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//4`
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: Factor the equation of the pair of straight lines
The equation \(y^2 - 9xy + 18x^2 = 0\) can be factored. We can rewrite it as:
\[ y^2 - 6xy - 3xy + 18x^2 = 0 \]
Now, we can factor by grouping:
\[ y(y - 6x) - 3x(y - 6x) = 0 \]
This gives us:
\[ (y - 3x)(y - 6x) = 0 \]
Thus, the lines are:
\[ y = 3x \quad \text{and} \quad y = 6x \]

### Step 2: Find the points of intersection with the line \(y = 9\)
Next, we find the points where these lines intersect the line \(y = 9\).

1. For \(y = 3x\):
\[ 9 = 3x \implies x = 3 \]
So, the point of intersection is \(A(3, 9)\).

2. For \(y = 6x\):
\[ 9 = 6x \implies x = \frac{3}{2} \]
So, the point of intersection is \(B\left(\frac{3}{2}, 9\right)\).

### Step 3: Identify the vertices of the triangle
The vertices of the triangle formed by the lines and the x-axis are:
- \(O(0, 0)\) (the origin)
- \(A(3, 9)\)
- \(B\left(\frac{3}{2}, 9\right)\)

### Step 4: Use the formula for the area of a triangle
The area \(A\) of a triangle formed by the points \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be calculated using the determinant 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 points \(O(0, 0)\), \(A(3, 9)\), and \(B\left(\frac{3}{2}, 9\right)\):
\[ A = \frac{1}{2} \left| 0(9 - 9) + 3(9 - 0) + \frac{3}{2}(0 - 9) \right| \]
This simplifies to:
\[ A = \frac{1}{2} \left| 0 + 27 - \frac{27}{2} \right| \]
Calculating further:
\[ A = \frac{1}{2} \left| 27 - 13.5 \right| = \frac{1}{2} \left| 13.5 \right| = \frac{13.5}{2} = 6.75 \]

### Step 5: Final calculation of the area
To express this in fractional form:
\[ 6.75 = \frac{27}{4} \]

Thus, the area of the triangle formed by the lines is:
\[ \boxed{\frac{27}{4}} \]