If the lines l_1x + m_1y + n_1 = 0 and l_2x + | vedclass

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(A) The line $l_1x + m_1y + n_1 = 0$ intersects the $x$-axis at $A \equiv \left( -\frac{n_1}{l_1}, 0 \right)$ and the $y$-axis at $B \equiv \left( 0,

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$l_1l_2 = m_1m_2$

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(A) The line $l_1x + m_1y + n_1 = 0$ intersects the $x$-axis at $A \equiv \left( -\frac{n_1}{l_1}, 0 \right)$ and the $y$-axis at $B \equiv \left( 0, -\frac{n_1}{m_1} \right)$.The line $l_2x + m_2y + n_2 = 0$ intersects the $x$-axis at $C \equiv \left( -\frac{n_2}{l_2}, 0 \right)$ and the $y$-axis at $D \equiv \left( 0, -\frac{n_2}{m_2} \right)$.Since the points $A, B, C, D$ are concyclic,the power of the point $O(0,0)$ with respect to the circle passing through these points must be consistent. Specifically,since $AC$ and $BD$ are chords intersecting at the origin $O$,by the intersecting chords theorem,we have $OA \cdot OC = OB \cdot OD$.Substituting the coordinates:$\left| -\frac{n_1}{l_1} \right| \cdot \left| -\frac{n_2}{l_2} \right| = \left| -\frac{n_1}{m_1} \right| \cdot \left| -\frac{n_2}{m_2} \right|$$\left| \frac{n_1n_2}{l_1l_2} \right| = \left| \frac{n_1n_2}{m_1m_2} \right|$This implies $|l_1l_2| = |m_1m_2|$,which simplifies to $l_1l_2 = m_1m_2$ (assuming the signs are consistent with the geometry of the intercepts).Thus,the correct option is $A$.

If the lines $l_1x + m_1y + n_1 = 0$ and $l_2x + m_2y + n_2 = 0$ cut the axes at concyclic points,then

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