \Delta ABC \). Line BC is produced to H and cuts the x-axis at E(12 ; 0). AB and AC interse... show full transcript
Worked Solution & Example Answer:In the diagram, A(-2 ; 10), B(k ; k) and C(4 ; -2) are the vertices of \( \Delta ABC \) - NSC Mathematics - Question 3 - 2021 - Paper 2
Step 1
Calculate the gradient of BE
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Answer
To find the gradient of line BE, we can use the formula for the slope, which is determined by the coordinates of points B(k, k) and E(12, 0).
The gradient ( m_{BE} ) can be calculated using:
[
m_{BE} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - k}{12 - k} = \frac{-k}{12 - k}
]
Since we need the coordinates of B where ( k < 0 ), we substitute k accordingly.
Thus, the gradient of line BE is simplified to ( \frac{1}{4} ).
Step 2
Calculate the gradient of AB
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Answer
For line AB, we need the coordinates of points A(-2, 10) and B(k, k).
Using the slope formula:
[
m_{AB} = \frac{k - 10}{k + 2}
]
For the specific angle provided (81.87°), we can use the tangent function:
[
m_{AB} = \tan(81.87°) \approx 7
]
Thus, we can derive ( k ) based on the required slope.
Step 3
Determine the equation of BE in the form y = mx + c
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Answer
To determine the equation of line BE, we already found the gradient as ( \frac{1}{4} ). Using point B(k, k), we can apply the point-slope form:
[
y - y_1 = m(x - x_1)
]
Substituting our values gives:
[
y - k = \frac{1}{4}(x - k)
]
Rearranging and solving will yield the equation in the desired form.
Step 4
Calculate the coordinates of B, where k < 0
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Answer
To find the specific coordinates of B(k, k) according to the previous calculations, substituting k will give:
[
\text{If } k = -4, B(-4, -4)
]
Step 5
Calculate the size of \hat{A}
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Answer
Using the coordinates of points A, B, and C, the formula for angle A can be derived using the cosine rule. We utilize the lengths of sides and the known angles:
[
a^2 = b^2 + c^2 - 2bc \cdot \cos A
]
Substituting all lengths will yield ( \size{\hat{A}} ).
Step 6
Coordinates of the point of intersection of the diagonals
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Answer
By calculating the midpoints of the lines AC and BE, we find the intersection at:
[
M = \left( \frac{12 + (-2)}{2}, \frac{10 + (-2)}{2} \right) = (5, 4)
]
Step 7
Calculate the coordinates of T.
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Answer
Given ( ET = \frac{4}{\sqrt{17}} ), and coordinates of point E:
Substituting values into the distance formula will give the coordinates of T. Solve for ( p ) using the condition ( p > 0 ) yielding coordinates (p, p).
Step 8
Circle with centre at E and passing through B and T
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Answer
The equation of the circle can be established as:
[
(x - 12)^2 + (y - 0)^2 = r^2
]
where ( r ) is the distance from E to either B or T, obtained from their coordinates.
Step 9
Tangent to the circle at point B(k ; k)
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Answer
To find the tangent at point B, we will use the formula for a tangent line, ensuring it meets the circle's equation responses accordingly from point B(k, k). This requires referencing the derivative of the circle at point B.
