For what value of k are the graphs of 12y=−3x+8 and 6y=kr−5 parallel? For what value of k are they perpendicular?

akuzativo617

akuzativo617

Answered question

2022-11-15

For what value of k are the graphs of 12y=−3x+8 and 6y=kr−5 parallel? For what value of k are they perpendicular?

Answer & Explanation

Deanna Sweeney

Deanna Sweeney

Beginner2022-11-16Added 14 answers

the given equations are 12y=−3x+8 and 6y=kx−5
Writing them first in Slope-Intercept forms
y = - 3 12 x + 8 12 and y = k 6 x - 5 6
If they are parallel then slopes are equal,...therefore
- 3 12 = k 6
and k=-3/2
If they are perpendicular then the product of the slope =−1
- 3 12 k 6 = - 1
and k = 24

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