Question

求过程答案，初二

Answer 1

解
设A点坐标为A(x, 0),考虑一下三种情况
(1) ∠BAC为直角：AB 2 + AC 2 = BC 2 ,得方程
(x+5) 2 + 36 + (x-5) 2 + 4 = (5+5) 2 + (2-6) 2
整理得
x 2 = 13 ==> x = ±√13
即：A点坐标为 A(-√13, 0) 或 A( √13, 0)；(图中的A2,A3)
(2) ∠ABC为直角：AB 2 + BC 2 = AC 2 ,得方程
(x+5) 2 + 36 + (5+5) 2 + (2-6) 2 = (x-5) 2 + 4
整理得
20x = -148 ==> x = -7.4
即：A点坐标为 A( -7.4, 0)；(图中A1)
(3) ∠ACB为直角：AC 2 + BC 2 = AB 2 ,得方程
(x-5) 2 + 4 + (5+5) 2 + (2-6) 2 = (x+5) 2 + 36
整理得
20x = 80 ==> x = 4
即：A点坐标为 A( 4, 0)

追问

整理得
x 2 = 13 ==> x = ±√13
整理得
20x = -148 ==> x = -7.4
整理得
20x = 80 ==> x = 4

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