1. Let the equation of the straight line where AB is located be y = x.

With x? +3y？ =4 Simultaneously

x？ - 1=0

Let the coordinates of A and B be (x 1, y 1) and (x2, y2) respectively.

ab=√( 1+k^2)[(x 1+x2)^2-4x 1x2]

AB=2√2

And because the height h on the side of AB is equal to the distance from the origin to the straight line L, h=|0-2|/√2=√2.

S△ABC= 1/2 |AB|？ h=2

2. Let the equation of the straight line where AB is located be y = x+m.

With x? +3y？ =4 Simultaneously

4x？ +6mx+3m？ -4=0

x 1+x2=-3m/2

x 1x2=(3m？ -4)/4

So |AB|=√2|x 1-x2|=√(32-6m? )/2

Because the length of BC is equal to the distance from (0, m) point to straight line L, that is |BC|=|2-m|/√2.

So |AC|? =|AB|？ +|BC|？ =-m？ -2m+ 10=-(m+ 1)？ + 1 1

So when m=- 1, the AC side is the longest (at this time △ =- 12+64 > 0).

At this time, the equation of the straight line where AB is located is y=x- 1.

Hope to adopt. If you have any questions, just ask.