Equation of Lines3/2

[MacaroniMarconi]

Hi, I've got this somewhat down...except when there are fractions involved which seems to be a large majority of the time. For example:

find the equation of each line with the given slope that passes through the given point. Write the equation in the form Ax + By = C

m= 3/2; (5, -6)

I start the equation like so:

y-(-6)= 3/2(x-5)

y- (-6)= 3/2x - 15/2

And then from here on I'm stumped.

 

[soroban]

Hello, MacaroniMarconi!

Find the equation of the line with slope \(\displaystyle \tfrac{3}{2}\) that passes through \(\displaystyle (5,-6)\)
Write the equation in the form \(\displaystyle Ax + Bx \:=\:C.\)

I start the equation like so:

y - (-6) = 3/2(x-5)

y- (-6) = 3/2x - 15/2

And then from here on I'm stumped.

You have: .\(\displaystyle y + 6 \:=\:\tfrac{3}{2}x - \tfrac{15}{2}\)

Multiply by 2: .\(\displaystyle 2y + 12 \:=\:3x - 15\)

And we have: .\(\displaystyle 3x - 2y \:=\:27\)

 

[HallsofIvy]

Hi, I've got this somewhat down...except when there are fractions involved which seems to be a large majority of the time. For example:

find the equation of each line with the given slope that passes through the given point. Write the equation in the form Ax + By = C

m= 3/2; (5, -6)

I start the equation like so:

y-(-6)= 3/2(x-5)

y- (-6)= 3/2x - 15/2

And then from here on I'm stumped.

If you don't mind fractions, simply subtract (3/2)x from both sides and subtract 6 from both sides:
y- (3/2)x= -6- 15/2

y- (3/2)x= -27/2 which is of the form Ax+ By= C with A= -3/2, B= 1, and C= -27/2. If you don't like fractions (or negative numbers) then multiply on both sides by -2, as Soroban suggests: 3x- 2y= 27