A(7,2) and C(1,4) are two vertices of a square ABCD
find the equation of the diagonal BD
and find the coordinates of B and of D
我搵個midpoint 係M(4,3 ) 同埋equation of BD 係y=3x-9
distance between AM 係root 10
跟住我諗假設個square 係一個circle 入面
4個既vertices of a square 會intercept 個circle. 個radius = AM 仲有BD 肯定係
lays on y=3x-9 所以我用條circle of equation (x-a)^2 + (y-b)^2= r^2
咁我sub 入circle equation ( x-4)^2 + (y-3)^2 = 10
跟住sub y= 3x-9..
得x^2 -8x+16 =0
跟住搵唔到答案
當中方法有咩錯, 點解唔岩!
急問數!!!!!!!!!!!!!!!!急PLEASE
2010-04-05 2:53 pm
回答 (2)
2010-04-05 5:03 pm
✔ 最佳答案
You forgot about the 10 on the right hand side of the circle equation. So the correct equation should be x^2 - 8x + 15 = 0, x = 3 or 5.2010-04-05 10:29:55 補充：
(3x - 12)^2 = [3(x - 4)]^2 = 3^2(x - 4)^2 = 9(x - 4)^2, NOT (x - 4)^2.
2010-04-05 10:33:07 補充：
You may consider the following way of solving:
(x - 4)^2 + (3x - 12)^2 = 10
(x - 4)^2 + 9(x - 4)^2 = 10
10(x - 4)^2 = 10
(x - 4)^2 = 1
so x - 4 = +/- 1, that is x = 4 + 1 or 4 - 1, that is 5 or 3.
2010-04-05 5:11 pm
( x-4)^2 + (y-3)^2 = 10
( x-4)^2 + (3x-9-3)^2 = 10
( x-4)^2 + (3x-12)^2 = 10
x^2 - 8x + 16 + 9x^2 - 72x + 144 = 10
10x^2 - 80x + 150 = 0
x^2 - 8x + 15 = 0
運算錯左, 減少個 10, 你都幾醒, 自己可以做埋佢, 係大意者
2010-04-05 09:12:15 補充：
唔夠樓上快, 不過我好似清楚 d, 請投我一票 .... 謝
( x-4)^2 + (3x-9-3)^2 = 10
( x-4)^2 + (3x-12)^2 = 10
x^2 - 8x + 16 + 9x^2 - 72x + 144 = 10
10x^2 - 80x + 150 = 0
x^2 - 8x + 15 = 0
運算錯左, 減少個 10, 你都幾醒, 自己可以做埋佢, 係大意者
2010-04-05 09:12:15 補充：
唔夠樓上快, 不過我好似清楚 d, 請投我一票 .... 謝
收錄日期: 2021-04-15 19:10:36
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