algebra/geometry questions?

Question

1)distance formula?
2)midpoint formula?
3)how do u make an equation of a perpindicular bisector of segment AB?(slope intercept form and standard)
4)how do u make an equation of a line parallel to AB passing thru point C? (slope intercept form and standard)
5)what is the slope intercept form of an equation of a line connecting A and B?
6)how do u find the equation of a line parallel to 4x-y=7, passing thru (2,1)
7)how do u write the standard form of an equation of a line that passes thru (6,2) with "m"=2

i need the steps on how to solve the problems, its part of a summer assignment so i dont have a textbook...

Answer 1

1. Square root of :(x2-x1)^2 + (y2-y1)^2

2.[ (x1+x2)/2, (y1+y2)/2]

3. Find the midpoint. Find the slope of the segment. Use the negative reciprocal to get the slope of the perpendicular. Then substitute the midpoint and the perpendicular slope into point-slope formula which is y2-y1=m(x2-x1) Then do what is necessary to get it set = y for slope intercept and Ax+By=C for standard.

4.Find the slope of AB. Use that with point C in the point-slope formula which I wrote out in #3.

5.Find the slope of AB. Choose either A or B and again use that point-slope formula and proceed to get it in the right form.

6. The slope of the line is 4. Using the point-slope formula from #3, y-1=4(x-2). I'll give you the answer in slope intercept. Distribute the 4 and you get y-1=4x-8. then add 1 to each side of the equatio and you get y = 4x-7

7.Using the ever popular point slope:

y-2=2(x-6)
y-2=2x-12
-2x + y = -12 + 2
2x-y=10

Answer 2

1) sq. root of ((y2-y1)^2 -(x2-x1)^2)

2) x= (x2-x1)-2; y=(y2-y1)/2

3) (y-y1)=m(x-x1) + b perp: opposite sign reciprocal of slope .
ex 3/4 --> -4/3, and b stays the same

4) you keep the same slope but change the y-intercept

5)

6) y=4x-7 so parallel line would be: y=4x+10

7) (6,2) with m=2
use the equation in #3
(y-2)= 2(x-6)
(y-2)= 2x -12
y= 2x + 10

This is all i could help you with, i hope it is clear enough for you
God Bless