The equations of the perpendicular bisectors | vedclass

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(A) Let $A = (1, -2)$. The perpendicular bisector of $AB$ is $L_1: x-y+5=0$. The slope of $L_1$ is $1$,so the slope of $AB$ is $-1$. The equation of $

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$14x+23y-40=0$

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(A) Let $A = (1, -2)$. The perpendicular bisector of $AB$ is $L_1: x-y+5=0$. The slope of $L_1$ is $1$,so the slope of $AB$ is $-1$. The equation of $AB$ is $y - (-2) = -1(x - 1) \Rightarrow x+y+1=0$. Intersection of $AB$ and $L_1$: $x - (-x-1) + 5 = 0$ $\Rightarrow 2x = -6$ $\Rightarrow x = -3, y = 2$. Since $E$ is the midpoint of $AB$,$\frac{x_B+1}{2} = -3 \Rightarrow x_B = -7$ and $\frac{y_B-2}{2} = 2 \Rightarrow y_B = 6$. Thus $B = (-7, 6)$. The perpendicular bisector of $AC$ is $L_2: x+2y=0$. The slope of $L_2$ is $-1/2$,so the slope of $AC$ is $2$. The equation of $AC$ is $y - (-2) = 2(x - 1) \Rightarrow 2x-y-4=0$. Intersection of $AC$ and $L_2$: $x + 2(2x-4) = 0$ $\Rightarrow 5x = 8$ $\Rightarrow x = 8/5, y = -4/5$. Since $F$ is the midpoint of $AC$,$\frac{x_C+1}{2} = 8/5 \Rightarrow x_C = 11/5$ and $\frac{y_C-2}{2} = -4/5 \Rightarrow y_C = 2/5$. Thus $C = (11/5, 2/5)$. The equation of line $BC$ passing through $(-7, 6)$ and $(11/5, 2/5)$ is: $y - 6 = \frac{2/5 - 6}{11/5 - (-7)} (x - (-7)) $ $\Rightarrow y - 6 = \frac{-28/5}{46/5} (x + 7) $ $\Rightarrow y - 6 = -\frac{14}{23} (x + 7) $ $\Rightarrow 23y - 138 = -14x - 98 $ $\Rightarrow 14x + 23y - 40 = 0$.

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

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