If the coordinates of points A, B, C are (-1, | vedclass

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(C) Given points are $A(-1, 5), B(0, 0), C(2, 2)$.$D$ is the midpoint of $BC$,so $D = (\frac{0+2}{2}, \frac{0+2}{2}) = (1, 1)$.The slope of line $AD$

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$x - 2y = 0$

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(C) Given points are $A(-1, 5), B(0, 0), C(2, 2)$.$D$ is the midpoint of $BC$,so $D = (\frac{0+2}{2}, \frac{0+2}{2}) = (1, 1)$.The slope of line $AD$ passing through $A(-1, 5)$ and $D(1, 1)$ is $m_{AD} = \frac{1-5}{1-(-1)} = \frac{-4}{2} = -2$.The line perpendicular to $AD$ passing through $B(0, 0)$ will have a slope $m = -\frac{1}{m_{AD}} = -\frac{1}{-2} = \frac{1}{2}$.The equation of this line is $y - 0 = \frac{1}{2}(x - 0)$,which simplifies to $2y = x$ or $x - 2y = 0$.

If the coordinates of points $A, B, C$ are $(-1, 5), (0, 0)$ and $(2, 2)$ respectively,and $D$ is the midpoint of $BC$,then the equation of the perpendicular drawn from $B$ to the line $AD$ is:

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