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Line passing through (7,-9) and perpendicular to the line whose equation is x-8y-5=0

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Final answer:

To find the equation of the line passing through (7, -9) and perpendicular to x - 8y - 5 = 0, determine the negative reciprocal of the slope of the given line, and then use point-slope form to find the perpendicular line's equation, which is y = -8x + 47.

Step-by-step explanation:

To find the line that passes through the point (7, -9) and is perpendicular to the line whose equation is x - 8y - 5 = 0, we need to follow these steps:

  1. Rewrite the given equation in slope-intercept form, y = mx + b, to find the slope m.
  2. Since perpendicular lines have slopes that are negative reciprocals, calculate the negative reciprocal of the slope from step 1 to get the slope of the line we're looking for.
  3. Use the point-slope form, y - y1 = m(x - x1), with the slope from step 2 and the given point (7, -9) to write the equation of the line.

First, let's find the slope. Rearranging the given equation:

x - 8y - 5 = 0

-8y = -x + 5

y = 1/8x - 5/8

The slope of the given line is 1/8. The negative reciprocal of 1/8 is -8.

Now, use the point-slope form with the slope -8 and point (7, -9):

y - (-9) = -8(x - 7)

y + 9 = -8x + 56

The equation of the line perpendicular to the given line and passing through (7, -9) is y = -8x + 47.

User Diego Vieira
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