Orthocentre of Equilateral Triangle with Given Vertex and Base Line

In an equilateral triangle, all four main triangle centers (centroid, orthocentre, circumcentre, and incenter) coincide. Therefore, the orthocentre $H(\alpha, \beta)$ is the same as the centroid $G(\alpha, \beta)$ of $\triangle PQR$.

Step 1: Find the foot of the perpendicular from $P(3, 5)$ to the line $QR$.

The base line $QR$ has the equation: $x + y - 4 = 0$

Let $M(x_0, y_0)$ be the foot of the perpendicular (and the midpoint of $QR$) dropped from vertex $P(3, 5)$ to the line $x + y - 4 = 0$.

Using the formula for the foot of the perpendicular from a point $(x_1, y_1)$ to a line $ax + by + c = 0$: $\frac{x_0 - x_1}{a} = \frac{y_0 - y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2}$

Substituting $x_1 = 3, y_1 = 5, a = 1, b = 1, c = -4$: $\frac{x_0 - 3}{1} = \frac{y_0 - 5}{1} = -\frac{1(3) + 1(5) - 4}{1^2 + 1^2}$ $\frac{x_0 - 3}{1} = \frac{y_0 - 5}{1} = -\frac{4}{2} = -2$

Solving for $x_0$ and $y_0$: $x_0 = 3 - 2 = 1$ $y_0 = 5 - 2 = 3$

Thus, the foot of the altitude $M$ is $(1, 3)$.

Step 2: Find the coordinates of the orthocentre $(\alpha, \beta)$.

The altitude $PM$ is also the median of the equilateral triangle. The centroid $H(\alpha, \beta)$ divides the median $PM$ internally in the ratio $2 : 1$ from the vertex $P(3, 5)$ to $M(1, 3)$.

Using the section formula: $\alpha = \frac{2 \cdot x_M + 1 \cdot x_P}{2 + 1} = \frac{2(1) + 1(3)}{3} = \frac{5}{3}$ $\beta = \frac{2 \cdot y_M + 1 \cdot y_P}{2 + 1} = \frac{2(3) + 1(5)}{3} = \frac{11}{3}$

Step 3: Calculate $9(\alpha + \beta)$.

$\alpha + \beta = \frac{5}{3} + \frac{11}{3} = \frac{16}{3}$

Therefore, $9(\alpha + \beta) = 9 \times \frac{16}{3} = 48$