课时分层作业(十四)(建议用时:40分钟)一、选择题1.到点A(1,3),B(-5,1)的距离相等的动点P满足的方程是()A.3x-y-8=0B.3x+y+4=0C.3x-y+6=0D.3x+y+2=0B[设P(x,y),由条件知=整理得3x+y+4=0
]2.已知点A与点B(1,2)关于直线x+y+3=0对称,则点A的坐标为()A.(3,4)B.(4,5)C.(-4,-3)D.(-5,-4)D[设A(x,y),则∴选D
]3.直线l经过原点,且经过另两条直线2x+3y+8=0,x-y-1=0的交点,则直线l的方程为()A.2x+y=0B.2x-y=0C.x+2y=0D.x-2y=0B[设所求直线方程为2x+3y+8+λ(x-y-1)=0,即(2+λ)x+(3-λ)y+8-λ=0,因为l过原点,所以λ=8
则所求直线方程为2x-y=0
]4.已知点M(0,-1),点N在直线x-y+1=0上,若直线MN垂直于直线x+2y-3=0,则N点的坐标是()A.(2,3)B.(-2,-1)C.(-4,-3)D.(0,1)A[由题意知,直线MN过点M(0,-1)且与直线x+2y-3=0垂直,其方程为2x-y-1=0
直线MN与直线x-y+1=0的交点为N,联立方程组解得即N点坐标为(2,3).]5.直线ax+4y-2=0与直线2x-5y+b=0垂直,垂足为(1,c),则a+b+c=()A.-2B.-4C.-6D.-8B[ 直线ax+4y-2=0与直线2x-5y+b=0垂直,∴-×=-1,∴a=10,∴直线ax+4y-2=0方程即为5x+2y-1=0
将点(1,c)的坐标代入上式可得5+2c-1=0,解得c=-2
将点(1,-2)的坐标1代入方程2x-5y+b=0得2-5×(-2)+b=0,解得b=-12
∴a+b+c=10-12-2=-4
]二、填空题6.过点A(4,a)和B(5,b)的