數學-直角坐標幾何

距離公式:
兩點 A(x1, y1), B(x2, y2)
兩點距離 = SQRT [(x2 – x1)^2 +(y2 – y1)^2]

1.
A (4,5). B(-3,2) C(x,0)
AC = SQRT[(x - 4)^2 + (0 – 5)^2]
BC = SQRT[(x – (-3))^2 + (0 – 2)^2]
AC = BC
SQRT[(x - 4)^2 + (0 – 5)^2] = SQRT[(x – (-3))^2 + (0 – 2)^2]
(x - 4)^2 + (0 – 5)^2 = [(x – (-3))^2 + (0 – 2)^2
(x - 4)^2 + (– 5)^2 = [(x – (-3))^2 + (– 2)^2
(x - 4)^2 + 25 = (x +3)^2 + 4
x^2 – 8x +16 +25 = x^2 +6x + 9 + 4
-8x + 41 = 6x +13
14x = 28
x = 28/14

x = 2 答案

2.
D (-5,6), E(2,1), F(x,0)
DF = SQRT[(x – (-5))^2 + (0 – 6)^2]
EF = SQRT[(x – 2)^2 + (0 – 1)^2]
DF = EF
SQRT[(x – (-5))^2 + (0 – 6)^2]= SQRT[(x – 2)^2 + (0 – 1)^2]
(x – (-5))^2 + (0 – 6)^2 = (x – 2)^2 + (0 – 1)^2
(x +5)^2 + ( – 6)^2 = (x – 2)^2 + ( – 1)^2
(x +5)^2 + 36 = (x – 2)^2 + 1
x^2 +10x +25 +36 = x^2 -4x + 4 + 1
+10x + 61 = -4x +5
14x = -56
x = -56/14

x = - 4 答案

3.
P (6,-1), Q(-4,-3), R(0,y)
PR = SQRT[(0 – 6)^2 + (y – (-1))^2]
QR = SQRT[(0 – (-4))^2 + (y – (-3))^2]
PR = QR
SQRT[(0 – 6)^2 + (y – (-1))^2] = SQRT[(0 – (-4))^2 + (y – (-3))^2]
(0 – 6)^2 + (y – (-1))^2 = (0 – (-4))^2 + (y – (-3))^2
( – 6)^2 + (y + 1)^2 = (4)^2 + (y +3)^2
36 + y^2 + 2y + 1 = 16 + y^2 + 6y +9
37 + 2y = + 6y +25
6y – 2y = 37 - 25
4y =12
y = 12/4

y = 3 答案

Comment:
兩點距離 = SQRT [(x2 – x1)^2 +(y2 – y1)^2]
第一條錯在 difference of coordinates, 不是 sum of coordinates
A(4,5) . B(-3,2) C(x,0)
sqrt(4+x)^2+25 = sqrt(-3+x)^2+4 ~ wrong
sqrt(4-x)^2+25 = sqrt(-3-x)^2+4 ~ right
sqrt(x- 4)^2+25 = sqrt(x-(-3))^2+4 ~ right

公式用錯了，
尾三步也是錯的，
16+8x+x^2+21 = 9-6x+x^2
37=9-14x
14x=-28
x=-2

就係計唔到--
以第一條:
AC=BC
sqrt(4+)^2+(5+0)^2 = sqrt (-3+x)^2+(2-0)^2
sqrt(4+x)^2+25 = sqrt(-3+x)^2+4
(4+x)^2+25 = (-3+x)^2+4
(4+x)^2+21 = (-3+x)^2
16+8x+x^2+21 = 9-6x+x^2
37 = 9-2x
2x = -28
x = 14
錯係邊到....... ?

use 距離公式

1. x=2

2012-03-20 23:18:15 補充：
2. x= -4

3. y=3