Find the area of the triangular region bounded by the sides `y = 2x + 1, y = 3x + 1 and x = 4 `

Answer

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

### Step 1: Find the points of intersection of the lines.

1. **Intersection of \(y = 2x + 1\) and \(y = 3x + 1\)**:
\[ 2x + 1 = 3x + 1 \]
Rearranging gives:
\[ 2x - 3x = 1 - 1 \implies -x = 0 \implies x = 0 \]
Substituting \(x = 0\) into \(y = 2x + 1\):
\[ y = 2(0) + 1 = 1 \]
So, the intersection point is \(A(0, 1)\).

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

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

### Step 2: Identify the vertices of the triangle.

The vertices of the triangle are:
- \(A(0, 1)\)
- \(B(4, 13)\)
- \(C(4, 9)\)

### Step 3: Calculate the area of the triangle.

The area \(A\) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Substituting the coordinates of points \(A(0, 1)\), \(B(4, 13)\), and \(C(4, 9)\):
\[ A = \frac{1}{2} \left| 0(13 - 9) + 4(9 - 1) + 4(1 - 13) \right| \]
Calculating each term:
\[ = \frac{1}{2} \left| 0 + 4(8) + 4(-12) \right| \]
\[ = \frac{1}{2} \left| 32 - 48 \right| \]
\[ = \frac{1}{2} \left| -16 \right| = \frac{1}{2} \times 16 = 8 \]

### Final Answer:
The area of the triangular region is \(8\) square units.

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