记y=3x－2为l1，其倾斜角为θ，所求直线为l2，其倾斜角为2θtanθ=k1=3tan2θ=2tanθ/(1－tan²θ)=2×3/(1－3²)=-3/4设l2为y=k2x＋b∵l2过原点且tan2θ=-3/4∴l2为y=-3/4x
y=-9/8x
