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Find the equation of the line that passes through the point (10, -13) and is perpendicular to the line with the equation (x - 5y)/4 = (x - 1)/2.

User Jamesdlin
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Final answer:

To find the equation of the line perpendicular to the given line and passing through the point (10, -13), calculate the negative reciprocal of the given line’s slope and use the point-slope form. The equation is y = -5x + 37.

Step-by-step explanation:

The student is asking for the equation of a line that is perpendicular to another line, and that passes through a given point (10, -13). First, we need to find the slope of the given line by rearranging its equation into slope-intercept form (y = mx + b). The equation given is (x - 5y)/4 = (x - 1)/2. Solving for y, we get:

  • 2(x - 5y) = 4(x - 1)
  • 2x - 10y = 4x - 4
  • 10y = 2x - 4
  • y = (2x/10) - (4/10)
  • y = (1/5)x - 2/5

The slope of this line is 1/5. Perpendicular lines have slopes that are negative reciprocals. Therefore, the slope of the perpendicular line is -5 (the negative reciprocal of 1/5). Using the point-slope form of the equation of a line, which is (y - y1) = m(x - x1), and the point (10, -13), the equation of the line is:

y - (-13) = -5(x - 10)

Simplifying this, we get:

y + 13 = -5x + 50

y = -5x + 37

This is the equation of the line that is perpendicular to the given line and passes through the point (10, -13).

User Dothem
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