Q1. Using Section Formula: $x = \frac{m_1 x_2 + m_2 x_1}{m_1+m_2} = \frac{3(8) + 1(4)}{4} = 7, $y = \frac{3(5) + 1(-3)}{4} = 3. Point is $(7, 3)$.
Q2. $PA = PB \rightarrow (x-7)^2 + (y-1)^2 = (x-3)^2 + (y-5)^2. Expand and simplify to get relation: $x - y = 2$.
Q3. Area formula: $\frac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| = \frac{1}{2}|2(0+4) - 1(-4-3) + 2(3-0)| = \frac{1}{2}|8 + 7 + 6| = 10.5\text{ sq units}$.
Q4. Diagonals of parallelogram bisect each other, so midpoint of AC = midpoint of BD. $\frac{6+9}{2} = \frac{8+p}{2} \rightarrow p = 7$.
Q5. Trisection points divide in ratios $1:2$ and $2:1. Coordinates are $(2, -5/3)$ and $(0, -7/3)$.
