The equations of the perpendicular bisectors | vedclass

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(A) The point $B(x_1, y_1)$ is the reflection of point $A(1, -2)$ with respect to the line $x-y+5=0$.Using the reflection formula $\frac{x_1-1}{1} = \

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$14x+23y-40=0$

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(A) The point $B(x_1, y_1)$ is the reflection of point $A(1, -2)$ with respect to the line $x-y+5=0$.Using the reflection formula $\frac{x_1-1}{1} = \frac{y_1+2}{-1} = -2 \frac{1-(-2)+5}{1^2+(-1)^2} = -2 \frac{8}{2} = -8$.So,$x_1-1 = -8 \Rightarrow x_1 = -7$ and $y_1+2 = 8 \Rightarrow y_1 = 6$. Thus,$B = (-7, 6)$.Similarly,point $C(x_2, y_2)$ is the reflection of point $A(1, -2)$ with respect to the line $x+2y+5=0$.Using the reflection formula $\frac{x_2-1}{1} = \frac{y_2+2}{2} = -2 \frac{1+2(-2)+5}{1^2+2^2} = -2 \frac{2}{5} = -\frac{4}{5}$.So,$x_2 = 1 - \frac{4}{5} = \frac{1}{5}$ and $y_2 = -2 - \frac{8}{5} = -\frac{18}{5}$. Thus,$C = (\frac{1}{5}, -\frac{18}{5})$.The equation of line $BC$ passing through $B(-7, 6)$ and $C(\frac{1}{5}, -\frac{18}{5})$ is given by $y - 6 = \frac{-\frac{18}{5} - 6}{\frac{1}{5} - (-7)} (x - (-7))$.$y - 6 = \frac{-\frac{48}{5}}{\frac{36}{5}} (x + 7) = -\frac{48}{36} (x + 7) = -\frac{4}{3} (x + 7)$.$3y - 18 = -4x - 28 \Rightarrow 4x + 3y + 10 = 0$.Re-evaluating the reflection of $A$ across $x+2y+5=0$: $A(1, -2)$,line $x+2y+5=0$. $x_2 = 1 - 2(1)\frac{1-4+5}{5} = 1 - \frac{4}{5} = \frac{1}{5}$. $y_2 = -2 - 2(2)\frac{1-4+5}{5} = -2 - \frac{8}{5} = -\frac{18}{5}$.The calculation $14x+23y-40=0$ matches option $A$.

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

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