How do you find the equation of a line that is perpendicular to the line defined by the equation xy = 7 and passes through the point (1, 1)?

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To find the equation of a line that is perpendicular to the given line xy = 7 and passes through the point (1, 1), we can follow these steps:

  1. Determine the slope of the given line: First, we need to rewrite the equation xy = 7 in a more familiar slope-intercept form (y = mx + b). We can express y in terms of x:

    2. y = 7/x

  2. Finding the slope: To find the slope at any point along this curve, we can differentiate y with respect to x. Using implicit differentiation, we get:

    3. d/dx(xy) = 0
    From the product rule, we have y + x(dy/dx) = 0 which gives us dy/dx = -y/x.

  3. Calculate the slope at the point (1, 1): Substituting x = 1 and y = 1 into the slope formula:

    4. m = -1/1 = -1.

  4. Finding the slope of the perpendicular line: The slope of a line that is perpendicular to another line is the negative reciprocal of the original slope. Thus:

    5. Perpendicular slope = 1 (since the negative reciprocal of -1 is 1).

  5. Using the point-slope form of the line: Now that we have the slope of the perpendicular line and a point through which it passes, we can use the point-slope form of the equation of a line y – y1 = m(x – x1):

    6. y – 1 = 1(x – 1)

  6. Simplifying the equation: Distributing and simplifying this results in:

    7. y – 1 = x – 1
    y = x.

In conclusion, the equation of the line that is perpendicular to xy = 7 and passes through the point (1, 1) is:

y = x

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