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 (1, 1)? (2022)

Options:

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

Exam Year: 2022

Solution:

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

What is the equation of the line that is perpendicular to y = 3x + 2 and passes through the point (1, 1)? (2022)

What is the equation of the line that is perpendicular to y = 3x + 2 and passes through the point (1, 1)? (2022)

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 a line equation, which is y - y1 = m(x - x1). Here, (x1, y1) is the point (1, 1) and m is -1/3.
Step 4: Substitute the values into the point-slope form: y - 1 = -1/3(x - 1).
Step 5: Simplify the equation. Distribute -1/3: y - 1 = -1/3x + 1/3.
Step 6: Add 1 to both sides to solve for y: y = -1/3x + 1/3 + 1.
Step 7: Combine the constants: 1/3 + 1 = 1/3 + 3/3 = 4/3. So, y = -1/3x + 4/3.

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