Question

what is the equation of a line that passes through point (6, 3) and is perpendicular to a line with a slope of -3/2?

Answer 1

For this case we have that by definition, the slope point equation of a line is given by:

$y = mx + b$

Where:

m: It's the slope

b: It is the cutoff point with the y axis

By definition, if two lines are perpendicular, the product of their slopes is -1. That is to say:

$m_ {1} * m_ {2} = - 1\\If\ it\ tells\ us: m_ {1} = - \frac {3} {2}:\\- \frac {3} {2} * m_ {2} = - 1\\m_ {2} = \frac {2} {3}$

Substituting:

$y = \frac {2} {3} x + b$

We substitute the point to find "b":

$3 = \frac {2} {3} 6 + b\\3 = 4 + b\\b = 3-4\\b = -1$

Finally:

$y = \frac {2} {3} x-1$

Answer:

$y = \frac {2} {3} x-1$