In an equilateral triangle $$PQR$$,let the vertex $$P = (3, 5)$$ and the side $$QR$$ be along the line $$x + y = 4$$. If the orthocentre of the triangle $$PQR$$ is $$(\alpha, \beta)$$, then $$9(\alpha + \beta)$$ is equal to :

To solve this problem, we need to use a special property of equilateral triangles: the orthocenter, circumcenter, and centroid all coincide at the same point.

In an equilateral triangle, the orthocenter $$H(\alpha, \beta)$$ is the same as the centroid. The centroid divides the median (the altitude from $$P$$ to side $$QR$$) in a ratio of $$2:1$$.

Let $$M(x_1, y_1)$$ be the foot of the perpendicular from vertex $$P(3, 5)$$ to the line $$x + y - 4 = 0$$. This point $$M$$ is the midpoint of side $$QR$$.

Using the formula for the foot of the perpendicular:

$$\frac{x_1 - x}{a} = \frac{y_1 - y}{b} = -\frac{(ax + by + c)}{a^2 + b^2}$$

$$\frac{x_1 - 3}{1} = \frac{y_1 - 5}{1} = -\frac{(3 + 5 - 4)}{1^2 + 1^2} = -\frac{4}{2} = -2$$

Solving for $$x_1$$ and $$y_1$$:

• $$x_1 - 3 = -2 \implies x_1 = 1$$
• $$y_1 - 5 = -2 \implies y_1 = 3$$

So, $$M = (1, 3)$$.

The orthocenter $$H(\alpha, \beta)$$ lies on the segment $$PM$$ and divides it in the ratio $$2:1$$ (measured from the vertex $$P$$).

Using the section formula for $$P(3, 5)$$ and $$M(1, 3)$$ with ratio $$m:n = 2:1$$:

$$\alpha = \frac{m x_1 + n x}{m + n} = \frac{2(1) + 1(3)}{2 + 1} = \frac{5}{3}$$

$$\beta = \frac{m y_1 + n y}{m + n} = \frac{2(3) + 1(5)}{2 + 1} = \frac{11}{3}$$

First, find the sum of $$\alpha$$ and $$\beta$$:

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

Now, multiply by 9:

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

Final Answer: 48 (Option D)