Question:

Step 1: Calculate the side length and midpoint of AB. Let $A \equiv (3, 2)$ and $B \equiv (5, 1)$. The side length $s$ of AB is: $$s = \sqrt{(5-3)^2 + (1-2)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}$$ The midpoint $M$ of AB is: $$M = \left(\frac{3+5}{2}, \frac{2+1}{2}\right) = \left(\frac{8}{2}, \frac{3}{2}\right) = \left(4, \frac{3}{2}\right)$$ Step 2: Determine the direction for constructing vertex P. The line passing through A and B has a direction vector $\vec{AB} = (5-3, 1-2) = (2, -1)$. A vector perpendicular to $\vec{AB}$ is $\vec{n} = (1, 2)$. The equation of the line AB is $x + 2y - 7 = 0$. To determine the side remote from the origin, we evaluate the line equation at the origin $(0,0)$: $0 + 2(0) - 7 = -7$. For a point P to be on the side remote from the origin, it must satisfy $x_P + 2y_P - 7 > 0$. Moving from the midpoint M in the direction of $\vec{n} = (1, 2)$ will lead to points where $x+2y-7 > 0$. The unit vector in this direction is $\hat{n} = \frac{(1, 2)}{\sqrt{1^2+2^2}} = \frac{(1, 2)}{\sqrt{5}}$. Step 3: Find the coordinates of the third vertex P. For an equilateral triangle with side length $s = \sqrt{5}$, the height $h$ is: $$h = \frac{\sqrt{3}}{2} s = \frac{\sqrt{3}}{2} \sqrt{5} = \frac{\sqrt{15}}{2}$$ The coordinates of P are found by adding the height vector to the midpoint M: $$P = M + h \cdot \hat{n}$$ $$P = \left(4, \frac{3}{2}\right) + \frac{\sqrt{15}}{2} \cdot \frac{(1, 2)}{\sqrt{5}}$$ $$P = \left(4, \frac{3}{2}\right) + \frac{\sqrt{3}\sqrt{5}}{2} \cdot \frac{(1, 2)}{\sqrt{5}}$$ $$P = \left(4, \frac{3}{2}\right) + \left(\frac{\sqrt{3}}{2} \cdot 1, \frac{\sqrt{3}}{2} \cdot 2\right)$$ $$P = \left(4 + \frac{\sqrt{3}}{2}, \frac{3}{2} + \sqrt{3}\right)$$ Step 4: Find the orthocentre of triangle ABP. For an equilateral triangle, the orthocentre coincides with the centroid. The centroid $H$ is given by the average of the coordinates of the vertices A, B, and P: $$H = \left(\frac{x_A + x_B + x_P}{3}, \frac{y_A + y_B + y_P}{3}\right)$$ The x-coordinate of H is: $$H_x = \frac{3 + 5 + \left(4 + \frac{\sqrt{3}}{2}\right)}{3} = \frac{12 + \frac{\sqrt{3}}{2}}{3} = \frac{12}{3} + \frac{\sqrt{3}}{6} = 4 + \frac{\sqrt{3}}{6}$$ The y-coordinate of H is: $$H_y = \frac{2 + 1 + \left(\frac{3}{2} + \sqrt{3}\right)}{3} = \frac{3 + \frac{3}{2} + \sqrt{3}}{3} = \frac{\frac{6+3}{2} + \sqrt{3}}{3} = \frac{\frac{9}{2} + \sqrt{3}}{3} = \frac{9}{6} + \frac{\sqrt{3}}{3} = \frac{3}{2} + \frac{\sqrt{3}}{3}$$ Thus, the orthocentre of triangle ABP is: $$H = \left(4 + \frac{\sqrt{3}}{6}, \frac{3}{2} + \frac{\sqrt{3}}{3}\right)$$