若直线ax+2y-6=0与x+(a-1)y-(a-1)=0平行,则他们之间的距离等于?
问题描述：
若直线ax+2y-6=0与x+(a-1)y-(a-1)=0平行,则他们之间的距离等于?
答
两直线平行,斜率相等.- a/2 = -1/(a-1) ==> 2 = a(a-1) ==> a^2 - a - 2 = 0 ==> (a -2)(a+1) = 0 ==> a = 2,- 1 (1) 当 a = 2 时,ax+2y-6=0 ==> x + y - 3 = 0 x+(a-1)y-(a-1)=0 ==> x + y - 3 = 0 此时两线重合.所以a= 2 不符合题意.(2) a = -1 ax+2y-6=0 ==> -x + 2y - 6 = 0 ==> x - 2y + 6 = 0 x+(a-1)y-(a-1)=0 ==> x - 2y = 0 tg apha = 6/x = 1/2 ==> x = 12 S = 12 * 6 /2 = √(6^2 + 12^2) * h / 2 ==> 36 = √180 * h /2 ==> 36 = 6√5 * h /2 ==> h = 12 / √5 ==> h = 12√5 / 5