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: $y - 7 = -2x + 6$ so $y = -2x + 13$. Step 4: Set $y=17$: $17 = -2x + 13$. Step 5: Solve for $x$: $-2x = 4$ so $x = -2$. Answer: (b) $-2$. 2. Problem: Determine the relationship between lines $2x - 3y + 1 = 0$ and $4x - 6y + 1 = 0$. Step 1: Find slopes: slope $m_1 = \frac{2}{3}$, slope $m_2 = \frac{4}{6} = \frac{2}{3}$. Step 2: Slopes are equal, check constants: lines differ in constant term, so they are parallel but not coincident. Answer: (a) parallel. 3. Problem: Find $c$ so lines $2x + cy - 3 = 0$ and $4x - 9y + 1 = 0$ are perpendicular. Step 1: Slopes: $m_1 = -\frac{2}{c}$, $m_2 = \frac{4}{9}$. Step 2: For perpendicular lines, $m_1 \times m_2 = -1$. Step 3: $-\frac{2}{c} \times \frac{4}{9} = -1$. Step 4: Simplify: $-\frac{8}{9c} = -1$ so $\frac{8}{9c} = 1$. Step 5: Solve for $c$: $9c = 8$ so $c = \frac{8}{9}$. Answer: (c) $\frac{8}{9}$. 4. Problem: Find equation of line crossing x-axis at $-3$ and parallel to $7y=11$. Step 1: $7y=11$ is horizontal line $y=\frac{11}{7}$. Step 2: Parallel line is also horizontal, so $y = k$. Step 3: Crossing x-axis at $-3$ means point $(-3,0)$ lies on line. Step 4: Horizontal line through $(-3,0)$ is $y=0$. Answer: (d) none of the given options. 5. Problem: Equation of line through $(3,4)$ parallel to x-axis. Step 1: Lines parallel to x-axis are horizontal: $y = k$. Step 2: Passes through $(3,4)$ so $y=4$. Answer: (c) $y=4$. 6. Problem: Solve system $x - 2y + \frac{1}{2} = 0$ and $2x - 4y + 1 = 0$. Step 1: Multiply first equation by 2: $2x - 4y + 1 = 0$. Step 2: Second equation is same: system is dependent. Step 3: Infinite solutions. 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: Rewrite first line: $2y = x - 5$ so $y = \frac{x}{2} - \frac{5}{2}$, slope $m_1=\frac{1}{2}$. Step 2: Rewrite second line: $y = -(1+c)x - 2$, slope $m_2 = -(1+c)$. Step 3: For parallel lines, $m_1 = m_2$. Step 4: $\frac{1}{2} = -(1+c)$ so $1+c = -\frac{1}{2}$. Step 5: $c = -\frac{3}{2}$. Answer: (b) $-\frac{3}{2}$. 8. Problem: Determine relation of lines $3x - y + 2=0$ and $2x - y - 3=0$. Step 1: Slopes: $m_1 = 3$, $m_2 = 2$. Step 2: Slopes not equal, product $3 \times 2 = 6 \neq -1$. Step 3: Lines are neither parallel nor perpendicular nor identical. Answer: (d) neither. 9. Problem: Graph of $y = -2$ is line? Step 1: $y = -2$ is horizontal line. Step 2: Horizontal lines are 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: Lines parallel to y-axis are vertical: $x = k$. Step 2: Passes through $(4,3)$ so $x=4$. Answer: (b) $x=4$. 11. Problem: Points $(-1,4)$, $(1,5)$, $(3,y)$ collinear, find $y$. Step 1: Find slope between first two points: $m = \frac{5-4}{1+1} = \frac{1}{2}$. Step 2: Use slope to find $y$: $m = \frac{y-5}{3-1} = \frac{y-5}{2}$. Step 3: Set equal: $\frac{1}{2} = \frac{y-5}{2}$. Step 4: Multiply both sides by 2: $1 = y - 5$. Step 5: $y = 6$. Answer: (c) 6.