Find the area of the triangle formed by the straight lines `7x-2y+10=0, 7x+2y-10=0` and `9x+y+2=0` (without sloving the vertices of the triangle).

Answer

To find the area of the triangle formed by the lines \(7x - 2y + 10 = 0\), \(7x + 2y - 10 = 0\), and \(9x + y + 2 = 0\) without solving for the vertices, we can use the formula for the area of a triangle given by three lines in the form \(Ax + By + C = 0\).

### Step-by-step Solution:

1. **Identify coefficients**:
For the lines given, we can identify the coefficients as follows:
- For the line \(7x - 2y + 10 = 0\):
- \(A_1 = 7\), \(B_1 = -2\), \(C_1 = 10\)
- For the line \(7x + 2y - 10 = 0\):
- \(A_2 = 7\), \(B_2 = 2\), \(C_2 = -10\)
- For the line \(9x + y + 2 = 0\):
- \(A_3 = 9\), \(B_3 = 1\), \(C_3 = 2\)

2. **Use the area formula**:
The area \(A\) of the triangle formed by these three lines can be calculated using the formula:
\[ A = \frac{1}{2} \left| \frac{C_1 A_2 B_3 + C_2 A_3 B_1 + C_3 A_1 B_2 - (A_1 B_2 C_3 + A_2 B_3 C_1 + A_3 B_1 C_2)}{A_1 B_2 - A_2 B_1} \right| \]

3. **Calculate the determinant**:
First, we need to calculate the determinant part:
\[ D = A_1 B_2 - A_2 B_1 = 7 \cdot 2 - 7 \cdot (-2) = 14 + 14 = 28 \]

4. **Calculate the numerator**:
Now, we compute the numerator:
\[ N = C_1 A_2 B_3 + C_2 A_3 B_1 + C_3 A_1 B_2 - (A_1 B_2 C_3 + A_2 B_3 C_1 + A_3 B_1 C_2) \]
- \(C_1 A_2 B_3 = 10 \cdot 7 \cdot 1 = 70\)
- \(C_2 A_3 B_1 = -10 \cdot 9 \cdot (-2) = 180\)
- \(C_3 A_1 B_2 = 2 \cdot 7 \cdot 2 = 28\)
- \(A_1 B_2 C_3 = 7 \cdot 2 \cdot 2 = 28\)
- \(A_2 B_3 C_1 = 7 \cdot 1 \cdot 10 = 70\)
- \(A_3 B_1 C_2 = 9 \cdot (-2) \cdot (-10) = 180\)

Now substituting these values into the numerator:
\[ N = 70 + 180 + 28 - (28 + 70 + 180) = 278 - 278 = 0 \]

5. **Calculate the area**:
Now substituting \(N\) and \(D\) into the area formula:
\[ A = \frac{1}{2} \left| \frac{0}{28} \right| = 0 \]

### Final Area:
The area of the triangle formed by the given lines is \(0\) square units, indicating that the lines are concurrent (they meet at a single point).