Let $$C(\alpha, \beta)$$ be the circumcentre of the triangle formed by the lines $$4x + 3y = 69$$, $$4y - 3x = 17$$, and $$x + 7y = 61$$. Then $$(\alpha - \beta)^2 + \alpha + \beta$$ is equal to

To solve this efficiently, notice the relationship between the slopes of the lines.

• Line 1 ($$L_1$$): $$4x + 3y = 69 \implies m_1 = -4/3$$
• Line 2 ($$L_2$$): $$-3x + 4y = 17 \implies m_2 = 3/4$$
• Since $$m_1 \cdot m_2 = (-4/3) \cdot (3/4) = -1$$, $$L_1$$ and $$L_2$$ are perpendicular. This is a right-angled triangle.

In a right-angled triangle, the circumcentre is the midpoint of the hypotenuse. The hypotenuse is the third line ($$L_3$$): $$x + 7y = 61$$.

Find the vertices by intersecting $$L_3$$ with $$L_1$$ and $$L_2$$:

• Vertex $$A$$ ($$L_1 \cap L_3$$): $$4x + 3y = 69$$ and $$x + 7y = 61$$. Solving gives $$(9, 11)$$.
• Vertex $$B$$ ($$L_2 \cap L_3$$): $$-3x + 4y = 17$$ and $$x + 7y = 61$$. Solving gives $$(5, 8)$$.

Circumcentre $$C(\alpha, \beta)$$ = Midpoint of $$AB$$:

$$\alpha = \frac{9+5}{2} = 7, \quad \beta = \frac{11+8}{2} = 9.5$$

We need $$(\alpha - \beta)^2 + \alpha + \beta$$:

$$(7 - 9.5)^2 + 7 + 9.5$$

$$(-2.5)^2 + 16.5 = 6.25 + 16.5 = \mathbf{22.75}$$