1. **Problem 1: Write the equation of line (AB) given points A and B.** The formula for the slope $m$ of the line through points $A(x_1,y_1)$ and $B(x_2,y_2)$ is: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ The equation of the line in point-slope form is: $$y - y_1 = m(x - x_1)$$ **Case 1:** $A(2,3)$, $B(5,7)$ Calculate slope: $$m = \frac{7 - 3}{5 - 2} = \frac{4}{3}$$ Equation: $$y - 3 = \frac{4}{3}(x - 2)$$ Simplify: $$y - 3 = \frac{4}{3}x - \frac{8}{3}$$ $$y = \frac{4}{3}x - \frac{8}{3} + 3 = \frac{4}{3}x + \frac{1}{3}$$ **Case 2:** $A(-5,2)$, $B(0,1)$ Calculate slope: $$m = \frac{1 - 2}{0 - (-5)} = \frac{-1}{5} = -\frac{1}{5}$$ Equation: $$y - 2 = -\frac{1}{5}(x + 5)$$ Simplify: $$y - 2 = -\frac{1}{5}x - 1$$ $$y = -\frac{1}{5}x - 1 + 2 = -\frac{1}{5}x + 1$$ **Case 3:** $A(3,4)$, $B(-7,-3)$ Calculate slope: $$m = \frac{-3 - 4}{-7 - 3} = \frac{-7}{-10} = \frac{7}{10}$$ Equation: $$y - 4 = \frac{7}{10}(x - 3)$$ Simplify: $$y - 4 = \frac{7}{10}x - \frac{21}{10}$$ $$y = \frac{7}{10}x - \frac{21}{10} + 4 = \frac{7}{10}x + \frac{19}{10}$$ 2. **Problem 2: Write the equation of line (d) passing through point A with direction vector $\vec{V}(v_x,v_y)$.** Parametric form: $$\begin{cases} x = x_0 + v_x t \\ y = y_0 + v_y t \end{cases}$$ where $A(x_0,y_0)$ and $t$ is a parameter. **Case 1:** $A(2,-3)$, $\vec{V}(3,-1)$ $$\begin{cases} x = 2 + 3t \\ y = -3 - t \end{cases}$$ **Case 2:** $A(0,2)$, $\vec{V}(-1,4)$ $$\begin{cases} x = 0 - t = -t \\ y = 2 + 4t \end{cases}$$ **Case 3:** $A(1,-2)$, $\vec{V}(0,-5)$ $$\begin{cases} x = 1 + 0 \cdot t = 1 \\ y = -2 - 5t \end{cases}$$ 3. **Problem 3: Write the equation of line (d') passing through A and parallel to line (d).** Parallel lines have the same slope. **Case 1:** $A(2,-3)$, $(d): 4x - 2y + 1 = 0$ Rewrite $(d)$ in slope-intercept form: $$4x - 2y + 1 = 0 \Rightarrow -2y = -4x - 1 \Rightarrow y = 2x + \frac{1}{2}$$ Slope $m = 2$. Equation of $(d')$ through $A$: $$y - (-3) = 2(x - 2)$$ $$y + 3 = 2x - 4$$ $$y = 2x - 7$$ **Case 2:** $A(-1,-1)$, $(d): y = 3x - 4$ Slope $m = 3$. Equation of $(d')$: $$y - (-1) = 3(x + 1)$$ $$y + 1 = 3x + 3$$ $$y = 3x + 2$$ **Case 3:** $A(5,3)$, $(d): \begin{cases} x = 2m - 1 \\ y = -m + 2 \end{cases}$ Express $y$ in terms of $x$: From $x = 2m - 1 \Rightarrow m = \frac{x + 1}{2}$ Substitute into $y$: $$y = -\frac{x + 1}{2} + 2 = -\frac{x}{2} - \frac{1}{2} + 2 = -\frac{x}{2} + \frac{3}{2}$$ Slope $m = -\frac{1}{2}$. Equation of $(d')$ through $A$: $$y - 3 = -\frac{1}{2}(x - 5)$$ $$y - 3 = -\frac{1}{2}x + \frac{5}{2}$$ $$y = -\frac{1}{2}x + \frac{5}{2} + 3 = -\frac{1}{2}x + \frac{11}{2}$$ 4. **Problem 4: Write the equation of line (d') passing through A and perpendicular to line (d).** If slope of $(d)$ is $m$, slope of perpendicular line is $m' = -\frac{1}{m}$. **Case 1:** $A(-1,2)$, $(d): y = -2x + 4$ Slope $m = -2$. Slope of perpendicular $m' = -\frac{1}{-2} = \frac{1}{2}$. Equation of $(d')$: $$y - 2 = \frac{1}{2}(x + 1)$$ $$y - 2 = \frac{1}{2}x + \frac{1}{2}$$ $$y = \frac{1}{2}x + \frac{5}{2}$$ **Case 2:** $A(3,-1)$, $(d): 2x - 4y + 1 = 0$ Rewrite $(d)$: $$2x - 4y + 1 = 0 \Rightarrow -4y = -2x - 1 \Rightarrow y = \frac{1}{2}x + \frac{1}{4}$$ Slope $m = \frac{1}{2}$. Slope of perpendicular $m' = -2$. Equation of $(d')$: $$y - (-1) = -2(x - 3)$$ $$y + 1 = -2x + 6$$ $$y = -2x + 5$$ **Case 3:** $A(0,3)$, $(d): \begin{cases} x = 4t - 2 \\ y = 6t + 2 \end{cases}$ Express $y$ in terms of $x$: From $x = 4t - 2 \Rightarrow t = \frac{x + 2}{4}$ Substitute into $y$: $$y = 6 \cdot \frac{x + 2}{4} + 2 = \frac{3}{2}x + 3 + 2 = \frac{3}{2}x + 5$$ Slope $m = \frac{3}{2}$. Slope of perpendicular $m' = -\frac{2}{3}$. Equation of $(d')$: $$y - 3 = -\frac{2}{3}(x - 0)$$ $$y - 3 = -\frac{2}{3}x$$ $$y = -\frac{2}{3}x + 3$$