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 formed by three lines given in the general form \(Ax + By + C = 0\).

### Step-by-Step Solution:

1. **Identify the coefficients of the lines:**
- For the first line \(7x - 2y + 10 = 0\):
- \(A_1 = 7\), \(B_1 = -2\), \(C_1 = 10\)
- For the second line \(7x + 2y - 10 = 0\):
- \(A_2 = 7\), \(B_2 = 2\), \(C_2 = -10\)
- For the third line \(9x + y + 2 = 0\):
- \(A_3 = 9\), \(B_3 = 1\), \(C_3 = 2\)

2. **Use the area formula for the triangle formed by three lines:**
\[ \text{Area} = \frac{1}{2} \left| \frac{C_1(A_2B_3 - A_3B_2) + C_2(A_3B_1 - A_1B_3) + C_3(A_1B_2 - A_2B_1)}{A_1B_2 - A_2B_1} \right| \]

3. **Calculate the determinant in the numerator:**
- First term: \(C_1(A_2B_3 - A_3B_2) = 10(7 \cdot 1 - 9 \cdot 2) = 10(7 - 18) = 10(-11) = -110\)
- Second term: \(C_2(A_3B_1 - A_1B_3) = -10(9 \cdot -2 - 7 \cdot 1) = -10(-18 - 7) = -10(-25) = 250\)
- Third term: \(C_3(A_1B_2 - A_2B_1) = 2(7 \cdot 2 - 7 \cdot -2) = 2(14 + 14) = 2(28) = 56\)

Now, summing these:
\[ -110 + 250 + 56 = 196 \]

4. **Calculate the denominator:**
\[ A_1B_2 - A_2B_1 = 7 \cdot 2 - 7 \cdot -2 = 14 + 14 = 28 \]

5. **Substitute back into the area formula:**
\[ \text{Area} = \frac{1}{2} \left| \frac{196}{28} \right| = \frac{1}{2} \cdot 7 = \frac{7}{2} \]

6. **Final area calculation:**
\[ \text{Area} = \frac{7}{2} \text{ square units} \]

### Final Answer:
The area of the triangle formed by the given lines is \(\frac{7}{2}\) square units.