Parallel and perpendicular lines through a point (SAT). Game on.

Parallel lines have the same slope, so $m = \dfrac{1}{2}$. Substitute $(6, 1)$ into $y = \dfrac{1}{2}x + b$: $1 = \dfrac{1}{2}(6) + b = 3 + b$, so $b = -2$. The equation is $y = \dfrac{1}{2}x - 2\textsf{.}$

Parallel lines have the same slope, so $m = \dfrac{2}{3}$. Substitute $(-3, 4)$ into $y = \dfrac{2}{3}x + b$: $4 = \dfrac{2}{3}(-3) + b = -2 + b$, so $b = 6$. The equation is $y = \dfrac{2}{3}x + 6\textsf{.}$

Parallel lines have the same slope, so $m = -1$. Substitute $(4, -2)$ into $y = -x + b$: $-2 = -4 + b$, so $b = 2$. The equation is $y = -x + 2\textsf{.}$

Perpendicular lines have slopes that are negative reciprocals, so $m = -\dfrac{1}{2}$. Substitute $(4, 3)$ into $y = -\dfrac{1}{2}x + b$: $3 = -\dfrac{1}{2}(4) + b = -2 + b$, so $b = 5$. The equation is $y = -\dfrac{1}{2}x + 5\textsf{.}$

Perpendicular lines have slopes that are negative reciprocals, so $m = \dfrac{1}{3}$. Substitute $(3, 1)$ into $y = \dfrac{1}{3}x + b$: $1 = \dfrac{1}{3}(3) + b = 1 + b$, so $b = 0$. The equation is $y = \dfrac{1}{3}x\textsf{.}$

Perpendicular lines have slopes that are negative reciprocals, so $m = -2$. Substitute $(2, -1)$ into $y = -2x + b$: $-1 = -2(2) + b = -4 + b$, so $b = 3$. The equation is $y = -2x + 3\textsf{.}$

Perpendicular lines have slopes that are negative reciprocals, so $m = -3$. Substitute $(3, 2)$ into $y = -3x + b$: $2 = -3(3) + b = -9 + b$, so $b = 11$. The equation is $y = -3x + 11\textsf{.}$

Perpendicular lines have slopes that are negative reciprocals, so $m = \dfrac{3}{2}$. Substitute $(6, -1)$ into $y = \dfrac{3}{2}x + b$: $-1 = \dfrac{3}{2}(6) + b = 9 + b$, so $b = -10$. The equation is $y = \dfrac{3}{2}x - 10\textsf{.}$