If the area of the triangle formed by the pair of lines `8x^2 - 6xy + y^2 = 0` and the line `2x+3y=a` is 7 then `a=`

Options

A. 14
B. `14sqrt2`
C. 28
D. none of these

Answer

To solve the problem, we need to find the value of \( a \) such that the area of the triangle formed by the lines represented by the equation \( 8x^2 - 6xy + y^2 = 0 \) and the line \( 2x + 3y = a \) is equal to 7.

### Step-by-step Solution:

1. **Identify the pair of lines from the quadratic equation:**
The equation \( 8x^2 - 6xy + y^2 = 0 \) represents a pair of straight lines. We can factor this equation to find the individual lines.

\[ 8x^2 - 6xy + y^2 = 0 \]

We can rewrite this as:

\[ (4x - y)(2x - y) = 0 \]

Thus, the two lines are:
\[ L_1: 4x - y = 0 \quad \text{(or } y = 4x\text{)} \]
\[ L_2: 2x - y = 0 \quad \text{(or } y = 2x\text{)} \]

2. **Write down the third line:**
The third line given in the problem is:
\[ L_3: 2x + 3y = a \]

3. **Find the intersection points of the lines:**
We will find the points of intersection of the lines \( L_1 \), \( L_2 \), and \( L_3 \).

- **Intersection of \( L_1 \) and \( L_2 \):**
Setting \( y = 4x \) in \( y = 2x \):
\[ 4x = 2x \implies 2x = 0 \implies x = 0 \implies y = 0 \]
Thus, the first point of intersection is \( (0, 0) \).

- **Intersection of \( L_2 \) and \( L_3 \):**
Substitute \( y = 2x \) into \( 2x + 3y = a \):
\[ 2x + 3(2x) = a \implies 2x + 6x = a \implies 8x = a \implies x = \frac{a}{8} \]
Then, substituting back for \( y \):
\[ y = 2\left(\frac{a}{8}\right) = \frac{a}{4} \]
Thus, the second point of intersection is \( \left(\frac{a}{8}, \frac{a}{4}\right) \).

- **Intersection of \( L_1 \) and \( L_3 \):**
Substitute \( y = 4x \) into \( 2x + 3y = a \):
\[ 2x + 3(4x) = a \implies 2x + 12x = a \implies 14x = a \implies x = \frac{a}{14} \]
Then, substituting back for \( y \):
\[ y = 4\left(\frac{a}{14}\right) = \frac{2a}{7} \]
Thus, the third point of intersection is \( \left(\frac{a}{14}, \frac{2a}{7}\right) \).

4. **Calculate the area of the triangle:**
The area \( A \) of the triangle formed by the points \( (0, 0) \), \( \left(\frac{a}{8}, \frac{a}{4}\right) \), and \( \left(\frac{a}{14}, \frac{2a}{7}\right) \) can be calculated using the 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:
\[ A = \frac{1}{2} \left| 0\left(\frac{a}{4} - \frac{2a}{7}\right) + \frac{a}{8}\left(\frac{2a}{7} - 0\right) + \frac{a}{14}\left(0 - \frac{a}{4}\right) \right| \]
Simplifying, we get:
\[ A = \frac{1}{2} \left| \frac{2a^2}{56} - \frac{a^2}{56} \right| = \frac{1}{2} \left| \frac{a^2}{56} \right| = \frac{a^2}{112} \]

5. **Set the area equal to 7:**
We know the area is 7, so:
\[ \frac{a^2}{112} = 7 \]
Multiplying both sides by 112:
\[ a^2 = 7 \times 112 = 784 \]
Taking the square root:
\[ a = \sqrt{784} = 28 \]

### Final Answer:
Thus, the value of \( a \) is \( 28 \).