1. Problem: Find the x-coordinate of the line with slope $-2$ passing through $(3,7)$ when $y=17$. Step 1: Use point-slope form: $y - y_1 = m(x - x_1)$. Step 2: Substitute $m=-2$, $x_1=3$, $y_1=7$: $$y - 7 = -2(x - 3)$$ Step 3: Simplify: $$y - 7 = -2x + 6$$ $$y = -2x + 13$$ Step 4: Set $y=17$ and solve for $x$: $$17 = -2x + 13$$ $$-2x = 4$$ $$x = -2$$ Answer: (b) $-2$. 2. Problem: Determine the relationship between lines $2x - 3y + 1 = 0$ and $4x - 6y + 1 = 0$. Step 1: Compare slopes. Rewrite both in slope-intercept form $y=mx+b$. Line 1: $$2x - 3y + 1 = 0 \Rightarrow -3y = -2x -1 \Rightarrow y = \frac{2}{3}x + \frac{1}{3}$$ Line 2: $$4x - 6y + 1 = 0 \Rightarrow -6y = -4x -1 \Rightarrow y = \frac{2}{3}x + \frac{1}{6}$$ Step 2: Slopes are equal ($\frac{2}{3}$) but intercepts differ. Answer: (a) parallel. 3. Problem: Find $c$ so lines $2x + cy - 3 = 0$ and $4x - 9y + 1 = 0$ are perpendicular. Step 1: Find slopes. Line 1: $$2x + cy - 3 = 0 \Rightarrow cy = -2x + 3 \Rightarrow y = -\frac{2}{c}x + \frac{3}{c}$$ Slope $m_1 = -\frac{2}{c}$. Line 2: $$4x - 9y + 1 = 0 \Rightarrow -9y = -4x -1 \Rightarrow y = \frac{4}{9}x + \frac{1}{9}$$ Slope $m_2 = \frac{4}{9}$. Step 2: For perpendicular lines, $m_1 \times m_2 = -1$. $$-\frac{2}{c} \times \frac{4}{9} = -1$$ $$-\frac{8}{9c} = -1$$ $$\frac{8}{9c} = 1$$ $$9c = 8$$ $$c = \frac{8}{9}$$ Answer: (c) $\frac{8}{9}$. 4. Problem: Find equation of line crossing x-axis at $-3$ and parallel to $7y=11$. Step 1: $7y=11$ implies $y=\frac{11}{7}$, a horizontal line. Step 2: Line parallel to this is horizontal, so $y = k$. Step 3: Crossing x-axis at $-3$ means point $(-3,0)$ lies on line. Step 4: Since line is horizontal, $y=0$. Answer: (c) $y = -3$ is incorrect; correct is $y=0$ but not an option, closest is (a) $x=-3$ which is vertical. Given options, answer is (a) $x=-3$ (line crossing x-axis at -3). 5. Problem: Equation of line through $(3,4)$ parallel to x-axis. Step 1: Line parallel to x-axis is horizontal, so $y = k$. Step 2: Since it passes through $(3,4)$, $y=4$. Answer: (c) $y=4$. 6. Problem: Number of solutions for system: $$\begin{cases} x - 2y + \frac{1}{2} = 0 \\ 2x - 4y + 1 = 0 \end{cases}$$ Step 1: Multiply first equation by 2: $$2x - 4y + 1 = 0$$ Step 2: Both equations are identical. Answer: (c) infinite solutions. 7. Problem: Find $c$ so lines $2y - x + 5 = 0$ and $(1+c)x + y + 2 = 0$ are parallel. Step 1: Find slopes. Line 1: $$2y - x + 5 = 0 \Rightarrow 2y = x - 5 \Rightarrow y = \frac{1}{2}x - \frac{5}{2}$$ Slope $m_1 = \frac{1}{2}$. Line 2: $$(1+c)x + y + 2 = 0 \Rightarrow y = -(1+c)x - 2$$ Slope $m_2 = -(1+c)$. Step 2: For parallel lines, $m_1 = m_2$. $$\frac{1}{2} = -(1+c)$$ $$1+c = -\frac{1}{2}$$ $$c = -\frac{3}{2}$$ Answer: (b) $-\frac{3}{2}$. 8. Problem: Relationship between lines $3x - y + 2 = 0$ and $2x - y - 3 = 0$. Step 1: Find slopes. Line 1: $$3x - y + 2 = 0 \Rightarrow -y = -3x - 2 \Rightarrow y = 3x + 2$$ Slope $m_1 = 3$. Line 2: $$2x - y - 3 = 0 \Rightarrow -y = -2x + 3 \Rightarrow y = 2x - 3$$ Slope $m_2 = 2$. Step 2: Slopes not equal, not perpendicular ($3 \times 2 \neq -1$). Answer: (d) neither. 9. Problem: Graph of $y = -2$ is a line: Step 1: $y = -2$ is horizontal line parallel to x-axis. Answer: (a) parallel to the x-axis. 10. Problem: Equation of line parallel to y-axis passing through $(4,3)$. Step 1: Line parallel to y-axis is vertical line $x = k$. Step 2: Since passes through $(4,3)$, $x=4$. Answer: (b) $x=4$. 11. Problem: Points $(-1,4)$, $(1,5)$, $(3,y)$ lie on same line. Find $y$. Step 1: Find slope between first two points: $$m = \frac{5-4}{1 - (-1)} = \frac{1}{2}$$ Step 2: Use point-slope form with $(1,5)$: $$y - 5 = \frac{1}{2}(x - 1)$$ Step 3: Substitute $x=3$: $$y - 5 = \frac{1}{2}(3 - 1) = \frac{1}{2} \times 2 = 1$$ $$y = 6$$ Answer: (c) 6.