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What is the equation of the line perpendicular to 4x+5y=11 that passes through the point (-4 -2)?

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Final answer:

The equation of the line perpendicular to 4x+5y=11 that passes through the point (-4, -2) is y = 5/4x + 3.

Step-by-step explanation:

To find the equation of the line perpendicular to 4x+5y=11 and passing through the point (-4, -2), we first need to determine the slope of the given line. The slope-intercept form of a line is y = mx + b, where m is the slope. For the given equation, 4x + 5y = 11, we can rewrite it in slope-intercept form:

5y = -4x + 11

y = (-4/5)x + 11/5

Therefore, the slope of the given line is -4/5. Perpendicular lines have slopes that are negative reciprocals of each other so the slope of the line we are seeking would be 5/4. Using the point-slope form of a line, which is y - y1 = m(x - x1), where (x1, y1) is the point the line passes through and m is the slope, we get:

y - (-2) = 5/4(x - (-4))

y + 2 = 5/4(x + 4)

The equation of the perpendicular line in point-slope form is y + 2 = 5/4(x + 4). To write this in slope-intercept form, we distribute and isolate y:

y = 5/4x + 5 - 2

y = 5/4x + 3

So, the slope-intercept form of the equation for the perpendicular line is y = 5/4x + 3.

User Mark Kennedy
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