Here are the solutions to the questions posed:
3.1
3.1.1 The length of AB.
To find the distance between points \( A(-2, 0) \) and \( B(2, -4) \), use the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Plugging in the coordinates: \[ d = \sqrt{(2 - (-2))^2 + (-4 - 0)^2} = \sqrt{(2 + 2)^2 + (-4)^2} \] \[ d = \sqrt{(4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \]
Answer: \( 4\sqrt{2} \)
3.1.2 The coordinates of C, the midpoint of AB.
The coordinates of the midpoint \( C \) between points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) can be calculated as:
\[ C \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
Using the coordinates of points \( A \) and \( B \): \[ C \left( \frac{-2 + 2}{2}, \frac{0 + (-4)}{2} \right) = C \left( \frac{0}{2}, \frac{-4}{2} \right) = C(0, -2) \]
Answer: \( (0, -2) \)
3.1.3 The gradient of AB.
The gradient (slope) \( m \) of line \( AB \) is calculated as:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Using the coordinates of points \( A \) and \( B \): \[ m = \frac{-4 - 0}{2 - (-2)} = \frac{-4}{2 + 2} = \frac{-4}{4} = -1 \]
Answer: \( -1 \)
3.1.4 The equation of the perpendicular bisector of AB.
The perpendicular bisector has a gradient that is the negative reciprocal of the line segment \( AB \). Since the gradient of \( AB \) is \( -1 \), the gradient of the perpendicular bisector \( m_p \) is:
\[ m_p = 1 \]
Now, using point \( C(0, -2) \) to find the equation in point-slope form:
\[ y - y_1 = m_p(x - x_1) \] \[ y - (-2) = 1(x - 0) \] \[ y + 2 = x \] \[ y = x - 2 \]
Answer: \( y = x - 2 \)
3.2
To find the equation of the line parallel to \( 2x + y - 6 = 0 \) and passing through the point \( (-2, 1) \):
First, rearranging into slope-intercept form: \[ y = -2x + 6 \] The slope of this line is \( -2 \). A parallel line will have the same slope.
Using point-slope form with point \( (-2, 1) \): \[ y - 1 = -2(x - (-2)) \] \[ y - 1 = -2(x + 2) \] \[ y - 1 = -2x - 4 \] \[ y = -2x - 3 \]
Answer: \( y = -2x - 3 \)
3.3
3.3.1 Determine the coordinates of point D to create parallelogram ABCD.
To find point D such that \( ABCD \) is a parallelogram, we can use the midpoints of diagonals \( AC \) and \( BD \) being the same.
Let \( D(x, y) \).
Since \( C(3, -2) \) is the third point, we find \( D \) such that: \[ \text{Midpoint of } AC = \text{Midpoint of } BD \]
Calculating the midpoint of \( AC \): \[ \text{Midpoint of } AC = \left( \frac{1 + 3}{2}, \frac{2 + (-2)}{2} \right) = \left( \frac{4}{2}, \frac{0}{2} \right) = (2, 0) \]
For diagonal \( BD \): \[ \text{Midpoint of } BD = \left( \frac{5 + x}{2}, \frac{1 + y}{2} \right) \]
Setting them equal: \[ (2, 0) = \left( \frac{5 + x}{2}, \frac{1 + y}{2} \right) \]
From the equations:
- \( 2 = \frac{5 + x}{2} \Rightarrow 4 = 5 + x \Rightarrow x = -1 \)
- \( 0 = \frac{1 + y}{2} \Rightarrow 0 = 1 + y \Rightarrow y = -1 \)
Thus, \( D = (-1, -1) \).
Answer: \( D(-1, -1) \)
3.3.2 Prove that ABCD is not a rhombus.
A rhombus requires that all sides are equal.
Calculating the lengths of \( AB \), \( BC \), \( CD \), and \( AD \):
- Length of \( AB \): \[ AB = \sqrt{(5 - 1)^2 + (1 - 2)^2} = \sqrt{16 + 1} = \sqrt{17} \]
- Length of \( BC \): \[ BC = \sqrt{(3 - 5)^2 + (-2 - 1)^2} = \sqrt{4 + 9} = \sqrt{13} \]
- Length of \( CD \): \[ CD = \sqrt{(-1 - 3)^2 + (-1 - (-2))^2} = \sqrt{16 + 1} = \sqrt{17} \]
- Length of \( AD \): \[ AD = \sqrt{(-1 - 1)^2 + (-1 - 2)^2} = \sqrt{4 + 9} = \sqrt{13} \]
Since \( AB \) and \( CD \) are equal, as well as \( BC \) and \( AD \) being equal but \( AB \neq BC \), \( ABCD \) is not a rhombus.
Answer: \( ABCD \) is not a rhombus because sides \( AB \) and \( BC \) are not equal.
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