What is the equation of the line that is perpendicular to the line defined by the equation 2y = 3x + 1 and goes through the point (3, 2)?
options:
y = - 1/3 x + 3
y = -2/3x + 4
y = 1/3x – 1
y = 2x – 4
y = 6x + 5

In order to determine the perpendicular slope to the given line, we first need to determine the slope of the given line, which is not apparent in its current form. We need to solve the equation for y to determine the slope:

2y = 3x + 1 in slope-intercept form is

[tex]y=\frac{3}{2}x+\frac{1}{2}[/tex] . Therefore, the perpendicular slope, the opposite reciprocal of the given slope of 3/2 is -2/3. Passing through the point (3, 2):

[tex]y-2=-\frac{2}{3}(x-3)[/tex] and

[tex]y-2=-\frac{2}{3}x+2[/tex] so

[tex]y=-\frac{2}{3}x+4[/tex], the second choice down.

What is the equation of the line that is perpendicular to the line defined by the equation 2y = 3x + 1 and goes through the point (3, 2)? options: y = - 1/3 x + 3 y = -2/3x + 4 y = 1/3x – 1 y = 2x – 4 y = 6x + 5

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What is the equation of the line that is perpendicular to the line defined by the equation 2y = 3x + 1 and goes through the point (3, 2)?
options:
y = - 1/3 x + 3
y = -2/3x + 4
y = 1/3x – 1
y = 2x – 4
y = 6x + 5