24. (12 分)已知$A(a,0),B(0,b),C(m,n)$，且$|a + b + 1| + \sqrt{b - a - 7} = 0$.
(1)直接写出$a =$,$b =$;
(2)如图1,若$m + n = 0$且$m ＞ 0$,$D$为$x$轴正半轴上一点,$∠ BAO = ∠ CDO$,三角形$ABD$的面积为13,求点$C$的坐标;
(3)如图2,若$n = 5$,三角形$ABC$的面积小于7且不小于5,求$m$的取值范围.
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24. 【点拨】本题考查坐标与图形性质,非负数的性质以及算术平方根,平行线的判定,三角形的面积,解题的关键是利用三角形的面积公式求得相关线段的长度.
【解析】(1)
∵ $|a + b + 1| + \sqrt{b - a - 7} = 0$,
∴ $\begin{cases}a + b + 1 = 0, \\ b - a - 7 = 0,\end{cases}$ 解得b = 3,
∴ a = -4,
∴ 点A的坐标为(-4,0),点B的坐标为(0,3). 故答案为-4,3.
(2)如题图1,连接AC,BC,OC,过点C作CM ⊥ y轴于点M,CN ⊥ x轴于点N. 由(1)得,A(-4,0),B(0,3),
∴ OA = 4,OB = 3,
∵ m + n = 0且m ＞ 0,
∴ n = -m.
∴ CM = CN = m,
∴ C(m, -m).
∵ ∠BAO = ∠CDO,
∴ AB // CD,
∴ $S_{△ABC} = S_{△ABD} = 13$.
∵ $S_{△ABC} = S_{△AOB} + S_{△AOC} + S_{△BOC} = \frac{1}{2}OA·OB + \frac{1}{2}OA·CN + \frac{1}{2}OB·CM = \frac{1}{2}×4×3 + \frac{1}{2}×4×m + \frac{1}{2}×3×m = 13$,解得m = 2,
∴ 点C的坐标为(2, -2).
(3)如图,过点C作直线CH ⊥ y轴于点H,延长AB交直线CH于点D,过点D作DG ⊥ x轴于点G,则四边形BOGD是直角梯形,四边形OGDH是长方形,
∴ DH = OG.
∵ C(m,n),n = 5,
∴ 设点D(x,5),则DH = OG = x,DG = 5.
∵ A(-4,0),B(0,3),
∴ OA = 4,OB = 3,
∴ AG = OA + OG = 4 + x.
∵ $S_{△ADG} = S_{△AOB} + S_{梯形BOGD}$,
∴ $\frac{1}{2}AG·DG = \frac{1}{2}OA·OB + \frac{1}{2}(OB + DG)·OG$,即$\frac{1}{2}(4 + x)×5 = \frac{1}{2}×4×3 + \frac{1}{2}(3 + 5)x$,解得 $x = \frac{8}{3}$,
∴ $D(\frac{8}{3},5)$.
∵ $S_{△ABC} = S_{△ACD} - S_{△BCD} = \frac{1}{2}CD×5 - \frac{1}{2}CD×(5 - 3) = \frac{3}{2}CD$,三角形ABC的面积小于7且不小于5.
∴ $5 ≤ \frac{3}{2}CD ＜ 7$,
∴ $\frac{10}{3} ≤ CD ＜ \frac{14}{3}$,
∴ $\frac{10}{3} ≤ |m - \frac{8}{3}| ＜ \frac{14}{3}$,解得 $m ≤ -\frac{2}{3}$ 或m ≥ 6 或 $-2 ＜ m ＜ \frac{22}{3}$ 且 $m ≠ \frac{8}{3}$,
∴ m的取值范围为 $6 ≤ m ＜ \frac{22}{3}$ 或 $-2 ＜ m ≤ -\frac{2}{3}$.