The area of the region bounded by the lines y=2x+1y=3x+1 and x=4 is

Options

A. 16 sq unit
B. `(121)/(3)` sq unit
C. `(121)/(6)` sq unit
D. 8 sq unit

Explanation

To find the area of the region bounded by the lines \( y = 2x + 1 \), \( y = 3x + 1 \), and \( x = 4 \), we will follow these steps:

### Step 1: Find the Intersection Points
We need to find the points where the lines intersect.

1. **Intersection of \( y = 2x + 1 \) and \( y = 3x + 1 \)**:
\[ 2x + 1 = 3x + 1 \]
Subtract \( 2x \) from both sides:
\[ 1 = x + 1 \]
Subtract 1 from both sides:
\[ x = 0 \]
Now substitute \( x = 0 \) back into either equation to find \( y \):
\[ y = 2(0) + 1 = 1 \]
So, the first intersection point is \( (0, 1) \).

2. **Intersection of \( y = 2x + 1 \) and \( x = 4 \)**:
Substitute \( x = 4 \) into \( y = 2x + 1 \):
\[ y = 2(4) + 1 = 8 + 1 = 9 \]
So, the second intersection point is \( (4, 9) \).

3. **Intersection of \( y = 3x + 1 \) and \( x = 4 \)**:
Substitute \( x = 4 \) into \( y = 3x + 1 \):
\[ y = 3(4) + 1 = 12 + 1 = 13 \]
So, the third intersection point is \( (4, 13) \).

### Step 2: Identify the Vertices of the Triangle
The vertices of the triangle formed by the intersection points are:
- \( A(0, 1) \)
- \( B(4, 9) \)
- \( C(4, 13) \)

### Step 3: Calculate the Area of the Triangle
We can use the formula for the area of a triangle given vertices \( (x_1, y_1) \), \( (x_2, y_2) \), \( (x_3, y_3) \):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates:
- \( (x_1, y_1) = (0, 1) \)
- \( (x_2, y_2) = (4, 9) \)
- \( (x_3, y_3) = (4, 13) \)

Now, plug in the values:
\[ \text{Area} = \frac{1}{2} \left| 0(9 - 13) + 4(13 - 1) + 4(1 - 9) \right| \]
Calculating each term:
\[ = \frac{1}{2} \left| 0 + 4(12) + 4(-8) \right| \]
\[ = \frac{1}{2} \left| 48 - 32 \right| \]
\[ = \frac{1}{2} \left| 16 \right| = \frac{16}{2} = 8 \]

Thus, the area of the region bounded by the lines is \( 8 \) square units.