过点p(0,1)作直线l,使它被两直线l1 2x+y-8=0和l1 2x-3y+10=0所截得的线段被点...
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设这个直线方程为Y=aX+b.

将P代入:b=1,直线方程为Y=aX+1.

L与L1的交点:aX+1=8-2X,(a+2)X=7,

X1=7/(a+2).y1=7a/(a+2)+1.

L与L2的交点:aX+1=(2X+10)/3,3aX+3=2X+10,(3a-2)X=7,

X2=7/(3a-2).y2=7a/(3a-2)+1

由于P到两个交点的距离相等.所以:

X1+X2=0 7/(a+2)=7/(3a-2) a=2.

y1+y2=2 7a/(a+2)+1+7a/(3a-2)+1=2 a=0.

l:y=2x+1,或Y=1