How do you write an equation of the line when it goes through (5,5) and parallel to the line 4x+3y=-2?

Question

How do you write an equation of the line when it goes through (5,5) and parallel to the line 4x+3y=-2?

Answer 1

The parallel line shows you the slope.

Assuming you want y=mx+b form, you'd convert to:

3y = -4x - 2

y = -4/3 x - 2/3

You can discard the -2/3, that's the offset of that particular line.  You take the -4/3 as the slope for your new line.

y = -4/3 x + b

Solve for b by plugging in the original point:

5 = (-4/3) * 5 + b

5 = -20/3 + b

15/3 = -20/3 + b

b = 35/3 = 11 2/3

y = -4/3 x + 11 2/3

Answer 2

First of all, the given equation

4x + 3y = -2

should be re-written as (slope and y-intercept form)

y = (-4x/3) - (2/3)  (1)

Since the new line is parallel to the given (1) one:

Therefore the new line should have the same slope

that is y = (-4x/3) + b  (2)

Subtitue x = 5 and y = 5 into equtaion (2), we have:

5 = (-4*5/3) + b

5 + 20/3  = b

35/3 = b

Therefore the new line has the equation: y = (-4/3) + 35/3

or can be written as 4x + 3y = 35