1 Answer

Pablo Abdelhay · Answer 1 · 2020-07-30T04:09:32+0000

For this case we have that by definition, the equation of a line in the slope-intersection form is given by:

y = mx + b

Where:

m: It's the slope

b: It is the cut-off point with the y axis

On the other hand we have that if two lines are perpendicular, then the product of their slopes is -1. So:

$m_ {1} * m_ {2} = - 1$

The given line is:
$3x-4y = 17\\-4y = 17-3x\\4y = 3x-17\\y = \frac {3} {4} x- \frac {17} {4}$

So we have:

$m = \frac {3} {4}$

We find
$m_ {2}$ :

$m_ {2} = \frac {-1} {\frac {3} {4}}\\m = - \frac {4} {3}$

So, a line perpendicular to the one given is of the form:

$y = - \frac {4} {3} x + b$

We substitute the given point to find "b":
$4 = - \frac {4} {3} (6) + b\\4 = -8 + b\\4 + 8 = b\\b = 12$

Finally, the line is:

$y = - \frac {4} {3} x + 12$

Answer:

$y = - \frac {4} {3} x + 12$

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