Consider the lines y = 3x , y = 6x , y = 9
What is the area of the triangle formed by these lines ?

Options

A. `(27)/(4)` sq units
B. `(27)/(9)` sq units
C. `(19)/(4)` sq units
D. `(19)/(2)` sq units

Answer

To find the area of the triangle formed by the lines \(y = 3x\), \(y = 6x\), and \(y = 9\), we will follow these steps:

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

1. **Intersection of \(y = 3x\) and \(y = 6x\)**:
- Set \(3x = 6x\).
- This gives \(0 = 3x\), so \(x = 0\).
- Substitute \(x = 0\) into \(y = 3x\) to find \(y\):
\[ y = 3(0) = 0 \]
- Thus, the intersection point \(A\) is \((0, 0)\).

2. **Intersection of \(y = 3x\) and \(y = 9\)**:
- Set \(3x = 9\).
- This gives \(x = 3\).
- Substitute \(x = 3\) into \(y = 3x\) to find \(y\):
\[ y = 3(3) = 9 \]
- Thus, the intersection point \(B\) is \((3, 9)\).

3. **Intersection of \(y = 6x\) and \(y = 9\)**:
- Set \(6x = 9\).
- This gives \(x = \frac{3}{2}\).
- Substitute \(x = \frac{3}{2}\) into \(y = 6x\) to find \(y\):
\[ y = 6\left(\frac{3}{2}\right) = 9 \]
- Thus, the intersection point \(C\) is \(\left(\frac{3}{2}, 9\right)\).

### Step 2: Calculate the area of the triangle formed by points \(A\), \(B\), and \(C\)

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

Substituting the points:
- \(A(0, 0)\) \(\Rightarrow (x_1, y_1) = (0, 0)\)
- \(B(3, 9)\) \(\Rightarrow (x_2, y_2) = (3, 9)\)
- \(C\left(\frac{3}{2}, 9\right)\) \(\Rightarrow (x_3, y_3) = \left(\frac{3}{2}, 9\right)\)

Substituting into the area formula:
\[ \text{Area} = \frac{1}{2} \left| 0(9 - 9) + 3(9 - 0) + \frac{3}{2}(0 - 9) \right| \]
\[ = \frac{1}{2} \left| 0 + 27 - \frac{27}{2} \right| \]
\[ = \frac{1}{2} \left| 27 - 13.5 \right| = \frac{1}{2} \left| 13.5 \right| = \frac{13.5}{2} = \frac{27}{4} \]

### Final Answer
Thus, the area of the triangle formed by the lines is \(\frac{27}{4}\).

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