已知三角形ABC三边所在直线方程为AB:3x+4y+12=0 BC:4x-3y+16=0 AC:2x+y-2=0求∠ABC的平分线所在直线的方
问题描述:
已知三角形ABC三边所在直线方程为AB:3x+4y+12=0 BC:4x-3y+16=0 AC:2x+y-2=0求∠ABC的平分线所在直线的方
答
没人做?我帮你吧3x+4y+12=0 3x+4y+12=0 4x-3y+16=0 4x-3y+16=0 2x+y-2=0 2x+y-2=0B (-4,0) A(4,-6) C(-1,4)BC=5 BA=10 ∠ABC的平分线交AC(x,y)(4-y)/(4--6)=5/10 y=-1(-1-x)/(-1-4)=5/10 x=3/2平分线上两点 (-4,0)...