What is the equation of the line that is perpendicular to y = 3x + 2 and passes

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Question: What is the equation of the line that is perpendicular to y = 3x + 2 and passes through the point (2, 3)?

Options:

Correct Answer: y = -1/3x + 4

Solution:

The slope of the perpendicular line is -1/3. Using point-slope form: y - 3 = -1/3(x - 2) gives y = -1/3x + 4.

What is the equation of the line that is perpendicular to y = 3x + 2 and passes through the point (2, 3)?

What is the equation of the line that is perpendicular to y = 3x + 2 and passes through the point (2, 3)?

Step 1: Identify the slope of the given line y = 3x + 2. The slope (m) is 3.
Step 2: Find the slope of the line that is perpendicular to the given line. The slope of a perpendicular line is the negative reciprocal of the original slope. So, the negative reciprocal of 3 is -1/3.
Step 3: Use the point-slope form of the equation of a line, which is y - y1 = m(x - x1), where (x1, y1) is the point the line passes through. Here, (x1, y1) is (2, 3) and m is -1/3.
Step 4: Substitute the values into the point-slope form: y - 3 = -1/3(x - 2).
Step 5: Simplify the equation to find the slope-intercept form (y = mx + b). Start by distributing -1/3: y - 3 = -1/3x + 2/3.
Step 6: Add 3 to both sides to isolate y: y = -1/3x + 2/3 + 3.
Step 7: Convert 3 to a fraction with a common denominator: 3 = 9/3. Now add: y = -1/3x + 2/3 + 9/3 = -1/3x + 11/3.
Step 8: The final equation of the line is y = -1/3x + 11/3.

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