两点A(3,1)B(-1,3)若C点满足向量OC=m向量OA+n向量OB,其中m+n=1,则C点的轨迹方程OC=a向量OA+b向量=(3a-b,a+3b)即x=3a-b,y=a+3b又a+b=1x+2y=(3a-b)+2(a+3b)=5a+5b=5即C点轨迹方程为x+2y=5 x+2y是哪来的
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![两点A(3,1)B(-1,3)若C点满足向量OC=m向量OA+n向量OB,其中m+n=1,则C点的轨迹方程OC=a向量OA+b向量=(3a-b,a+3b)即x=3a-b,y=a+3b又a+b=1x+2y=(3a-b)+2(a+3b)=5a+5b=5即C点轨迹方程为x+2y=5 x+2y是哪来的](/uploads/image/z/6833182-22-2.jpg?t=%E4%B8%A4%E7%82%B9A%283%2C1%29B%28-1%2C3%29%E8%8B%A5C%E7%82%B9%E6%BB%A1%E8%B6%B3%E5%90%91%E9%87%8FOC%3Dm%E5%90%91%E9%87%8FOA%2Bn%E5%90%91%E9%87%8FOB%2C%E5%85%B6%E4%B8%ADm%2Bn%3D1%2C%E5%88%99C%E7%82%B9%E7%9A%84%E8%BD%A8%E8%BF%B9%E6%96%B9%E7%A8%8BOC%3Da%E5%90%91%E9%87%8FOA%2Bb%E5%90%91%E9%87%8F%3D%283a-b%2Ca%2B3b%29%E5%8D%B3x%3D3a-b%2Cy%3Da%2B3b%E5%8F%88a%2Bb%3D1x%2B2y%3D%283a-b%29%2B2%28a%2B3b%29%3D5a%2B5b%3D5%E5%8D%B3C%E7%82%B9%E8%BD%A8%E8%BF%B9%E6%96%B9%E7%A8%8B%E4%B8%BAx%2B2y%3D5+x%2B2y%E6%98%AF%E5%93%AA%E6%9D%A5%E7%9A%84)
两点A(3,1)B(-1,3)若C点满足向量OC=m向量OA+n向量OB,其中m+n=1,则C点的轨迹方程OC=a向量OA+b向量=(3a-b,a+3b)即x=3a-b,y=a+3b又a+b=1x+2y=(3a-b)+2(a+3b)=5a+5b=5即C点轨迹方程为x+2y=5 x+2y是哪来的
两点A(3,1)B(-1,3)若C点满足向量OC=m向量OA+n向量OB,其中m+n=1,则C点的轨迹方程
OC=a向量OA+b向量
=(3a-b,a+3b)
即x=3a-b,y=a+3b
又a+b=1
x+2y=(3a-b)+2(a+3b)=5a+5b=5
即C点轨迹方程
为x+2y=5 x+2y是哪来的
两点A(3,1)B(-1,3)若C点满足向量OC=m向量OA+n向量OB,其中m+n=1,则C点的轨迹方程OC=a向量OA+b向量=(3a-b,a+3b)即x=3a-b,y=a+3b又a+b=1x+2y=(3a-b)+2(a+3b)=5a+5b=5即C点轨迹方程为x+2y=5 x+2y是哪来的
设C(x,y)
OC=a向量OA+b向量
=(3a-b,a+3b)=(x,y)
即3a-b=x (1)
a+3b=y (2)
(1)*3+(2) 10a=3x+y a=(3x+y)/10
代入(1) b=3a-x=(3y-x)/10
因为a+b=1
所以(3x+y)/10+(3y-x)/10=1
2x+4y=10
x+2y=5
即为所求
x=3a-b, y=a+3b
向量OC=(3a-b,a+3b)
得到C点的坐标是(3a-b,a+3b)
从而在在直角坐标系中
x=3a-b, y=a+3b
又因为a+b=1
所以要将上面的式子凑成a+b的倍数
那就只能是(3a-b)+2(a+3b)=x+2y=5(a+b)
就得到x+2y=5
x=3a-b, y=a+3b
x+2y=3a-b+2a+6b
=5a+5b
a+b=1
x+2y=5a+5b=5