PLEASE HELP! Due tomorrow!

1.

There may be several  ways to do this, but one of the easier ways  is through  the "formula" for finding the distance between a point and aline

If  (m, n)  is the point  and  Ax + BY + C = 0   is the equation of the line....the distance is given by

l  Am + Bn + C  l              l  4(3) + 3(4) + 7  l              l  31  l               31

______________  =  __________________  =    _______      =       ___  =  6.2    units

 √ [ A^2 + B^2 ]              √ [ 4^2 + 3^2  ]                    √  25                   5

2.

5x-12y+45=0   and  5x-12y-46=0

12y   = 5x  + 45     and      12y  = 5x  - 46

y  = (5/12)x + 45/12

y = (5/12) x  +  15/4

These lines have the same slope.......the first line has  a  y intercept at  15/4....

So the point  (0, 15/4)  is on the first line

The distance between this point and the second line is the perpendicular distance between both lines...so we have

l  5(0)  - 12 (15/4) - 46  l               l  -12 (15/4)  - 46  l            l  -91  l              91       

____________________    =  ________________  =          ______     =      ___    =   7 units

      √ [ 5^2  + 12^2  ]                      √ [ 169 ]                             13                   13

3. A circle centered at the origin is tangent to the line x-y+4=0. What is the area of the circle?

Since the circle is centered at  (0,0)  a radius drawn  from this point will be perpendicular to this line at the point of tangency....and this will be the shortest distance from (0,0) to this line....so.....this distance  is given by

l  1 (0) -  1 (0)   + 4  l                       4          4√2             2√2     =  √8  units  = radius  of the circle

________________       =            ___  =    ____   =

   √  [1^2  + 1^2 l                            √2            2

So....the area  of the circle  = pi * radius ^2    =    pi  *  ( √8)^2   =   8 pi units^2

THANK YOU SO MUCH!