Question 12

Explanation

To calculate the area of region $S$, we need to determine the vertices of the solution region for the system of inequalities. This region is bounded by three lines:

1. Line 1: $y = 2x + 4$
2. Line 2: $y + x - 4 = 0 \Rightarrow y = -x + 4$
3. Line 3: $x = 4$

Let's find the intersection points between these lines.

Intersection of Line 1 and Line 2

Equate the $y$ values from both equations:

Substitute $x = 0$ into one of the equations (e.g., Line 2):

So, the first intersection point is $A(0, 4)$.

Intersection of Line 1 and Line 3

Substitute $x = 4$ into the equation of Line 1:

So, the second intersection point is $B(4, 12)$.

Intersection of Line 2 and Line 3

Substitute $x = 4$ into the equation of Line 2:

So, the third intersection point is $C(4, 0)$.

Calculating the Area

Region $S$ is a triangle with vertices $A(0, 4)$, $B(4, 12)$, and $C(4, 0)$.

We can calculate the area using the base and height.

• Base: The vertical line connecting point $B$ and $C$ (since both have $x = 4$). Length of base $= y_B - y_C = 12 - 0 = 12$.

• Height: The horizontal distance from point $A$ to the line $x = 4$. Height $= x_{BC} - x_A = 4 - 0 = 4$.

Area of the triangle:

Thus, the area of region $S$ is $24$ square units.