General

[quote=DrHye]No prob. I'm only "excited" to help out with math on basil because it's my current job in real life
But I edited my previous post regarding the next problem. I'll let you handle it[/quote]

Sounds like a steal, receiving free math help on basil.

[quote=alex3650]Thanks, i'm so dumb e.e Every single time after summer, I forget what I had learned previously[/quote]

No prob. I'm only "excited" to help out with math on basil because it's my current job in real life
But I edited my previous post regarding the next problem. I'll let you handle it

[quote=DrHye]Edit your OP because that's not how you wrote it. Anyway,

You're given an equation y=2/7x - 8 and need an equation for a line parallel to it, but goes through (14,-3). So you need to find the y-intercept for your new equation. This can be done by simply plugging in your new x and y values. Note that lines are parallel when their slopes are equal, so m doesn't change

y=mx+b
-3=(2/7)*14 + b.... We are solving for b, the y-intercept
-3=4+b
-7=b

So we have m=2/7 and b=-7. Plug these into your general equation for the parallel line
y=mx+b
y=(2/7)x+(-7)
y=(2/7)x - 7

Will edit or post again with the perpendicular equation[/quote]

Thanks, i'm so dumb e.e Every single time after summer, I forget what I had learned previously

[quote=alex3650]The 2 equations that I make, have to pass through the point (14,-3) when your trying to graph it.[/quote]

Edit your OP because that's not how you wrote it. Anyway,

You're given an equation y=2/7x - 8 and need an equation for a line parallel to it, but goes through (14,-3). So you need to find the y-intercept for your new equation. This can be done by simply plugging in your new x and y values. Note that lines are parallel when their slopes are equal, so m doesn't change

y=mx+b
-3=(2/7)*14 + b.... We are solving for b, the y-intercept
-3=4+b
-7=b

So we have m=2/7 and b=-7. Plug these into your general equation for the parallel line
y=mx+b
y=(2/7)x+(-7)
y=(2/7)x - 7

Will edit or post again with the perpendicular equation

Edit: You're not given a point that lies on the perpendicular line, so b can be anything. Unless they gave you a point to use (or if it's (14,-3) again, then my first post and this one should be enough for you to solve this part on your own

y=2/7x - 8

Parallel is 2/7
y=2/7x +b for the parallel equation
Passes through the points 14 and -13, assuming this is what you mean.
(-13)=2/7(14) +b
Solve for b

Perpendicular slope is -7/2
y=-2/7x +b for the perp equation
Repeat steps

[quote=DrHye]What do you mean 14/-3? As a y-intercept, or did you mistype a coordinate?
You do have the slopes right for both parts, by the way. A slope for a perpendicular line is found by taking the slope of the original line, flipping it (aka taking the reciprocal) and switching the sign (- to + or + to minus).[/quote]
The 2 equations that I make, have to pass through the point (14,-3) when your trying to graph it.
@Fiercerain I already know about the slope, but need help with finding b
@HolyDragon meant -3 not -13 xD

@DrHye: Negative inverse, what he said! I'm editing my earlier post.

[quote=Fiercerain]Given that:
y=(2/7)x - 8
The slope is: (2/7)
A parallel line will have the same slope, and what you can do is use the slope-intercept form where you have:
y - y1 = (2/7)*(x - x1)

A perpendicular line is the inverse of a line's slow so.. 1/m meaning 1/(2/7) gives you a perpendicular slope of (7/2)

and again you can use the slope-intercept form to pass through any specific line.[/quote]

Negative inverse, not inverse
Otherwise they're both still traveling up or both down (from left to right)

Or maybe instead of saying "negative inverse," it's just the inverse with the opposite sign

Given that:
y=(2/7)x - 8
The slope is: (2/7)
A parallel line will have the same slope, and what you can do is use the slope-intercept form where you have:
y - y1 = (2/7)*(x - x1)

A perpendicular line is the negative inverse of a line's slow so.. -(1/m) meaning -1/(2/7) gives you a perpendicular slope of -(7/2)

and again you can use the slope-intercept form to pass through any specific line.

Will not judge if below 7th grade

[quote=alex3650]Line L - y=2/7x - 8

How can I make an equation so that line L is parallel to line M, but passes through the point 14/-3. And a perpendicular equation also.
I know that for parallel, the slope would be the same so, y=2/7x (- or +) b. I need help for finding b.
For the perpendicular equation, I think it is y=-7/2x (- or +) b. Still need help on finding b.[/quote]

plug the point 14/-3 into the point slope form of the perpendicular equation. then solve into mx+b.

What do you mean 14/-3? As a y-intercept, or did you mistype a coordinate?
You do have the slopes right for both parts, by the way. A slope for a perpendicular line is found by taking the slope of the original line, flipping it (aka taking the reciprocal) and switching the sign (- to + or + to minus).

What math is this and what grade are you in.