1.求頂點坐標為A(1,1) , B(-2,5)及C(5,4)的三角形周界
2.証明頂點坐標為P(-1,-2) , Q(6,2)及R(2,4)的三角形為直角三角形
3.証明三點P(2,-2) , Q(3,3)及R(5,13)共線
4.如果通過M(-3 , a)及N(12 , -1)兩點的直線的傾角是57°。求a , 準確至一位小數
我想問下直線的坐標幾何...有冇人識做牙?!
2009-11-20 7:14 am
回答 (2)
2009-11-20 7:40 am
✔ 最佳答案
1. AB = √[(1 + 2)^2 + (1 - 5)^2] = 5AC = √[(1 - 5)^2 + (1 - 4)^2] = 5
BC = √[(-2 - 5)^2 + (5 - 4)^2] = 5√2
三角形周界 = 10 + 5√2
2.PR斜率 = (4 + 2)/(2 + 1) = 2
QR斜率 = (4 - 2)/(2 - 6) = -1/2
PR斜率 x QR斜率 = (2)(-1/2) = -1 => PR 垂直QR => 三角形為直角三角形
3.PR斜率 = (13 + 2)/(5 - 2) = 5
QR斜率 = (13 - 3)/(5 - 3) = 5
因此PQR 共線
4. MN斜率 = (-1 - a)/(12 + 3) = -(1 + a)/15 = tan 57
-(1 + a) = 15 * 1.54 = 23.1
a + 1 = -23.1
a = -24.1
2009-11-23 6:50 pm
收錄日期: 2021-04-23 23:20:17
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091119000051KK01306