若三直线2x+3y+8=8,x-y-1=0和x+ky=0相交于一点,求k的值

∵2x+3y+8=8 => y= -2/3x
x-y-1=0 => y=x-1
-2/3x=x-1 => x=3/5
∴三直线交点为(3/5,-2/5)
故x+ky=0 过点(3/5,-2/5) => k=3/2