In a triangle ABC, side AB has the equation 2 | vedclass

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(C) Let the coordinates of $B$ be $(x_1, y_1)$ and $C$ be $(x_2, y_2).$Since $(5, 6)$ is the mid-point of $BC,$$\frac{x_1 + x_2}{2} = 5 \implies x_1 +

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$x + y = 11$

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(C) Let the coordinates of $B$ be $(x_1, y_1)$ and $C$ be $(x_2, y_2).$Since $(5, 6)$ is the mid-point of $BC,$$\frac{x_1 + x_2}{2} = 5 \implies x_1 + x_2 = 10 \implies x_1 = 10 - x_2$$\frac{y_1 + y_2}{2} = 6 \implies y_1 + y_2 = 12 \implies y_1 = 12 - y_2$Point $B(x_1, y_1)$ lies on $2x + 3y = 29 \implies 2x_1 + 3y_1 = 29.$Point $C(x_2, y_2)$ lies on $x + 2y = 16 \implies x_2 + 2y_2 = 16.$Substitute $x_1$ and $y_1$ into the equation of $AB:$$2(10 - x_2) + 3(12 - y_2) = 29$$20 - 2x_2 + 36 - 3y_2 = 29$$2x_2 + 3y_2 = 27.$Now solve the system:$x_2 + 2y_2 = 16 \implies x_2 = 16 - 2y_2.$Substitute into $2x_2 + 3y_2 = 27:$$2(16 - 2y_2) + 3y_2 = 27$$32 - 4y_2 + 3y_2 = 27 \implies y_2 = 5.$Then $x_2 = 16 - 2(5) = 6.$So $C = (6, 5).$Then $x_1 = 10 - 6 = 4$ and $y_1 = 12 - 5 = 7.$So $B = (4, 7).$The equation of line $BC$ passing through $(4, 7)$ and $(6, 5)$ is:$y - 7 = \frac{5 - 7}{6 - 4}(x - 4)$$y - 7 = \frac{-2}{2}(x - 4)$$y - 7 = -x + 4$$x + y = 11.$

In a triangle $ABC,$ side $AB$ has the equation $2x + 3y = 29$ and the side $AC$ has the equation $x + 2y = 16.$ If the mid-point of $BC$ is $(5, 6),$ then the equation of $BC$ is:

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