M为直线2X-Y+3=0上一动点,A(4,2)为一定点,又点P在直线AM上运动,且IAPI:IPMI=3,求P点轨迹方程?

问题描述:

M为直线2X-Y+3=0上一动点,A(4,2)为一定点,又点P在直线AM上运动,且IAPI:IPMI=3,求P点轨迹方程?

设P(x,y),M(x0,y0),由IAPI:IPMI=3,得(4+3x0)/(1+3)=x,(2+3y0)/(1+3)=y,则x0=(4x-4)/3,y0=(4y-2)/3,因为点M(x0,y0)在直线2X-Y+3=0上,所以2X0-Y0+3=0,即2(4x-4)/3-(4y-2)/3+3=0,化简得 8x-4y-3=0,即为...