The coordinates of C are (4,-5, 0) - Leaving Cert Mathematics - Question 8 - 2017

To show that the line segments are parallel, we calculate the slopes of both segments.

1. Slope of segment [AB]:
   The coordinates of A are (1, -2) and B are (4, 2). Slope formula:

   $m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - (-2)}{4 - 1} = \frac{4}{3}$

2. Slope of segment [DE]:
   The coordinates of D are (6, -6) and E are (15, 6). Slope formula:

   $m_{DE} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - (-6)}{15 - 6} = \frac{12}{9} = \frac{4}{3}$

Since both slopes are equal (i.e., $m_{AB} = m_{DE} = \frac{4}{3}$), it can be concluded that the segments are parallel.

To find the area of triangle CBA, we use the formula for the area of a triangle given vertices at coordinates (x1, y1), (x2, y2), (x3, y3):

$ext{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Using the points:

• C = (4, -5)
• B = (4, 2)
• A = (1, -2)

Substituting the values:

$ext{Area} = \frac{1}{2} | 4(2 - (-2)) + 4((-2) - (-5)) + 1((-5) - 2)|$

Calculating:

The distance |AB| can be calculated using the distance formula:

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

Using the coordinates A = (1, -2) and B = (4, 2): $|AB| = \sqrt{(4 - 1)^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ units}$

To find the scale factor, we compare the length of side DE to side AB.
We already found:

• |AB| = 5 units
• |DE| = 15 units

The scale factor (k) can be established as: $k = \frac{|DE|}{|AB|} = \frac{15}{5} = 3$

The area of triangle CDE can be calculated using the scale factor squared. Given the area of triangle CBA is 4 square units and the found scale factor k = 3: $ext{Area}_{CDE} = k^2 \times \text{Area}_{CBA} = 3^2 \times 4 = 9 \times 4 = 36 \text{ square units}$