The co-ordinates of three points A, B, and C are: A(2, 2), B(6, –6), C(–2, –3) - Leaving Cert Mathematics - Question 3 - 2012

To find where line AB intersects the y-axis, we set x = 0 in the equation of AB:

$2(0) + y - 6 = 0 \Rightarrow y - 6 = 0 \Rightarrow y = 6$

Thus, the coordinates of D are (0, 6).

To find the perpendicular distance from point C(–2, –3) to line AB, we can use the perpendicular distance formula:

$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$

From the equation of line AB, we have a = 2, b = 1, and c = -6. Using the coordinates of C:

$d = \frac{|2(-2) + 1(-3) - 6|}{\sqrt{2^2 + 1^2}} = \frac{|-4 - 3 - 6|}{\sqrt{4 + 1}} = \frac{|-13|}{\sqrt{5}} = \frac{13}{\sqrt{5}}$

To find the area of triangle ADC, we can use the formula:

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$

From part (b), we have:

• Base AD = 6 units (distance from A(2, 2) to D(0, 6))
• Height = \frac{13}{\sqrt{5}} (the perpendicular distance from C to line AB)

Thus, the area becomes:

$\text{Area} = \frac{1}{2} (6) \left( \frac{13}{\sqrt{5}} \right) = \frac{39}{\sqrt{5}}$

Simplifying further, the area of triangle ADC is approximately 13 square units.