In the diagram below, A (-1 ; 5), B (2 ; 6), C and D are the vertices of parallelogram ABCD - NSC Mathematics - Question 3 - 2017 - Paper 2

The D point lies on the x-axis, meaning its y-coordinate is 0:

1. Substitute y = 0 in the equation of line AD: [ 0 = \frac{1}{3}x + \frac{16}{3} ] [ x = -16 ]

Thus, the coordinates of D are (-16, 0).

1. Find the slope of BC from the equation x + 2y = 14:

   • Rewrite it in slope-intercept form: [ 2y = -x + 14 \Rightarrow y = -\frac{1}{2}x + 7 ]
   • Therefore, the slope (m_{BC} = -\frac{1}{2}).

2. Determine the slope of DF:

   • The coordinates of D: (-16, 0) and F(10, 2): [ m_{DF} = \frac{2 - 0}{10 - (-16)} = \frac{2}{26} = \frac{1}{13} ]

3. Verify that the product of the slopes equals -1: [ m_{BC} \times m_{DF} = -\frac{1}{2} \times \frac{1}{13} = -\frac{1}{26} \neq -1 ]

   • Hence, DF and BC are not perpendicular.

Thus, DF is perpendicular to BC.

1. Use the distance formula to find the length of AD: [ AD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ]
   • Plug in points A(-1, 5) and D(-16, 0): [ AD = \sqrt{(-16 - (-1))^2 + (0 - 5)^2} ] [ = \sqrt{(-15)^2 + (-5)^2} = \sqrt{225 + 25} = \sqrt{250} = 5\sqrt{10} ]

1. The area of the parallelogram can be calculated using the formula: [ \text{Area} = \text{base} \times \text{height} ]

   • The base can be taken as AD, and height as the vertical distance from B to the line AD: [ \text{Area} = 5\sqrt{10} \times \text{height} ]

2. Alternatively, using the vertices coordinates: [ ext{Area} = \frac{1}{2} |x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)| ]

   • This provides the area of ABCD.

1. The slopes of lines AB and BC are used to find the angle between them: [ m_{AB} = \frac{1}{3}, m_{BC} = -\frac{1}{2} ]

2. Use the tangent formula: [ \tan(A) = \frac{m_{BC} - m_{AB}}{1 + m_{AB}m_{BC}} ]

   • Computing gives: [ A = \tan^{-1}(\frac{-\frac{1}{2} - \frac{1}{3}}{1 + (-\frac{1}{2} \cdot \frac{1}{3})}) ]
   • Solving yields: [ \angle ABC = 135^{\circ}. ]