Here are the steps to solve the problem. First, complete the table for the given relations. Step 1: Complete the table For the relation $x - 2y = -2$: If $x = -2$: $-2 - 2y = -2 \implies -2y = 0 \implies y = 0$ If $x = -1$: $-1 - 2y = -2 \implies -2y = -1 \implies y = 0.5$ If $x = 0$: $0 - 2y = -2 \implies -2y = -2 \implies y = 1$ If $x = 1$: $1 - 2y = -2 \implies -2y = -3 \implies y = 1.5$ If $x = 2$: $2 - 2y = -2 \implies -2y = -4 \implies y = 2$ If $x = 3$: $3 - 2y = -2 \implies -2y = -5 \implies y = 2.5$ For the relation $x - 2y = 2$: If $x = -2$: $-2 - 2y = 2 \implies -2y = 4 \implies y = -2$ If $x = -1$: $-1 - 2y = 2 \implies -2y = 3 \implies y = -1.5$ If $x = 0$: $0 - 2y = 2 \implies -2y = 2 \implies y = -1$ If $x = 1$: $1 - 2y = 2 \implies -2y = 1 \implies y = -0.5$ If $x = 2$: $2 - 2y = 2 \implies -2y = 0 \implies y = 0$ If $x = 3$: $3 - 2y = 2 \implies -2y = -1 \implies y = 0.5$ The completed table is: $$ \begin{array}{|c|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline x - 2y = -2 & 0 & 0.5 & 1 & 1.5 & 2 & 2.5 \\ \hline x - 2y = 2 & -2 & -1.5 & -1 & -0.5 & 0 & 0.5 \\ \hline \end{array} $$ b. (i) Drawing axes Draw two perpendicular lines, one horizontal (OX-axis) and one vertical (OY-axis), intersecting at the origin (0,0). Mark units on both axes such that every 2 cm represents 2 units. This means 1 cm represents 1 unit. Label the positive and negative directions for both axes. c. (ii) Drawing the graphs To draw the graphs, plot the points from the completed table for each relation and then draw a straight line through them. For the relation $x - 2y = -2$: Plot the points: $(-2, 0)$, $(-1, 0.5)$, $(0, 1)$, $(1, 1.5)$, $(2, 2)$, $(3, 2.5)$. Draw a straight line connecting these points. For the relation $x - 2y = 2$: Plot the points: $(-2, -2)$, $(-1, -1.5)$, $(0, -1)$, $(1, -0.5)$, $(2, 0)$, $(3, 0.5)$. Draw a straight line connecting these points. d. Conclusion from the graph To determine the conclusion, we can analyze the equations in slope-intercept form ($y = mx + c$). For the first relation: $x - 2y = -2$ $$2y = x + 2$$ $$y = \frac{1}{2}x + 1$$ The slope is $m_1 = \frac{1}{2}$ and the y-intercept is $c_1 = 1$. For the second relation: $x - 2y = 2$ $$2y = x - 2$$ $$y = \frac{1}{2}x - 1$$ The slope is $m_2 = \frac{1}{2}$ and the y-intercept is $c_2 = -1$. Since both lines have the same slope ($m_1 = m_2 = \frac{1}{2}$) but different y-intercepts ($c_1 \neq c_2$), the lines are parallel. They will never intersect. The conclusion from the graph is that the two lines are parallel. $$\boxed{\text{The two lines are parallel.}}$$