Let C(alpha, beta) be the circumcenter of the | vedclass

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(B) First,find the vertices of the triangle by solving the equations in pairs:$1$. $4x + 3y = 69$ and $4y - 3x = 17$: Multiplying the first by $4$ and

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$17$

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(B) First,find the vertices of the triangle by solving the equations in pairs:$1$. $4x + 3y = 69$ and $4y - 3x = 17$: Multiplying the first by $4$ and second by $3$,we get $16x + 12y = 276$ and $-9x + 12y = 51$. Subtracting gives $25x = 225$,so $x = 9$. Then $36 + 3y = 69 \implies 3y = 33 \implies y = 11$. Vertex $A = (9, 11)$.$2$. $4x + 3y = 69$ and $x + 7y = 61$: Multiplying the second by $4$,we get $4x + 28y = 244$. Subtracting the first gives $25y = 175$,so $y = 7$. Then $x + 49 = 61 \implies x = 12$. Vertex $B = (12, 7)$.$3$. $4y - 3x = 17$ and $x + 7y = 61$: Multiplying the second by $3$,we get $3x + 21y = 183$. Adding to the first gives $25y = 200$,so $y = 8$. Then $x + 56 = 61 \implies x = 5$. Vertex $C = (5, 8)$.Note that the lines $4x + 3y = 69$ and $4y - 3x = 17$ are perpendicular since $(4)(-3) + (3)(4) = 0$. Thus,the triangle is a right-angled triangle with the right angle at the intersection of these two lines,which is $(9, 11)$.The circumcenter of a right-angled triangle is the midpoint of the hypotenuse. The hypotenuse is the line $x + 7y = 61$ connecting $(12, 7)$ and $(5, 8)$.Circumcenter $(\alpha, \beta) = \left(\frac{12+5}{2}, \frac{7+8}{2}\right) = \left(\frac{17}{2}, \frac{15}{2}\right)$.Now,calculate $(\alpha - \beta)^2 + \alpha + \beta = \left(\frac{17}{2} - \frac{15}{2}\right)^2 + \frac{17}{2} + \frac{15}{2} = (1)^2 + \frac{32}{2} = 1 + 16 = 17$.

Let $C(\alpha, \beta)$ be the circumcenter 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 $.........$.

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