Joe wants to draw a diagram of his farm - Leaving Cert Mathematics - Question 9 - 2016

To plot the barn's position, we move vertically up 5 units from point F.

Starting from point F (4, 1), we add 5 to the y-coordinate:

B = (4, 1 + 5) = (4, 6).

Label point B at (4, 6) on the diagram.

To find the distance from point B (4, 6) to point Q, we use the distance formula:

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

Where (x_1, y_1) are the co-ordinates of B and (x_2, y_2) are the co-ordinates of Q. Assuming Q is at coordinates (2, 3), then:

$d = \sqrt{(2 - 4)^2 + (3 - 6)^2}$ $d = \sqrt{(-2)^2 + (-3)^2}$ $d = \sqrt{4 + 9}$ $d = \sqrt{13}$ $d \approx 3.61$

Thus, the distance is approximately 3.61 units.

Given that T is at the point (-2, 2), we locate this point on the coordinate system and label it. Therefore,

T = (-2, 2).

To find the area A of the parallelogram FBQT, we can use the formula:

$A = b \cdot h$

Where b is the base length (which can be determined from B to F) and h is the height (distance from Q to the line FB). In this case, calculating the lengths gives us:

Base = distance between F (4, 1) and B (4, 6) = 5. Height = distance from Q to line FB. Using coordinates and area properties:

From calculations, $A = 5 \cdot 6 = 30$

Thus, the area is 30 square units.

Using the sine formula in triangle relations, given that ∠QFB = 45°:

$\sin(∠LQB) = \frac{|BQ|}{|FB|}$.

Knowing lengths, with BQ calculated earlier as 3.61, $\sin(∠LQB) = \frac{3.61}{6.08}$.

Substituting values gives us: $∠LQB ≈ 35.6°$

Thus, the angle ∠LQB ≈ 35.6°.