过点P(0,1)作直线,使它被两条直线L1:x-3y+10=0,L2:2x+y-8=0
过点P(0,1)作直线,使它被两条直线L1:x-3y+10=0,L2:2x+y-8=0所截得的线段平分于点P,则此直线方程为___________.
过点P(0,1)作直线,使它被两条直线L1:x-3y+10=0,L2:2x+y-8=0所截得的线段平分于点P,则此直线方程为___________. 解: 设直线方程为y=kx+b 联立: y=kx+b x-3y+10=0求得交点A(xa,ya) xa=(3b-10)/(1-3k) ya=(b-10k)/(1-3k) 联立: y=kx+b 2x+y-8=0 求得交点B(xb,yb) xb=(8-b)/(k+2) yb=(8k-2b)/(k+2) ∵点P(0,1)是AB的中点 ∴xa+xb=0 ya+yb=2 带入即可
答:斜率k1=1/3,k2=-1/2,夹角a满足 tana=|(k2-k1)/(1+k2k1)|=|(-1/2-1/3)/(1-1/6)|=1.详情>>
答:详情>>