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Find the equation of the line perpendicular to y = -2/5x + 3 and passing through the point (7, -1).

User Wahkuna
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1 Answer

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Final answer:

The equation of the line perpendicular to y = -2/5x + 3 and passing through the point (7, -1) is y = 5/2x - 37/2.

Step-by-step explanation:

The student is asking for the equation of a line that is perpendicular to a given line and passes through a specific point. The given line is y = -2/5x + 3. Since the slope of this line is -2/5, the slope of the perpendicular line must be the negative reciprocal of this value, which is 5/2. Now, using the point-slope form of the equation of a line, which is y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point through which the line passes, we have:

Point: (7, -1)Slope: 5/2

Plugging these values into the point-slope form yields:

y - (-1) = 5/2(x - 7)

We simplify this to get the slope-intercept form y = mx + b:

y = 5/2x - 35/2 - 1

y = 5/2x - 37/2

Therefore, the equation of the line perpendicular to y = -2/5x + 3 and passing through the point (7, -1) is y = 5/2x - 37/2.

User Ramankingdom
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