1. Determine graphically, the coordinates of vertices of a triangle whose equations are $2x - 3y + 6 = 0$; $2x + 3y - 18 = 0$ and $x = 0$. Also, find the area of this triangle.

The three lines $2x - 3y + 6 = 0$, $2x + 3y - 18 = 0$ and $x = 0$ form a triangle with vertices $A(0, 2)$, $B(0, 6)$ and $C(3, 4)$, and its area is 6 square units.

To plot the lines, points are found for each equation: for $2x - 3y + 6 = 0$, the points $(0, 2), (3, 4) \text{ and } (-3, 0)$ lie on the line; for $2x + 3y - 18 = 0$, the points $(0, 6), (3, 4) \text{ and } (9, 0)$ lie on the line. The third line $x = 0$ is simply the y-axis. Plotting all three lines on graph paper shows that line 1 meets the y-axis at $A(0, 2)$, line 2 meets the y-axis at $B(0, 6)$, and the two slanting lines intersect each other at $C(3, 4)$ — this can be verified algebraically by adding the two equations to get $4x - 12 = 0$, so $x = 3$, and substituting back gives $y = 4$.

Since A and B both lie on the y-axis, side AB is vertical with length $|6-2| = 4\text{ units}$, taken as the base. The perpendicular (horizontal) distance from $C(3, 4)$ to the y-axis is simply its x-coordinate, 3 units, taken as the height. Hence $\text{Area} = \frac{1}{2}\times \text{base} \times \text{height} = \frac{1}{2}\times 4 \times 3 = 6\text{ square units}$.