Help!!

In rectangle ABCD, we have A=(6,-22), B=(2006,178) , and D=(8,y), for some integer y. What is the area of rectangle ABCD?

The slope from A to B is  [ 178 - -22] / [ 2006 -  6 ]    = 200/ 2000 =  1/10

So...the slope between  A and D will have a negative reciprocal slope....so we have

[ y - - 22 ] / [ 8 - 6 ] = -10

[ y + 22] [ 2]  = -10

y + 22 =   -20

y = -2

So D =  (8, - 2)

And the distance from A to B =   sqrt [ (2006 - 6)^2 + (-22 - 178)^2 ]  = sqrt [ 2000^2 + 200^2 ]  = sqrt (4040000)

And the distance from A to D is  sqrt [ (8 - 6)^2 + (-22 - - 2)^2 ]  = sqrt [ (4 + 400] = sqrt (404)

So....the area of ABCD =   sqrt (4040000)*sqrt (404) =  sqrt (404000 * 404 ) =   40,400 units^2

The number of distinct points in the xy-plane common to the graphs of (x + y - 5)(2x - 3y + 5) = 0 and (x -y + 1)(3x + 2y - 12) = 0

Notice that this will  be true whenever

x + y  - 5 =  0      and

x - y + 1 =   0                        add these

2x  - 4   =  0

 x = 2

And y = 3

And  note that these two values make  (2x - 3y + 5)  =  0    and ( 3x + 2y - 12)  = 0

So.....the common point to both graphs is  (2,3)

See the graphs here :  https://www.desmos.com/calculator/rwobs7w4hu