In the diagram, A, B, C and D are the vertices of a rhombus - NSC Mathematics - Question 3 - 2016 - Paper 2

To find the coordinates of point K (the intersection point of AC and BD), we solve the equations of both lines simultaneously.

Starting with:

1. AC: x + 3y = 10
2. BD: 3x - y = 0

From equation 2, we can express y in terms of x:

y = 3x

Substituting this into equation 1 gives:

$x + 3(3x) = 10$ $x + 9x = 10$ $10x = 10 \Rightarrow x = 1$

Now substituting x back into the equation for y:

y = 3(1) = 3

Thus, the coordinates of K are (1, 3).

Using K (1, 3) as the midpoint of diagonal BD and knowing that AB is a rhombus, we can compute the coordinates of B. Since K is the midpoint, and BD is horizontal, we know:

Let B be (x_B, y_B) and D is given as (3, 9).

The midpoint formula states:

o = (\frac{x_B + 3}{2}, \frac{y_B + 9}{2}) = (1, 3)

By equating the x-coordinates:

$\frac{x_B + 3}{2} = 1 \Rightarrow x_B + 3 = 2 \Rightarrow x_B = -1$

And the y-coordinates:

$\frac{y_B + 9}{2} = 3 \Rightarrow y_B + 9 = 6 \Rightarrow y_B = -3$

Thus, the coordinates of B are (-1, -3).

Given that AD = √50, we can derive the coordinates for points A and C. The distance formula will be utilized:

$D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Assuming we denote A as (x_A, y_A) and D as (3, 9), we can set up the equation:

$\sqrt{(x_A - 3)^2 + (y_A - 9)^2} = \sqrt{50}$

Squaring both sides yields:

$(x_A - 3)^2 + (y_A - 9)^2 = 50$

We already have the coordinates for B (-1, -3) and K (1, 3). Now, utilizing symmetry (in rhombus opposite sides are equal), we can conclude the coordinates of A and C can be derived accordingly, centered around; substituting known points will help solve:

Assuming A is symmetric to B, we find: A(x_A, y_A): (-1 + 21, -3 + 23) = (1, 3) C(x_C, y_C): (3 - 2, 9 - 2*3) = (1, 3) (0, 6) is also computed.

Finally, coordinates of A and C are (4, 6) and (1, 2) respectively.

The gradient (m) of line PQ can be calculated using the coordinates of points P(-3, 2) and Q(5, 8). The formula for the gradient is:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Plugging in the coordinates: $m = \frac{8 - 2}{5 - (-3)} = \frac{6}{8} = \frac{3}{4}$

Thus, the gradient of PQ is (\frac{3}{4}).

To calculate the angle θ that line PQ forms with the positive x-axis, we use the tangent function, where:

$\tan(θ) = \frac{\text{gradient}}{1} = \frac{3}{4}$

To find θ, we use the inverse tangent function:

$θ = \tan^{-1}\left(\frac{3}{4}\right)$

Calculating this gives us approximately θ = 36.87 degrees, rounding it to one decimal place yields: θ ≈ 36.9 degrees.

Since the line we need is parallel to PQ, its gradient will also be (\frac{3}{4}). We can use the point-slope form of a line equation:

$y - y_1 = m (x - x_1)$

Using the intercept point (8, 0) as (x_1, y_1):

y - 0 = \frac{3}{4}(x - 8)

Simplifying this gives us:

y = \frac{3}{4}x - 6\ This is the required equation of the line.