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(A) The perpendicular distance $d$ from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by $d = \left| \frac{Ax_1 + By_1 + C}{\sqrt{A^2 +

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$2, -18$

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(A) The perpendicular distance $d$ from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by $d = \left| \frac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}} \right|$.Given point $(4, 1)$ and line $3x - 4y + k = 0$,we have:$2 = \left| \frac{3(4) - 4(1) + k}{\sqrt{3^2 + (-4)^2}} \right|$$2 = \left| \frac{12 - 4 + k}{\sqrt{9 + 16}} \right|$$2 = \left| \frac{8 + k}{5} \right|$$|8 + k| = 10$This implies $8 + k = 10$ or $8 + k = -10$.If $8 + k = 10$,then $k = 2$.If $8 + k = -10$,then $k = -18$.Therefore,the values of $k$ are $2$ and $-18$.

If the length of the perpendicular drawn from the point $(4,1)$ to the line $3x - 4y + k = 0$ is $2$ units,then the values of $k$ are:

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