Let (alpha, beta) be the centroid of the tria | vedclass

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(B) Step $1$: Find the vertices of the triangle by solving the equations in pairs.Solving $15x - y = 82$ and $6x - 5y = -4$: Multiplying the first by

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$x^2 - 13x + 42 = 0$

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(B) Step $1$: Find the vertices of the triangle by solving the equations in pairs.Solving $15x - y = 82$ and $6x - 5y = -4$: Multiplying the first by $5$,we get $75x - 5y = 410$. Subtracting the second,$69x = 414$,so $x = 6$. Then $y = 15(6) - 82 = 8$. Vertex $A = (6, 8)$.Solving $6x - 5y = -4$ and $9x + 4y = 17$: Multiplying the first by $4$ and the second by $5$,we get $24x - 20y = -16$ and $45x + 20y = 85$. Adding them,$69x = 69$,so $x = 1$. Then $y = (6(1) + 4)/5 = 2$. Vertex $B = (1, 2)$.Solving $15x - y = 82$ and $9x + 4y = 17$: Multiplying the first by $4$,we get $60x - 4y = 328$. Adding the second,$69x = 345$,so $x = 5$. Then $y = 15(5) - 82 = -7$. Vertex $C = (5, -7)$.Step $2$: Calculate the centroid $(\alpha, \beta)$.$\alpha = \frac{6 + 1 + 5}{3} = 4$$\beta = \frac{8 + 2 - 7}{3} = 1$Step $3$: Find the roots of the quadratic equation.Roots are $\alpha + 2\beta = 4 + 2(1) = 6$ and $2\alpha - \beta = 2(4) - 1 = 7$.Step $4$: Form the equation.The equation is $(x - 6)(x - 7) = 0$,which is $x^2 - 13x + 42 = 0$.

Let $(\alpha, \beta)$ be the centroid of the triangle formed by the lines $15x - y = 82$,$6x - 5y = -4$,and $9x + 4y = 17$. Then $\alpha + 2\beta$ and $2\alpha - \beta$ are the roots of the equation $...........$.

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