Given the four lines with equations x + 2y = | vedclass

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(D) The equations of the lines are:$L_1: x + 2y - 3 = 0$$L_2: 3x + 4y - 7 = 0$$L_3: 2x + 3y - 4 = 0$$L_4: 4x + 5y - 6 = 0$First,check for parallel lin

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None of these

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(D) The equations of the lines are:$L_1: x + 2y - 3 = 0$$L_2: 3x + 4y - 7 = 0$$L_3: 2x + 3y - 4 = 0$$L_4: 4x + 5y - 6 = 0$First,check for parallel lines by comparing slopes $(m = -A/B)$:$m_1 = -1/2, m_2 = -3/4, m_3 = -2/3, m_4 = -4/5$.Since no slopes are equal,no two lines are parallel. Thus,they cannot form a rectangle.Next,check for concurrency for any three lines. For $L_1, L_2, L_3$:$\Delta = \begin{vmatrix} 1 & 2 & -3 \ 3 & 4 & -7 \ 2 & 3 & -4 \end{vmatrix} = 1(-16 + 21) - 2(-12 + 14) - 3(9 - 8) = 5 - 4 - 3 = -2 
eq 0$.Since the determinant is not zero,the lines are not concurrent.Therefore,the correct option is $D$.

Given the four lines with equations $x + 2y = 3,$ $3x + 4y = 7,$ $2x + 3y = 4,$ and $4x + 5y = 6,$ these lines are:

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