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

Since D lies on the x-axis, its y-coordinate is 0. To find x-coordinate, we solve for where line AD intersects the x-axis (y=0).

Setting y = 0 in the equation of line AD:

0 = \frac{1}{3}x + \frac{16}{3}

Multiplying through by 3 gives:

0 = x + 16 => x = -16

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

First, we find the slope of BC from its equation x + 2y = 14. Rearranging gives:

$2y = -x + 14 \rightarrow y = -\frac{1}{2}x + 7$

Thus, the slope of BC (m_{BC}) is -(\frac{1}{2}).

To find the slope of DF, we calculate based on D(-16, 0) and F(10, 2):

$m_{DF} = \frac{y_F - y_D}{x_F - x_D} = \frac{2 - 0}{10 - (-16)} = \frac{2}{26} = \frac{1}{13}$

For two lines to be perpendicular, the product of their slopes must equal -1:

$m_{DF} * m_{BC} = \frac{1}{13} * -\frac{1}{2} = -\frac{1}{26} \neq -1$

Thus, DF is not perpendicular to BC.

To find the length of AD, we use the coordinates of A (-1, 5) and D (-16, 0):

Length (AD) = (\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(-16 - (-1))^2 + (0 - 5)^2} = \sqrt{(-15)^2 + (-5)^2} = \sqrt{225 + 25} = \sqrt{250} = 5\sqrt{10} ).

The area of a parallelogram can be calculated using the formula:

$\text{Area} = base \times height$

Using AD as the base and the length of the perpendicular height from B to AD:

Height can be derived from the slope of AD. For the slope of (\frac{1}{3}), the height (h) using AB as height would be equal to 5 units (height):

$\text{Area} = 5\sqrt{10} \cdot 5 = 25\sqrt{10}.$

Using the slopes of AB and BC to find the angle:

Let ( m_1 = \frac{1}{3} ) (slope of AB) and ( m_2 = -\frac{1}{2} ) (slope of BC), then:

$\tan(\theta) = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| = \left| \frac{-\frac{1}{2} - \frac{1}{3}}{1 + (\frac{1}{3})(-\frac{1}{2})} \right|$

Computing gives:

$= \left| \frac{-\frac{3}{6} - \frac{2}{6}}{1 - \frac{1}{6}} \right| = \left| \frac{-\frac{5}{6}}{\frac{5}{6}} \right| = 1$

Thus, ( \angle ABC = 45^\circ. )