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(D) Let the equation of the line passing through $(0, 2)$ be $y-2 = mx$,or $mx - y + 2 = 0$.The perpendicular distance from $(4, 4)$ to this line is $

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$y+2x=10$

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(D) Let the equation of the line passing through $(0, 2)$ be $y-2 = mx$,or $mx - y + 2 = 0$.The perpendicular distance from $(4, 4)$ to this line is $2$ units.$\frac{|m(4) - 4 + 2|}{\sqrt{m^2 + (-1)^2}} = 2$$\frac{|4m - 2|}{\sqrt{m^2 + 1}} = 2$$|2m - 1| = \sqrt{m^2 + 1}$Squaring both sides:$4m^2 - 4m + 1 = m^2 + 1$$3m^2 - 4m = 0$$m(3m - 4) = 0$So,$m = 0$ or $m = \frac{4}{3}$.Case $1$: $m = 0$. The line is $y = 2$. The foot of the perpendicular from $(4, 4)$ to $y = 2$ is $D(4, 2)$.Case $2$: $m = \frac{4}{3}$. The line is $y - 2 = \frac{4}{3}x$,or $4x - 3y + 6 = 0$. The foot of the perpendicular $(x_1, y_1)$ from $(4, 4)$ satisfies $\frac{x_1 - 4}{4} = \frac{y_1 - 4}{-3} = -\frac{4(4) - 3(4) + 6}{4^2 + (-3)^2} = -\frac{10}{25} = -\frac{2}{5}$.$x_1 = 4 - \frac{8}{5} = \frac{12}{5}$,$y_1 = 4 + \frac{6}{5} = \frac{26}{5}$. So $C(\frac{12}{5}, \frac{26}{5})$.The line joining $D(4, 2)$ and $C(\frac{12}{5}, \frac{26}{5})$ has slope $\frac{\frac{26}{5} - 2}{\frac{12}{5} - 4} = \frac{16/5}{-8/5} = -2$.The equation is $y - 2 = -2(x - 4)$,which simplifies to $y + 2x = 10$.

Two straight lines are drawn through the point $(0, 2)$ such that the lengths of the perpendiculars from the point $(4, 4)$ to these lines are each equal to $2$ units. The equation of the line joining the feet of these perpendiculars is

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