The lines are: B) Perpendicular
Step-by-step explanation:
we have to convert both lines in slope-intercept form to find slopes
So,
[tex]3x - 6y = -13\\-6y = -3x-13\\\frac{-6y}{-6} = \frac{-3x-13}{-6}\\\frac{-6y}{-6} = \frac{-3x}{-6}+\frac{-13}{-6}\\y = \frac{1}{2} +\frac{13}{6}[/tex]
Let m1 be the slope of first line
[tex]m_1 = \frac{1}{2}[/tex]
For the second line:
[tex]18x + 9y = 5\\9y = -18x+5\\\frac{9y}{9} = \frac{-18x+5}{9}\\\frac{9y}{9} = \frac{-18}{9}x+\frac{5}{9}\\y = -2x+\frac{5}{9}[/tex]
Let m2 be the slope of line 2
So,
If the lines are parallel, their slopes are equal
If the lines are perpendicular, product of their slopes is -1
We can see that
[tex]\frac{1}{2} * -2 = -1[/tex]
Hence,
The lines are: B) Perpendicular
Keywords: Slopes, Parallel lines
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