How would i find the equations of the following lines?

Question

I know the formulas y-y1=m(x-x1)
and
y2-y1
-------
x2-x1

The questions are

1. Is parallel to 2x+3y=8 and passes through (5,4)

2. Is perpendicular to y= - 1/2x+3
and has a y-intercept
of -2

I can do other question using this formula but these are giving me a hard time.

Answer 1

1) Parallel means they have the same slope. You need to find the slope of 2x + 3y = 8
To find the slope, solve for y.
2x + 3y = 8
Subtract 2x from each side
3y = -2x + 8
Divide both sides by 3
y = (-2/3)x + 8/3
The slope is m = -2/3
Now use your point.
(y-4) = (-2/3)(x-5)

2) Perpendicular means the slopes are negative reciprocals.
The slope of y = (-1/2)x + 3 is -1/2
The negative reciprocal of -1/2 is 2.
Since you are given that -2 is the y-intercept, you can use the y-intercept equation.
y = mx + b
y = 2x - 2

Answer 2

For 1: Write the eqn in the form of y= mx+b, then use y-y1 =m(x-x1) with the point given (5,4) to produce the new eqn.
Perpendicular: y = 2x -2 ( remember to flip the coefficient and change the sign, b is the y intercept.

Answer 3

1.) Parallels have the same slopes but different y-intercepts.
2x+3y=8
3y=-2x+8
y=-2/3x+8/3
Using point-slope form, put this slope and the point in the problem.
y-4=-2/3(x-5)
3y-12=-2x+10
y=-2/3+22 is the answer.
2.) Perpendiculars have opposite reciprocals.
Change the slope from -1/2 to 2. Their product must equal -1.
Put the information into a problem.
y=2x-2 is the answer.