Area of triangle formed by straight lines `3x - y = 3, x - 2y + 4 =0 and y =0` is

Options

A. `(15)/(4)` sq. unit
B. `(15)/(2)` sq. unit
C. 15 sq. unit
D. Cannot be determined

Answer

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

### Step 1: Find the intersection points of the lines
We need to find the points where these lines intersect.

1. **Intersection of \(3x - y = 3\) and \(y = 0\)**:
- Substitute \(y = 0\) into \(3x - y = 3\):
\[ 3x - 0 = 3 \implies 3x = 3 \implies x = 1 \]
- So, the point is \(A(1, 0)\).

2. **Intersection of \(x - 2y + 4 = 0\) and \(y = 0\)**:
- Substitute \(y = 0\) into \(x - 2y + 4 = 0\):
\[ x - 0 + 4 = 0 \implies x + 4 = 0 \implies x = -4 \]
- So, the point is \(B(-4, 0)\).

3. **Intersection of \(3x - y = 3\) and \(x - 2y + 4 = 0\)**:
- From \(3x - y = 3\), we can express \(y\) in terms of \(x\):
\[ y = 3x - 3 \]
- Substitute \(y\) into \(x - 2y + 4 = 0\):
\[ x - 2(3x - 3) + 4 = 0 \implies x - 6x + 6 + 4 = 0 \implies -5x + 10 = 0 \implies 5x = 10 \implies x = 2 \]
- Now substitute \(x = 2\) back to find \(y\):
\[ y = 3(2) - 3 = 6 - 3 = 3 \]
- So, the point is \(C(2, 3)\).

### Step 2: Calculate the area of the triangle
The vertices of the triangle are \(A(1, 0)\), \(B(-4, 0)\), and \(C(2, 3)\).

Using the formula for the area of a triangle given vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates:
\[ \text{Area} = \frac{1}{2} \left| 1(0 - 3) + (-4)(3 - 0) + 2(0 - 0) \right| \]
\[ = \frac{1}{2} \left| 1(-3) + (-4)(3) + 0 \right| \]
\[ = \frac{1}{2} \left| -3 - 12 \right| = \frac{1}{2} \left| -15 \right| = \frac{15}{2} \]

Thus, the area of the triangle is \(\frac{15}{2}\) square units.

### Final Answer:
The area of the triangle formed by the given lines is \(\frac{15}{2}\) square units.

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