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Find the equation of the line that is perpendicular to y=2/3x-13 and passes through the point (4,-1)

User Sorak
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Final answer:

The equation of the line that is perpendicular to y = 2/3x - 13 and passes through the point (4, -1) is y = -3/2x + 5.

Step-by-step explanation:

To find the equation of the line that is perpendicular to y = 2/3x - 13 and passes through the point (4, -1), we first need to determine the slope of the given line. The slope is 2/3, so the slope of the line perpendicular to it will be the negative reciprocal, which is -3/2. Next, we use the point-slope form of a line's equation, which is y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point the line passes through.

Substituting the known slope and point into the point-slope equation gives us:

y - (-1) = -3/2(x - 4)

Simplifying this equation, we get:

y + 1 = -3/2(x - 4)

y + 1 = -3/2x + 6

y = -3/2x + 5

Thus, the equation of the line perpendicular to y = 2/3x - 13 that passes through the point (4, -1) is y = -3/2x + 5.

User Mieliespoor
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