Write the equation of a line that is perpendicular to the line 2x-3y= 0 and contains the point (3,2).?

Question

solve

Answer 1

the slope of the given line is -A/B = -2/-3 = 2/3

So the slope of the perpendicular line is -3/2

And since the point (3,2) needs to be there, just use the point-slope form to get the equation of the line:

y - 2 = -3/2 ( x-3)

If you wanted to convert to y-intercept:
y-2 = -3/2x +9/2
y = -3/2x + 9/2 + 4/2
y = -3/2x +13/2

Answer 2

Hello,

The slope of the first line = 2/3 y = 2/3x

So the slope of the line perpendicular is -(3/2) then
y = (-3/2)x + b then put in (3,2) and we have 2 = (-3/2)*3 + b so

2 = -9/2 + b and b = 6.5 so y = (-3/2)x + 6.5


Hope This Helps!