Find the equation of a line that is perpendicular to the line 2x-3y=9 and passes through the point (4,-1)

y = -3/2x + 5

Step-by-step explanation:

Step 1: find the slope

2x - 3y = 9

2x - 9 = 3y

Make y the subject of the formula by dividing both sides by 3

(2x - 9)/3 = 3y/3

( 2x - 9)/3 = y

y = 2x/3 - 9/3

y = 2x/3 - 3

m = 2x/3

Note: if two lines are perpendicular to the other, both lines are negative reciprocal of each other

m = -3/2x

Step 2 : using the point slope equation

y - y1 = m( x - x1)

( 4 , -1)

x1 = 4

y1 = -1

y -(-1) = m( x - 4)

y + 1 = -3/2( x - 4)

y + 1 = (-3x + 12)/2

y = (-3x + 12)/2 - 1

LCM = 2

y = (-3x + 12 - 2)/2

y = ( -3x + 10)/2

y = -3x/2 + 10/2

y = -3x/2 + 5

The equation of the line is

y = -3/2x + 5

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