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What is the equation of a line that is perpendicular to the line y = 3x + 1 and passes through the point (9, 6)?

A. y = -1/3x + 9
B. y = 3x + 24
C. y = -1/3x + 11
D. y = 3x + 33

User Croll
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Final answer:

The equation of the line that is perpendicular to y = 3x + 1 and passes through (9, 6) is A. y = -1/3x + 9, which is found using the negative reciprocal slope of the given line and applying the point-slope form with the given point.

Step-by-step explanation:

The question involves finding the equation of a line that is perpendicular to another line and passes through a given point. The slope of any line perpendicular to the line y = 3x + 1 would be the negative reciprocal of 3, which is -1/3, because perpendicular lines have slopes that are negative reciprocals of each other. Since the line must pass through the point (9, 6), we can use the point-slope form to determine the equation of the line, which is y - y1 = m(x - x1), substituting in m = -1/3, and the given point (9, 6).

Our equation becomes: y - 6 = -1/3(x - 9). To find the slope-intercept form, we'll simplify the equation by distributing the slope and moving the 6 over to the other side:

y - 6 = -1/3x + 3

y = -1/3x + 3 + 6

y = -1/3x + 9

Therefore, the correct answer is A. y = -1/3x + 9.

User Psorensen
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