The equations of the perpendicular bisectors | vedclass

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(D) The perpendicular bisectors of the sides of a triangle intersect at the circumcenter $O$. Since the perpendicular bisectors of $AB$ and $AC$ are $

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$14x + 23y - 40 = 0$

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(D) The perpendicular bisectors of the sides of a triangle intersect at the circumcenter $O$. Since the perpendicular bisectors of $AB$ and $AC$ are $x - y + 5 = 0$ and $x + 2y = 0$,their intersection point $O$ is found by solving these equations: $x - y = -5$ and $x + 2y = 0$. Subtracting the equations,we get $-3y = -5$,so $y = \frac{5}{3}$. Then $x = -2y = -\frac{10}{3}$. Thus,the circumcenter $O$ is $(-\frac{10}{3}, \frac{5}{3})$. Since $O$ is the circumcenter,$OA = OB = OC$. Let $B = (x_1, y_1)$. The reflection of $A(1, -2)$ about the line $x - y + 5 = 0$ gives $B$. The formula for the reflection of $(x_0, y_0)$ about $ax + by + c = 0$ is $\frac{x - x_0}{a} = \frac{y - y_0}{b} = -2 \frac{ax_0 + by_0 + c}{a^2 + b^2}$. For $A(1, -2)$ and $x - y + 5 = 0$: $\frac{x_1 - 1}{1} = \frac{y_1 + 2}{-1} = -2 \frac{1 - (-2) + 5}{1^2 + (-1)^2} = -2 \frac{8}{2} = -8$. So,$x_1 - 1 = -8 \implies x_1 = -7$ and $y_1 + 2 = 8 \implies y_1 = 6$. Thus,$B = (-7, 6)$. Similarly,for $C(x_2, y_2)$,the reflection of $A(1, -2)$ about $x + 2y = 0$: $\frac{x_2 - 1}{1} = \frac{y_2 + 2}{2} = -2 \frac{1 + 2(-2)}{1^2 + 2^2} = -2 \frac{-3}{5} = \frac{6}{5}$. So,$x_2 - 1 = \frac{6}{5} \implies x_2 = \frac{11}{5}$ and $y_2 + 2 = \frac{12}{5} \implies y_2 = \frac{2}{5}$. Thus,$C = (\frac{11}{5}, \frac{2}{5})$. The equation of line $BC$ passing through $(-7, 6)$ and $(\frac{11}{5}, \frac{2}{5})$ is: $y - 6 = \frac{\frac{2}{5} - 6}{\frac{11}{5} - (-7)} (x - (-7)) = \frac{-\frac{28}{5}}{\frac{46}{5}} (x + 7) = -\frac{14}{23} (x + 7)$. $23(y - 6) = -14(x + 7) \implies 23y - 138 = -14x - 98 \implies 14x + 23y - 40 = 0$.

The equations of the perpendicular bisectors of the sides $AB$ and $AC$ of a triangle $ABC$ are $x - y + 5 = 0$ and $x + 2y = 0$ respectively. If the point $A$ is $(1, -2)$,then the equation of line $BC$ is

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