已知直线方程为（2+m)x+(1-2m)y+4-3m=0 (1）m为何值时,点（3,4）到直线的距离最大,最大值为?（2
1个回答

(1)

(2 + m)x + (1 -2m)y + 4 - 3m = 0可变为2x + y + 4 + m(x - 2y - 3) = 0,原式表示过二直线2x + y + 4 = 0,x - 2y - 3 = 0的交点A(-1,-2)的直线

B(3,4),与B距离最大的直线与AB垂直且过A

AB的斜率k = (-3-4)/(-1-3) = 3/2

垂线的斜率k' = -1/k = -2/3

直线的方程:y + 2= (-2/3)(x + 1) ,2x + 3y + 8 = 0

(2)

直线的斜率k (k < 0),方程:y + 2 = k(x + 1)

y = 0,x = 2/k - 1 < 0,A(2/k - 1,0)

x = 0,y = k - 2 0)

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