Step 1: Find the vertices by solving the equations algebraically, and find the area with the coordinate area formula, instead of reading the graph and using base times height.
Plotting points is one way to find where the lines meet, but we can get the exact same vertices faster by solving pairs of equations directly, and then use a formula that works for any triangle once we know its three corners.
Step 2: Find vertex A, the intersection of the two slanted lines.
The two lines are:
\[ 2x - 3y + 6 = 0 \quad \text{...(i)} \]
\[ 2x + 3y - 18 = 0 \quad \text{...(ii)} \]
Add equations (i) and (ii) together. The $y$ terms cancel because they have opposite signs:
\[ (2x - 3y + 6) + (2x + 3y - 18) = 0 \]
\[ 4x - 12 = 0 \implies x = 3 \]
Substitute $x = 3$ back into equation (i):
\[ 2(3) - 3y + 6 = 0 \implies 12 - 3y = 0 \implies y = 4 \]
So vertex $A = (3, 4)$.
Step 3: Find vertex B, where the first line meets the third line $x = 0$.
Substitute $x = 0$ into equation (i):
\[ 2(0) - 3y + 6 = 0 \implies -3y = -6 \implies y = 2 \]
So vertex $B = (0, 2)$.
Step 4: Find vertex C, where the second line meets $x = 0$.
Substitute $x = 0$ into equation (ii):
\[ 2(0) + 3y - 18 = 0 \implies 3y = 18 \implies y = 6 \]
So vertex $C = (0, 6)$.
Step 5: Apply the coordinate area formula to the three vertices.
For a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Using $A(3,4)$, $B(0,2)$, $C(0,6)$:
\[ \text{Area} = \frac{1}{2} \left| 3(2 - 6) + 0(6 - 4) + 0(4 - 2) \right| \]
\[ = \frac{1}{2} \left| 3(-4) + 0 + 0 \right| = \frac{1}{2} \left| -12 \right| = \frac{1}{2} \times 12 = 6 \]
Step 6: Final answer.
The vertices are $A(3,4)$, $B(0,2)$, $C(0,6)$, and the area of the triangle is $6$ square units.
\[ \boxed{\text{Area} = 6 \text{ sq. units}} \]