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.