The area of triangle formed by the straight lines whose equations are ` y= 4x + 2, 2y = x + 3 ` and `x=0` is :

Options

A. `(25)/(7sqrt(2))`
B. `(sqrt(2))/(28)`
C. `(1)/(28)`
D. `(15)/(7)`

Answer

To find the area of the triangle formed by the lines \( y = 4x + 2 \), \( 2y = x + 3 \), and \( x = 0 \), we will follow these steps:

### Step 1: Find the intersection points of the lines
1. **Find the intersection point of the lines \( y = 4x + 2 \) and \( 2y = x + 3 \)**.
- Substitute \( y = 4x + 2 \) into \( 2y = x + 3 \):
\[ 2(4x + 2) = x + 3 \]
\[ 8x + 4 = x + 3 \]
\[ 7x = -1 \implies x = -\frac{1}{7} \]
- Substitute \( x = -\frac{1}{7} \) back into \( y = 4x + 2 \):
\[ y = 4\left(-\frac{1}{7}\right) + 2 = -\frac{4}{7} + 2 = \frac{10}{7} \]
- Thus, the intersection point \( A \) is \( \left(-\frac{1}{7}, \frac{10}{7}\right) \).

2. **Find the intersection point of the lines \( 2y = x + 3 \) and \( x = 0 \)**.
- Substitute \( x = 0 \) into \( 2y = x + 3 \):
\[ 2y = 0 + 3 \implies y = \frac{3}{2} \]
- Thus, the intersection point \( B \) is \( (0, \frac{3}{2}) \).

3. **Find the intersection point of the lines \( y = 4x + 2 \) and \( x = 0 \)**.
- Substitute \( x = 0 \) into \( y = 4x + 2 \):
\[ y = 4(0) + 2 = 2 \]
- Thus, the intersection point \( C \) is \( (0, 2) \).

### Step 2: Calculate the area of the triangle
We have the vertices of the triangle:
- \( A \left(-\frac{1}{7}, \frac{10}{7}\right) \)
- \( B (0, \frac{3}{2}) \)
- \( C (0, 2) \)

Using the formula for the area of a triangle given by 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:
\[ \text{Area} = \frac{1}{2} \left| -\frac{1}{7}\left(\frac{3}{2} - 2\right) + 0\left(2 - \frac{10}{7}\right) + 0\left(\frac{10}{7} - \frac{3}{2}\right) \right| \]
Calculating the expression:
\[ = \frac{1}{2} \left| -\frac{1}{7}\left(-\frac{1}{2}\right) \right| = \frac{1}{2} \left| \frac{1}{14} \right| = \frac{1}{28} \]

### Final Answer
The area of the triangle is \( \frac{1}{28} \) square units.

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