find the equation of the line passing through (4,1) and perpendicular to the line 3x+2y-6=0

Question

find the equation of the line passing through (4,1) and perpendicular to the line 3x+2y-6=0

Answer 1

The equation of a line is y = mx + b, it can be written as Ax+By+C=0

m is the slope of the line and it is -A/B

3x+2y-6=0

m = -3/2

 

If 2 lines are perpendicular, then the multiplication of their slopes is -1. Let's suppose the new line has slope m1

 

m*m1 = -1

(-3/2)*m1 = -1

m1 = 2/3

 

Now, we can replace the slope gotten and the point given (4,1) to find the equation of the line.

y = mx + b

1 = 4m1 + b

1 = 4(2/3) + b

b = -5/3

The equation is:

y = (2/3)x - 5/3