Question

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∠AB

Answer 1

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3x+4y+12=0 3x+4y+12=0 4x-3y+16=0

4x-3y+16=0 2x+y-2=0 2x+y-2=0

B (-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) (-1,3/2)

k=1/2 b=2

平分线y=1/2*x+2