Parts of the lines AC and BC are shown in the co-ordinate diagram below (not to scale) - Leaving Cert Mathematics - Question 1 - 2022

To determine if AC is perpendicular to BC, we first need to find the slope of BC. Using points B (2, 0) and C (0, 3):

• Let (x_1, y_1) = (2, 0) and (x_2, y_2) = (0, 3).

Using the slope formula: $m_{BC} = \frac{3 - 0}{0 - 2} = -\frac{3}{2}$

Next, we check the relationship: $m_{AC} \times m_{BC} = \frac{3}{2} \times \left(-\frac{3}{2}\right) = -\frac{9}{4} \neq -1$

Since the product of the slopes is not -1, we conclude that AC is not perpendicular to BC.

The length |LM| can be calculated using the distance formula:

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

Given points L (x_L, y_L) and M (9, 1):

• The x-coordinate of L is the negative of that of M due to symmetry about the y-axis, thus x_L = -9.

Substituting the coordinates: $|LM| = \sqrt{(9 - (-9))^2 + (1 - 1)^2} = \sqrt{(9 + 9)^2} = \sqrt{18^2} = 18$

The horizontal line through point M (9, 1) has the equation: $y = 1$

To find the x-coordinate of point N, substitute y = 1 into the line equation: $x + 4(1) - 13 = 0$ This simplifies to: $x + 4 - 13 = 0$ $x - 9 = 0$ Thus, x = 9. Hence, point N is symmetric to M, giving N (-9, 1). The final answer is: $N = (-9, 1)$