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=
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![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=](/uploads/image/z/6976542-30-2.jpg?t=y%3Dkx%2Bb%E8%BF%87y%2B-2x%2B10%E7%9A%84%E4%BA%A4%E7%82%B9A%2Cy%3Dkx%2Bb%E4%BA%A4y%E8%BD%B4%E4%BA%8EB%2Cy%3D-2x%2B10%E4%BA%A4y%E8%BD%B4%E4%BA%8EC%E7%82%B9%2C%E8%8B%A5S%E2%96%B3ABC%3D12%2C%E5%88%99k%3D+b%3D)
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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