What is the equation of the line perpendicular to 3x + 2y = 8 passing through (6, -1)

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The equation of the line perpendicular to 3x + 2y = 8 and passing through (6, -1) has to be determined.

Use the property that the product of the slope of two perpendicular lines is equal to -1.

3x + 2y = 8 can be rewritten as `y = (-3/2)x...

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The equation of the line perpendicular to 3x + 2y = 8 and passing through (6, -1) has to be determined.

Use the property that the product of the slope of two perpendicular lines is equal to -1.

3x + 2y = 8 can be rewritten as `y = (-3/2)x + 4` . The slope of this line is -3/2. The slope of a line perpendicular to this line is `-3/2` .

A line that passes through (6, -1) and that has a slope `-3/2` is `(y +1)/(x - 6) = -3/2`

=> 2(y + 1) = -3(x - 6)

=> 2y + 2 = -3x + 18

=> 3x + 2y - 16 = 0

The equation of the required line is 3x + 2y - 16 = 0

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