Slope-point line equations (ACT/SAT/PSAT 8/9/10). Game on.

Point-slope form is $y - y_1 = m(x - x_1)$. With $m = 4$ and $(x_1, y_1) = (-1, 2)$, the equation is $y - 2 = 4(x - (-1))$, or $y - 2 = 4(x + 1)\textsf{.}$

Slope-intercept form is $y = mx + b$. Substituting $(4, 1)$ with $m = \dfrac{3}{2}$ gives $1 = \dfrac{3}{2}(4) + b = 6 + b$, so $b = -5$. The equation is $y = \dfrac{3}{2}x - 5\textsf{.}$

Point-slope form is $y - y_1 = m(x - x_1)$. With $m = -2$ and $(x_1, y_1) = (5, -3)$, the equation is $y - (-3) = -2(x - 5)$, or $y + 3 = -2(x - 5)\textsf{.}$

Point-slope form is $y - y_1 = m(x - x_1)$. With $m = -\dfrac{1}{2}$ and $(x_1, y_1) = (6, 4)$, the equation is $y - 4 = -\dfrac{1}{2}(x - 6)\textsf{.}$

First write slope-intercept form: $8 = 3(2) + b$ gives $b = 2$, so $y = 3x + 2$. Rearranging into standard form gives $3x - y = -2\textsf{.}$

First write slope-intercept form: $5 = -4(1) + b$ gives $b = 9$, so $y = -4x + 9$. Rearranging into standard form gives $4x + y = 9\textsf{.}$

Point-slope form is $y - y_1 = m(x - x_1)$. With $m = \dfrac{5}{4}$ and $(x_1, y_1) = (-4, 3)$, the equation is $y - 3 = \dfrac{5}{4}(x - (-4))$, or $y - 3 = \dfrac{5}{4}(x + 4)\textsf{.}$

First write slope-intercept form: $10 = \dfrac{2}{3}(9) + b$ gives $b = 4$, so $y = \dfrac{2}{3}x + 4$. Multiplying by $3$ and rearranging gives $-2x + 3y = 12\textsf{.}$