6. (a) Particles of weight 5 N, 9 N, 6 N and 1 N are placed at the points $(p, 5)$, $(7, q)$, $(-6, -q)$ and $(8, 5)$ respectively - Leaving Cert Applied Maths - Question 6 - 2015

To find the value of $q$, we consider the y-coordinate:

$q = \frac{5(5) + 9(q) + 6(-q) + 1(5)}{21}$

1. Calculate weighted sums:

   • $5(5) + 9(q) + 6(-q) + 1(5)$
   • $= 25 + 9q - 6q + 5$
   • $= 30 + 3q$.

2. Again,

   • $q = \frac{30 + 3q}{21}$

3. Multiplying both sides by 21:

   • $21q = 30 + 3q$

4. Isolate $q$:

   • $21q - 3q = 30$
   • $18q = 30$
   • $q = \frac{30}{18}$.
   • Thus, $q = 1.6667$. But according to the marking scheme, $q = 3$.

For the quadrilateral lamina, we calculate the center of gravity using the area method:

1. Area of triangle ABC:

   $Area_{ABC} = \frac{1}{2} \times 9 \times 6 = 54$

   Co-ordinate of center of gravity for triangle $ABC$ is $(4, 8)$.

2. Area of triangle ACD:

   $Area_{ACD} = \frac{1}{2} \times 18 \times 15 = 135$

   Co-ordinate of center of gravity for triangle $ACD$ is $(10, 5)$.

3. Total Area of the Lamina:

   Total area = Area of $ABC$ + Area of $ACD = 54 + 135 = 189$.

4. Now, using the areas to find coordinates:

   • For x-coordinate: $x_c = \frac{(189)(4) + (135)(10)}{189} = 8.3$

   • For y-coordinate: $y_c = \frac{(189)(8) + (135)(5)}{189} = 5.9$.

   Thus, the co-ordinates of the centre of gravity of the lamina are approximately $(8.3, 5.9)$.