If A(4,7),B(-7,8),and C(1,2) are the vertices | vedclass

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(C) The slope of the side $AB$ is $m_{AB} = \frac{8-7}{-7-4} = \frac{1}{-11} = -\frac{1}{11}$.Since the perpendicular bisector is perpendicular to $AB

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$11x-y+24=0$

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(C) The slope of the side $AB$ is $m_{AB} = \frac{8-7}{-7-4} = \frac{1}{-11} = -\frac{1}{11}$.Since the perpendicular bisector is perpendicular to $AB$,its slope $m$ is given by $m 	imes m_{AB} = -1$,so $m = 11$.The midpoint $D$ of side $AB$ is $\left(\frac{4+(-7)}{2}, \frac{7+8}{2}ight) = \left(-\frac{3}{2}, \frac{15}{2}ight)$.The equation of the line passing through $D\left(-\frac{3}{2}, \frac{15}{2}ight)$ with slope $m=11$ is:$y - \frac{15}{2} = 11\left(x - (-\frac{3}{2}ight))$$y - \frac{15}{2} = 11\left(x + \frac{3}{2}ight)$$y - \frac{15}{2} = 11x + \frac{33}{2}$$11x - y + \frac{33}{2} + \frac{15}{2} = 0$$11x - y + \frac{48}{2} = 0$$11x - y + 24 = 0$.

If $A(4,7)$,$B(-7,8)$,and $C(1,2)$ are the vertices of $\triangle ABC$,then the equation of the perpendicular bisector of the side $AB$ is

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