The equations of the perpendicular bisectors | vedclass

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(A) Let $B(x_1, y_1)$ and $C(x_2, y_2)$ be the vertices of the triangle.Since the perpendicular bisector of $AB$ is $x-y+5=0$,the reflection of $A(1,

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$14x+23y-40=0$

Vedclass · Answer

(A) Let $B(x_1, y_1)$ and $C(x_2, y_2)$ be the vertices of the triangle.Since the perpendicular bisector of $AB$ is $x-y+5=0$,the reflection of $A(1, -2)$ across this line gives the coordinates of $B$.Using the formula for reflection of a point $(x_0, y_0)$ across $ax+by+c=0$ as $\frac{x-x_0}{a} = \frac{y-y_0}{b} = -2\frac{ax_0+by_0+c}{a^2+b^2}$:$\frac{x_1-1}{1} = \frac{y_1+2}{-1} = -2\frac{1-(-2)+5}{1^2+(-1)^2} = -2\frac{8}{2} = -8$.$x_1-1 = -8 \Rightarrow x_1 = -7$.$y_1+2 = 8 \Rightarrow y_1 = 6$.So,$B = (-7, 6)$.Similarly,the reflection of $A(1, -2)$ across the perpendicular bisector of $AC$,which is $x+2y=0$,gives $C(x_2, y_2)$:$\frac{x_2-1}{1} = \frac{y_2+2}{2} = -2\frac{1+2(-2)}{1^2+2^2} = -2\frac{-3}{5} = \frac{6}{5}$.$x_2-1 = \frac{6}{5} \Rightarrow x_2 = \frac{11}{5}$.$y_2+2 = \frac{12}{5} \Rightarrow y_2 = \frac{2}{5}$.So,$C = (\frac{11}{5}, \frac{2}{5})$.The equation of line $BC$ passing through $(-7, 6)$ and $(\frac{11}{5}, \frac{2}{5})$ is:$y-6 = \frac{\frac{2}{5}-6}{\frac{11}{5}-(-7)}(x+7) = \frac{-28/5}{46/5}(x+7) = -\frac{14}{23}(x+7)$.$23(y-6) = -14(x+7) \Rightarrow 23y-138 = -14x-98$.$14x+23y-40=0$.

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

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