y=kx+b过y+-2x+10的交点A,y=kx+b交y轴于B,y=-2x+10交y轴于C点,若S△ABC=12,则k= b=

y=kx+b过y+-2x+10的交点A,y=kx+b交y轴于B,y=-2x+10交y轴于C点,若S△ABC=12,则k= b=
y=kx+b过y+-2x+10的交点A,y=kx+b交y轴于B,y=-2x+10交y轴于C点,若S△ABC=12,则k= b=

y=kx+b过y+-2x+10的交点A,y=kx+b交y轴于B,y=-2x+10交y轴于C点,若S△ABC=12,则k= b=
y=3x-5与y=-2x+10的交点为a,
联立
y=3x-5
y=-2x+10
得x=3,y=4
直线过点（3,4）
因为y=-2x+10交x轴于c,则c=5
因为
s△abc=12
而s△abc+s△boc=s△abo+s△aoc
s△abo+s△aoc
=1/2*b*3+1/2*5*4
=3b/2+10
s△boc=1/2*b*5
则有
1/2*b*5+12=3b/2+10
则b=-2
直线过点（3,4）则
3k-2=4
k=2
直线y=kx+b为
y=2x-2