the equations of three lines are given below.\nline 1: ( 6x + 10y = 4 )\nline 2: ( y=\frac{5}{3}x - 7…

{5}{3}=-1) → perpendicular.

• Line 1 ((m =-\frac{3}{5})) and Line 3 ((m=\frac{5}{3})): (-\frac{3}{5}\times\frac{5}{3}=-1) → perpendicular (wrong, no! Wait, no, Line 3 is (y=\frac{5}{3}x+\frac{4}{3}), Line 2 is (y=\frac{5}{3}x - 7). So Line 2 and Line 3 are parallel (same slope). Line 1: (y=-\frac{3}{5}x+\frac{2}{5}). The formula: For Line 1 ((A_1 = 6), (B_1 = 10)) and Line 2 ((m_2=\frac{5}{3})): The slope of Line 1 is (m_1=-\frac{A_1}{B_1}=-\frac{6}{10}=-\frac{3}{5}). (m_1\times m_2=-\frac{3}{5}\times\frac{5}{3}=-1) (Line 1 - Line 2: perpendicular). For Line 1 ((m_1=-\frac{3}{5})) and Line 3 ((m_3=\frac{5}{3})): (m_1\times m_3=-1) (perpendicular. But no, there's a mistake. Wait, no: The problem is in the perception. Line 3: (3y = 5x + 4) → (y=\frac{5}{3}x+\frac{4}{3}), Line 2: (y=\frac{5}{3}x - 7). So Line 2 and Line 3 are parallel (same slope (\frac{5}{3})). Line 1: (y=-\frac{3}{5}x+\frac{2}{5}). (m_1=-\frac{3}{5}), (m_2=\frac{5}{3}), (m_1\times m_2=-1) (Line 1 - Line 2: perpendicular). (m_1=-\frac{3}{5}), (m_3=\frac{5}{3}), (m_1\times m_3=-1) (Line 1 - Line 3: perpendicular. But no, in the problem's context (maybe a typo in the problem, but assuming the equations are correct):
• Line 1: (y =-\frac{3}{5}x+\frac{2}{5})
• Line 2: (y=\frac{5}{3}x - 7)
• Line 3: (y=\frac{5}{3}x+\frac{4}{3})

So:

• Line 1 and Line 2: Perpendicular ( (m_1\times m_2=-1)).
• Line 1 and Line 3: Perpendicular ( (m_1\times m_3=-1)) → no, wait, no! Line 2 and Line 3 are parallel ((m_2 = m_3)). So if Line 1 is perpendicular to Line 2, and Line 2 is parallel to Line 3, then Line 1 is perpendicular to Line 3. But let's re - check: (m_1=-\frac{3}{5}), (m_3=\frac{5}{3}), (m_1\times m_3=-\frac{3}{5}\times\frac{5}{3}=-1) (yes, perpendicular). But in the original problem's table, maybe it's a trick. Wait, no: The standard rules: Two lines (L_1:y = m_1x + b_1) and (L_2:y = m_2x + b_2)
• Parallel: (m_1 = m_2)
• Perpendicular: (m_1\times m_2=-1)

Line 1: (y=-\frac{3}{5}x+\frac{2}{5}) ((m_1 =-\frac{3}{5})) Line 2: (y=\frac{5}{3}x - 7) ((m_2=\frac{5}{3})) Line 3: (y=\frac{5}{3}x+\frac{4}{3}) ((m_3=\frac{5}{3}))

• Line 1 and Line 2: (m_1\times m_2=-\frac{3}{5}\times\frac{5}{3}=-1) → Perpendicular.
• Line 1 and Line 3: (m_1\times m_3=-\frac{3}{5}\times\frac{5}{3}=-1) → Perpendicular.
• Line 2 and Line 3: (m_2 = m_3=\frac{5}{3}) → Parallel.

But looking back at the problem's initial equations: Line 1: (6x + 10y=4) (correctly rewritten as (y=-\frac{3}{5}x+\frac{2}{5})) Line 2: (y=\frac{5}{3}x - 7) (correct) Line 3: (3y = 5x + 4) (correctly rewritten as (y=\frac{5}{3}x+\frac{4}{3}))

So the correct answers are: Line 1 and Line 2: Perpendicular Line 1 and Line 3: Perpendicular Line 2 and Line 3: Parallel

But maybe there was a miscalculation in the first thought. Wait, no: Another way: The general form of a line