Asked by Jane
Please show me how to solve:
For what value of k are the graphs of 12y = -3x + 8 and 6y =kx - 5 parallel? For what value of k are they perpendicular?
For what value of k are the graphs of 12y = -3x + 8 and 6y =kx - 5 parallel? For what value of k are they perpendicular?
Answers
Answered by
Steve
if lines are parallel, they have the same slope
so, the first step is to convert each equation to the slope-intercept form:
y = -1/4 x + 2/3
y = k/6 x - 5/6
If the slopes are the same, then
-1/4 = k/6
k = -3/2
If the lines are perpendicular, their slopes are negative reciprocals; or, their product is -1.
-1/4 * k/6 = -1
-k/24 = -1
k = 24
so, the first step is to convert each equation to the slope-intercept form:
y = -1/4 x + 2/3
y = k/6 x - 5/6
If the slopes are the same, then
-1/4 = k/6
k = -3/2
If the lines are perpendicular, their slopes are negative reciprocals; or, their product is -1.
-1/4 * k/6 = -1
-k/24 = -1
k = 24
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