[高職數學] 直線 4題

1. 下列各組點何者不可構成直角三角形?
(A) XY斜率 = (-7-9)/(10+2) = -4/3
XZ斜率 = (-5-9)/(12+2) = -1
YZ斜率 = (-5+7)/(12-10) = 1
(XZ斜率)(YZ斜率) = -1可成直角三角形
(B) XY斜率 = (-1+2)/(5-3) = 1/2
XZ斜率 = (0+2)/(10-7) = 2/3
YZ斜率 = (0+1)/(10-5) = 1/5
沒兩線為垂直,不可成直角三角形
(C) XY斜率 = (-2-1)/(3-2) = -3
XZ斜率 = (-1-1)/(-4-2) = 1/3
YZ斜率 = (-1+2)/(-4-3) = -1/7
(XY斜率)(XZ斜率) = -1可成直角三角形
(D) XY斜率 = (-4-1)/(3-0) = -5/3
XZ斜率 = (1-1)/(2-0) = 0
YZ斜率 = (1+4)/(2-3) = -5
沒兩線為垂直,不可成直角三角形
2. 設y^2 = 4x上一點為(a,b)
b^2 = 4a, a = b^2/4
該點為(b^2/4, b)和x + y + 2 = 0的距離為| b^2/4 + b + 2 | / √(1 + 1)
= | (1/4)(b^2 + 4b + 4) + 1 | / √2
= | (1/4)(b + 1)^2 + 1 | / √2
最小值 = 1 / √2
3. 設x + 2y = 6 一點為數(a,b)
a + 2b = 6 => a = 6 - 2b
矩形長為 a = 6 - 2b
濶為 b
面積 = b(6 - 2b)
= -2(b^2 - 3b)
= -2(b^2 - 3b + 9/4) + 9/2
= -2(b - 3/2)^2 + 9/2
最大面積時 b = 3/2, a = 6 - (2)(3/2) = 3
周長 = (3/2 + 3)*2 = 9
4. L1 : x - 2y + 3 = 0 ... (1)
L2 : 2x - y + 3 = 0 ... (2)
(1)*2 => 2x - 4y + 6 = 0 ...(3)
(2) - (3) => 3y - 3 =0 => y = 1
代入(1), x - 2 + 3 = 0 => x = -1
L1,L2交點為(-1,1)
不能圍成一三角型有三可能性
(1) L1, L3平行
L1斜率 = 1/2
L3斜率 = -a/2
1/2 = -a/2 => a = -1
(2) L2, L3平行
L2斜率 = 2
2 = -a/2 => a = -4
(3) L1, L2, L3交於一點(-1,1)
-a + 2 - 5 = 0 => a = -3
a之所有可能值之和= -1 - 4 - 3 = -8

1. 下列各組點 何者不可構成直角三角形?
(A) (-2,9)(10,-7)(12,-5)
(B) (3,-2)(5,-1)(10,0)
(C) (2,1)(3,-2)(-4,-1)
(D) (0,1)(3,-4)(2,1)
ANS:B
1.問題 要怎麼想比較好? 畫圖嗎?
用斜率 2線垂直 協率相乘=-1 以選項(A)為例
p(-2,9) q(10,-7) r(12,-5)
pq斜率 (-7-9)/(10-(-2))=-12/16=-3/4...(1)
qr (-5-(-7))/(12-10)=1....(2)
pr (-5-9)/(12-(-2))=-1....(3)
(2)*(3)=-1
2.試問在坐標平面上 曲線y^2=4x 與 x+y+2=0之間的最短距離為何?
ANS:1/2^1/2



3.坐標平面上一矩形的四個頂點分別在原點 X軸正向 Y軸正向 及x+2y=6上 當此舉行 有最大面積時 其周長為何?
ANS:9

4.三直線L1:x-2y+3=0 L2:2x-y+3=0 L3:ax+2y-5=0 不能圍成一三角型 則a之所有可能值之和為? ANS:-8
最好有詳解 因為我對圖形滿弱的 感謝

第二題答案應該是1吧?