Question

It is my last assignment for today!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
PLEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEASSSSEEEEEE!!!!!!!! HELP!!!!!!!!!

1. Which is the equation of the given line in point-slope form? (The points are, 1 1/2, 2 & -4,-1)

A. y+1=6/7 (x+4)
B. y+1=6/11 (x+4)
C. y+4=11/6 (x+1)
D. y-1=6/11(x-4)

2. Which is an equation of the given line in standard form? (The line is, 1 1/2, 2 & -4,-1)

A. -11x-6y=-13
B. -6x+11y=13
C. -6x+11y=-13
D. -6x+7y=-17

3. Which is an equation of the given line in standard form? (The line is, -4,-3 & 2,-3)

A. x = –3
B. x = 3
C. y = –3
D. y = –3x

4. Which is the equation of the given line in slope-intercept form? (The line is, -1,2 & 1,-4)

A. y=-3x-1
B. y=3x+1
C. y=3x-1
D. y=-1/3x-1

5. Which is the equation of the given line in standard form? (The line is, 3,2 & 3,-4)

A. x = 3
B. x = –3
C. y = 3
D. y = 3x

PLEASE EXPLAIN HOW YOU GOT IT.... Thank you!!! :)

Answer 1

*Cracks fingers* Kk

When dealing with (x,y) coordinates close them after its just (x,y) not (x,y and x,y) it'll be considered wrong.

So, to get them into slope-intercept form use the formula: M = Y2-y1/X2-x1

Then once you have your slope, you'll use y-y1=m(x-x1)

Answer 2

Okay thanks that helps me understand it better... will you give me the answers please. :)

Answer 3

Hello??

Answer 4

I'm not allowed to do that, its against the CoC.