Question

在平面直角坐标系中,放置一个如图所示的直角三角形纸片AOB,已知OA=2,∠AOB=30°.D、E两点同时从原点O出发,

Answer 1

1) 设C点坐标(0,c),三角形ABC底为AC=|3-c|,AC上的高=B的横坐标 = 5

面积 = |3-c|*5/2 = 10

|3-c| = 4

3-c = 4,c = -1

或3-c = -4,c = 7

C(0,-1)或C(0,7)

(2) 设P点坐标(p,0),三角形ABP底为AB=√[(5-0)²+(2-3)²] = √26

AB的方程(两点式):(y - 2)/(x - 5) = (2 -3)/(5-0) = -1/5

x + 5y -15 =0

AB上的高=P与AB的距离 d= |p + 5*0 -15|/√(1²+5²) = |p-15|/√26

三角形PAB面积 = |AB|*d/2 = √26 * ( |p-15|/√26) /2 = |p-15| = 8

|p-15| = 16

p-15 = 16,p = 31

或 p-15 = -16,p = -1

P(-1,0)或(31,0)