Write the equation of the line perpendicular to 3x + 2y = 8, passing through (6, 1)

Answer 1

So, there are two ways to do this. Either way, we start by getting our equation into slope-intercept form.

3x + 2y = 8

2y = -3x + 8

y = (-3/2)x + 4

Ok, we have a slope of -3/2. Perpendicular lines have opposite, reciprocal slopes, so our new line will have a slope of 2/3.

1. y = mx + b

1 = (2/3)(6) + b

1 = 4 + b

b = -3

y = (2/3)x - 3

2. Point-slope form

y - y₁ = m(x - x₁)

y - 1 = (2/3)(x - 6)

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

y = (2/3)x - 3

Answer 2

The slope of a perpendicular line is the negative reciprocal of the first line. Mathematically, that is m1 x m2 = -1. so first find the slope of the given line, and use that relationship to find the slope of the perpendicular line. Next use the point slope formula (y - y1) = m(x - x1) to find the equation of the perpendicular line.

Starting with 3x + 2y = 8, put the equation into standard form by subtracting 3x from both sides, and the dividing the new equation by 2.

Subtract 3x:

2y = 8 - 3x

Divide by 2:

y = 4 -( 3/2)x

So the slope of the first line is -3/2. Invert that and multiply by -1 to get the new slope which is 2/3, and use that to find the new equation using the point slope method with the point (6,1):

y - 1 = 2/3(x-6), distribute 2/3 across the binomial, (x-6):

y - 1 = (2/3)x -4, add 1 to both sides:

y = (2/3)x -3, the equation in standard form

Hope this helps