The three lines `l_(1)=4x-3y+2=0,l_(2)=3x+4y-4=0` and `l_(3)=x-7y+6=0`

Options

A. form a rt. angled isosceles triangle
B. form a rt. Angled triangle
C. are concurrent
D. None

Answer

To determine whether the three lines \( l_1 = 4x - 3y + 2 = 0 \), \( l_2 = 3x + 4y - 4 = 0 \), and \( l_3 = x - 7y + 6 = 0 \) are concurrent or not, we can use the determinant method. If the determinant is equal to zero, then the lines are concurrent; otherwise, they are not.

### Step-by-Step Solution:

1. **Identify the coefficients of the lines:**
- For line \( l_1: 4x - 3y + 2 = 0 \) → \( a_1 = 4, b_1 = -3, c_1 = 2 \)
- For line \( l_2: 3x + 4y - 4 = 0 \) → \( a_2 = 3, b_2 = 4, c_2 = -4 \)
- For line \( l_3: x - 7y + 6 = 0 \) → \( a_3 = 1, b_3 = -7, c_3 = 6 \)

2. **Set up the determinant:**
The determinant \( \Delta \) is given by:
\[ \Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = \begin{vmatrix} 4 & -3 & 2 \\ 3 & 4 & -4 \\ 1 & -7 & 6 \end{vmatrix} \]

3. **Calculate the determinant:**
We can expand the determinant using the first row:
\[ \Delta = 4 \begin{vmatrix} 4 & -4 \\ -7 & 6 \end{vmatrix} - (-3) \begin{vmatrix} 3 & -4 \\ 1 & 6 \end{vmatrix} + 2 \begin{vmatrix} 3 & 4 \\ 1 & -7 \end{vmatrix} \]

Now calculating each of the 2x2 determinants:
- \( \begin{vmatrix} 4 & -4 \\ -7 & 6 \end{vmatrix} = (4)(6) - (-4)(-7) = 24 - 28 = -4 \)
- \( \begin{vmatrix} 3 & -4 \\ 1 & 6 \end{vmatrix} = (3)(6) - (-4)(1) = 18 + 4 = 22 \)
- \( \begin{vmatrix} 3 & 4 \\ 1 & -7 \end{vmatrix} = (3)(-7) - (4)(1) = -21 - 4 = -25 \)

Substituting back into the determinant:
\[ \Delta = 4(-4) + 3(22) + 2(-25) = -16 + 66 - 50 = 0 \]

4. **Conclusion:**
Since \( \Delta = 0 \), the lines are concurrent and do not form a triangle.

### Final Answer:
The lines are concurrent and do not form a triangle.

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