1. **Problem 1: Express the equation of a line in standard form given slope and y-intercept.** The slope-intercept form of a line is given by: $$y = mx + b$$ where $m$ is the slope and $b$ is the y-intercept. To convert to standard form $Ax + By = C$, rearrange terms so that $x$ and $y$ are on the left and constant on the right. --- **a. Slope $m=0$, y-intercept $b=-1$** Start with: $$y = 0 \cdot x - 1 = -1$$ Rewrite as: $$y = -1$$ Move all terms to one side: $$0 \cdot x + y = -1$$ Standard form: $$0x + y = -1$$ --- **b. Slope $m=\frac{3}{4}$, y-intercept $b=-10$** Start with: $$y = \frac{3}{4}x - 10$$ Multiply both sides by 4 to clear fraction: $$4y = 3x - 40$$ Rearranged: $$-3x + 4y = -40$$ Multiply both sides by $-1$ to make $A$ positive: $$3x - 4y = 40$$ --- **c. Slope $m=4$, y-intercept $b=\frac{5}{2}$** Start with: $$y = 4x + \frac{5}{2}$$ Multiply both sides by 2 to clear fraction: $$2y = 8x + 5$$ Rearranged: $$-8x + 2y = 5$$ Multiply both sides by $-1$: $$8x - 2y = -5$$ --- 2. **Problem 2: Find the standard form of the equation of the line passing through a point with a given slope.** Use point-slope form: $$y - y_1 = m(x - x_1)$$ Then convert to standard form $Ax + By = C$. --- **a. Point $(6,0)$, slope $m = -\frac{1}{2}$** Start with: $$y - 0 = -\frac{1}{2}(x - 6)$$ Simplify: $$y = -\frac{1}{2}x + 3$$ Multiply both sides by 2: $$2y = -x + 6$$ Rearranged: $$x + 2y = 6$$ --- **b. Point $(3,-2)$, slope $m = -\frac{4}{7}$** Start with: $$y - (-2) = -\frac{4}{7}(x - 3)$$ Simplify: $$y + 2 = -\frac{4}{7}x + \frac{12}{7}$$ Multiply both sides by 7: $$7y + 14 = -4x + 12$$ Rearranged: $$4x + 7y = -2$$ --- **c. Point $(0,0)$, slope $m = \frac{1}{4}$** Start with: $$y - 0 = \frac{1}{4}(x - 0)$$ Simplify: $$y = \frac{1}{4}x$$ Multiply both sides by 4: $$4y = x$$ Rearranged: $$-x + 4y = 0$$ Multiply both sides by $-1$: $$x - 4y = 0$$