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Consider the line y=-x-3.

Find the equation of the line that is perpendicular to this line and passes through the point (6, 5).
Find the equation of the line that is parallel to this line and passes through the point (6, 5).

User BernatC
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2 Answers

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Final answer:

The equation of the line perpendicular to y = -x - 3 and passing through (6, 5) is y = x - 1. The equation of the line parallel to y = -x - 3 and passing through (6, 5) is y = -x + 11.

Step-by-step explanation:

To find the equation of a line that is perpendicular to the given line y = -x - 3, we first determine the slope of the given line, which is -1. The slope of any line perpendicular to this would be the negative reciprocal, which is 1. Using the point (6, 5) and the slope 1, we use the point-slope form to find the equation of the perpendicular line: y - 5 = 1(x - 6), simplifying to y = x - 1.

For the parallel line, we require a line with the same slope as the given line, which is -1. Using the point (6, 5) again, we write the point-slope form: y - 5 = -1(x - 6), simplifying to y = -x + 11 for the equation of the line parallel to the given line.

User Tharsan
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Answer:

perp: 1

y - 5 = x -6

y = x - 1

parallel: -1

y - 5 = -(x - 6)

y - 5 = -x + 6

y = -x + 11

User Nicholas Cardot
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