From the midpoint coordinate formula:

x0=(2+0)/2= 1，y0=(4+(-2))/2= 1

M( 1， 1)？ .

Let the equation of the straight line where the center line CM of AB side is located be:

y=kx+b，

Substitute c (-2,3) and M( 1, 1),

Solve the equation:

3=-2k+b？ ( 1)?

1=k+b？ (2)

Get k=-2/3.

b=5/3

∴ The equation of the straight line where the center line CM of AB side is located is:

y？ =? -? 2x/3？ +? 5/3。

2.？ In the same way.

The equation of straight line with AB edge is:

y-4=3(x-2)

3x-y-2=0。

Distance from point C (-2,3) to straight line AB: 3x-y-2 = 0.

d = | 3 *(2)-3-2 |/√？ 10

= 1 1√? 10/ 10.

Get S△ABC=|AB|*d/2.

=2√ 10*( 1 1√? 10/ 10)/2

= 1 1.