What is the equation of the line that is perpendicular to y = 3x + 1 and passes through the point (2, 3)?

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Q: What is the equation of the line that is perpendicular to y = 3x + 1 and passes through the point (2, 3)?

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

Steps: 8

Step 1: Identify the slope of the given line. The equation is y = 3x + 1, so the slope (m) is 3.
Step 2: Find the slope of the line that is perpendicular to the given line. The perpendicular slope is the negative reciprocal of 3, which is -1/3.
Step 3: Use the point-slope form of the equation of a line, which is y - y1 = m(x - x1). Here, (x1, y1) is the point (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. Distribute -1/3: y - 3 = -1/3x + 2/3.
Step 6: Add 3 to both sides to solve for y: y = -1/3x + 2/3 + 3.
Step 7: Convert 3 to a fraction with a common denominator: 3 = 9/3. So, y = -1/3x + 2/3 + 9/3.
Step 8: Combine the fractions: y = -1/3x + 11/3.