求满足下列条件的直线的方程:1)过点A(3,2),且与直线4x+y-2=0平行.2)过点C(2,-3),且平行于点M(-1,-5)的直线。3)过点B(3,0),且与直线2x+y-5=0垂直
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![求满足下列条件的直线的方程:1)过点A(3,2),且与直线4x+y-2=0平行.2)过点C(2,-3),且平行于点M(-1,-5)的直线。3)过点B(3,0),且与直线2x+y-5=0垂直](/uploads/image/z/6856247-47-7.jpg?t=%E6%B1%82%E6%BB%A1%E8%B6%B3%E4%B8%8B%E5%88%97%E6%9D%A1%E4%BB%B6%E7%9A%84%E7%9B%B4%E7%BA%BF%E7%9A%84%E6%96%B9%E7%A8%8B%EF%BC%9A1%29%E8%BF%87%E7%82%B9A%283%2C2%29%2C%E4%B8%94%E4%B8%8E%E7%9B%B4%E7%BA%BF4x%2By-2%3D0%E5%B9%B3%E8%A1%8C.2%29%E8%BF%87%E7%82%B9C%282%EF%BC%8C-3%29%EF%BC%8C%E4%B8%94%E5%B9%B3%E8%A1%8C%E4%BA%8E%E7%82%B9M%28-1%EF%BC%8C-5%29%E7%9A%84%E7%9B%B4%E7%BA%BF%E3%80%823%29%E8%BF%87%E7%82%B9B%283%EF%BC%8C0%29%EF%BC%8C%E4%B8%94%E4%B8%8E%E7%9B%B4%E7%BA%BF2x%2By-5%3D0%E5%9E%82%E7%9B%B4)
求满足下列条件的直线的方程:1)过点A(3,2),且与直线4x+y-2=0平行.2)过点C(2,-3),且平行于点M(-1,-5)的直线。3)过点B(3,0),且与直线2x+y-5=0垂直
求满足下列条件的直线的方程:1)过点A(3,2),且与直线4x+y-2=0平行.
2)过点C(2,-3),且平行于点M(-1,-5)的直线。3)过点B(3,0),且与直线2x+y-5=0垂直
求满足下列条件的直线的方程:1)过点A(3,2),且与直线4x+y-2=0平行.2)过点C(2,-3),且平行于点M(-1,-5)的直线。3)过点B(3,0),且与直线2x+y-5=0垂直
1)与直线4x+y-2=0平行,则
斜率相等,为-4
根据点斜式,可求出直线方程:
y-2=-4(x-3)
即4x+y-14=0
2)y+3=5(x-2) =>5x-y-13=0
3)y-0=1/2(x-3) =>x-2y-3=0
设4x+y +k =0再把a 点带进去,算出k =-12
因为平行,所K要一样,所以,设直线方程为4X+Y+Z=0
X=3 Y=2代入 Z=-14
方程为 4X+Y-14=0
3) 因垂直 所以斜率为负倒数,为1/4 所以设方程为 Y=1/4X+Z
X=3 Y=0代入 Z=-3/4 则 4Y-X+3=0
2题有误
第一个条件不够,第二个是2x+y-6=0
解⑴直线4x+y-2=0变形为y=-4x+2∵平行∴新直线方程可设y=-4x+b将A点坐标代人得:b=14∴y=-4x+14。⑵有问题⑶直线2x+y-5=0变形为y=-2x+5∵垂直∴新方程可设y=½x+b,B点坐标代人得b=-3/2∴y=½x-3/2
大家都理解错了
第二题应该是向量M=(-1,-5)
就ok