Draw a triangle ABC and mark the coordinates of A, B and C respectively. Vector AB = (-2,0)-(2,0) = (-4,0).

Then | vector AB | = √ (4 2-0 2) = 4, so the side length of AB is 4. Similarly, if | vector BC | = 2 √ 3 = | vector AC |, the triangle ABC is an isosceles triangle. Let CP be perpendicular to AB at point P, then CP is the middle vertical line of AB. AP=PB=2 .

In the right triangle CBP, PC = √ (BC 2-Pb 2) = √ (12-4) = 2 √ 2.

Therefore, s ⊿ ABC =1/2 * ab * PC =1/2 * 4 * 2 √ 2 = 4 √ 2.

2。 According to the meaning of the question, let the analytical formula of L 1 be y=a 1x+b 1, and substitute it into the coordinate values of point A and point B, then 0=-a+b, and 3 = 2a+b.

Then a 1= 1, b 1= 1, then the analytical formula of L 1 is y=x+ 1.

If the modulus of vector AB is 3 according to the coordinates of point A and point B, the length of side AB of triangle ABC has been determined to be 3.

Given S⊿ABC=3 and the height of the triangle is H, then h=6/3=2.

If L2 is just perpendicular to the X axis, the coordinate of point C is (2,0), and the analytical formula of L2 is y=3.

If L2 intersects the X axis at point C, which is exactly at the origin, let the analytical expression of L2 be y=a2x+b2.

Substituting the coordinates of point B and point C can be solved as y=-2/3x.

If L2 is not perpendicular to the X-axis and does not intersect with the origin, the vector AC of point P perpendicular to the X-axis cannot be determined, and the position of point C cannot be determined. L2 only determines a point B, and the analytical expression of L2 cannot be obtained.