The points A(2, 1), B(6, 3), C(5, 5), and D(1, 3) are the vertices of the rectangle ABCD as shown - Leaving Cert Mathematics - Question 3 - 2017

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

$ext{Area} = ext{length} imes ext{width}$

From our previous calculation, we found |AD| = √5. Now we need to find |AB|. Using points A(2, 1) and B(6, 3):

Using the distance formula again:

$|AB| = ext{sqrt}((6 - 2)^2 + (3 - 1)^2)$ $= ext{sqrt}((4)^2 + (2)^2)$ $= ext{sqrt}(16 + 4)$ $= ext{sqrt}(20)$ $= 2 ext{sqrt}(5)$

Now, substituting into the area formula:

$ext{Area} = |AD| imes |AB| = ext{sqrt}(5) imes 2 ext{sqrt}(5)$ $= 2 imes 5 = 10$

Thus, the area of rectangle ABCD is 10 square units.

To find the equation of the line BC, we can use the slope-intercept method. First, calculate the slope (m) using points B(6, 3) and C(5, 5):

m_{BC} = rac{y_2 - y_1}{x_2 - x_1} = rac{5 - 3}{5 - 6} = rac{2}{-1} = -2

Now, use point-slope form, y - y_1 = m(x - x_1):

Using point B (6, 3):

$y - 3 = -2(x - 6)$ $y - 3 = -2x + 12$ $y = -2x + 15$

Rearranging gives the equation in the required form:

$2x + y - 15 = 0$

To find angle ABD, we can use the sine, cosine, or tangent ratios based on the triangle ABD:

1. Calculate lengths:

   • |AB| = 2√5
   • |AD| = √5
   • |BD| can be found using the distance formula:

   Using points B(6, 3) and D(1, 3): $|BD| = 6 - 1 = 5$

2. Apply the cosine rule: $ext{cos}(ABD) = \frac{|AD|^2 + |AB|^2 - |BD|^2}{2 |AD| |AB|}$

   Substituting values: $= \frac{( ext{√5})^2 + (2 ext{√5})^2 - 5^2}{2 imes ext{√5} imes 2 ext{√5}}$ $= \frac{5 + 20 - 25}{8} = \frac{0}{8} = 0$

Thus, angle ABD is: $ABD = ext{cos}^{-1}(0) = 90^{ ext{o}}$