Particles of weight 9 N, 8 N, q N and 2 N are placed at the points (−4, 3), (8, 6), (p, 5) and (q, −p) respectively - Leaving Cert Applied Maths - Question 6 - 2016

Substituting $p = 2$ back into the second equation:

$2(2) - q = -1 \implies 4 - q = -1 \implies q = 5$

Thus, the values are:

• $p = 2$
• $q = 5$

To determine the centre of gravity of the remaining lamina, we first compute the area of triangle ABC:

$\text{Area}_{ABC} = \frac{1}{2} \times 45 \times 108 = 2430$

Next, we find the area of the incircle:

$\text{Area}_{Circle} = \pi (18)^2 \approx 1017.876$

Calculating the area of the remaining lamina:

$\text{Area}_{Lamina} = \text{Area}_{ABC} - \text{Area}_{Circle} = 2430 - 1017.876 \approx 1412.124$

Now we compute the x-coordinate of the centre of gravity:

$1412.124 \times x = 2430(5) - 1017.876(18)\implies x \approx 12.8$

Next, we calculate the y-coordinate:

$1412.124 \times y = 2430(36) - 1017.876(18)\implies y \approx 49.0$

Thus, the coordinates of the centre of gravity for the remaining lamina are approximately $(12.8, 49.0)$.