Area of the triangle formed by the lines `y^(2)-9xy+18x^(2)=0` and `y=9` is

Options

A. `27//4`
B. `0`
C. `9//3`
D. `27`

Answer

To find the area of the triangle formed by the lines given by the equation \(y^2 - 9xy + 18x^2 = 0\) and the line \(y = 9\), we can follow these steps:

### Step 1: Identify the lines from the quadratic equation
The equation \(y^2 - 9xy + 18x^2 = 0\) is a homogeneous quadratic equation in \(y\) and \(x\). We can factor it to find the slopes of the lines.

### Step 2: Factor the quadratic equation
We rewrite the equation as:
\[ y^2 - 9xy + 18x^2 = 0 \]
This can be factored as:
\[ (y - 3x)(y - 6x) = 0 \]
Thus, the two lines are:
\[ y = 3x \quad \text{and} \quad y = 6x \]

### Step 3: Find the points of intersection with the line \(y = 9\)
To find the points of intersection of these lines with \(y = 9\), we substitute \(y = 9\) into the equations of the lines.

1. For \(y = 3x\):
\[ 9 = 3x \implies x = 3 \implies \text{Point } A(3, 9) \]

2. For \(y = 6x\):
\[ 9 = 6x \implies x = \frac{3}{2} \implies \text{Point } B\left(\frac{3}{2}, 9\right) \]

### Step 4: Identify the third vertex of the triangle
The third vertex of the triangle is the origin, \(O(0, 0)\), since both lines pass through the origin.

### Step 5: Calculate the area of the triangle
The area \(A\) of the triangle formed by the points \(O(0, 0)\), \(A(3, 9)\), and \(B\left(\frac{3}{2}, 9\right)\) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base \(AB\) is the distance between points \(A\) and \(B\) and the height is the perpendicular distance from the origin to the line \(y = 9\).

### Step 6: Calculate the length of base \(AB\)
The distance \(AB\) can be calculated as:
\[ AB = \sqrt{\left(3 - \frac{3}{2}\right)^2 + (9 - 9)^2} = \sqrt{\left(\frac{3}{2}\right)^2} = \frac{3}{2} \]

### Step 7: Calculate the height
The height from the origin to the line \(y = 9\) is simply the y-coordinate of the line, which is 9.

### Step 8: Calculate the area
Now substituting the values into the area formula:
\[ \text{Area} = \frac{1}{2} \times AB \times \text{height} = \frac{1}{2} \times \frac{3}{2} \times 9 = \frac{27}{4} \]

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