再次讨论一元线性回归问题!!

前几天发了帖子,询问根据给出的一系列点求直线.<<狼行天下>>大哥给了一个算法,如下:
Private Sub Command1_Click()
Picture1.Scale (0, 20)-(12, 0) '设置坐标范围
Dim p(4, 1) As Double, i As Integer
For i = 0 To 4
p(i, 0) = Choose(i + 1, 1.2, 3.7, 4.1, 5.1, 8.3)
p(i, 1) = Choose(i + 1, 2.2, 6.4, 7.8, 10.1, 15.8)
Next ' 定义五个点
drawline Picture1, p '画出过五个点的直线
End Sub
Sub drawline(ByVal pic As PictureBox, ByRef p() As Double)
Dim sigmax As Double, sigmay As Double, sigmaxx As Double, sigmaxy As Double, n As Integer
Dim i As Long
Dim a As Double, b As Double '截距斜率
Dim x0 As Double, y0 As Double, x1 As Double, y1 As Double '定义两端点
n = UBound(p) - LBound(p) + 1 '点的个数
For i = LBound(p) To UBound(p)
Picture1.Circle (p(i, 0), p(i, 1)), Picture1.ScaleWidth / 200, vbRed '描点
Picture1.CurrentX = p(i, 0)
Picture1.CurrentY = p(i, 1)
Picture1.ForeColor = vbBlue
Picture1.Print "(" & p(i, 0) & ","; p(i, 1) & ")" '数据标志
sigmax = sigmax + p(i, 0) 'Σx
sigmay = sigmay + p(i, 1) 'Σy
sigmaxx = sigmaxx + p(i, 0) ^ 2 'Σx^2
sigmaxy = sigmaxy + p(i, 0) * p(i, 1) 'Σx*y
Next
a = (sigmaxx * sigmay - sigmax * sigmaxy) / (n * sigmaxx - sigmax ^ 2) '截距
b = (n * sigmaxy - sigmax * sigmay) / (n * sigmaxx - sigmax ^ 2) '斜率
x0 = Picture1.ScaleLeft
y0 = a + b * x0 '左端点
x1 = Picture1.ScaleLeft + Picture1.ScaleWidth
y1 = a + b * x1 '右端点
Picture1.Line (x0, y0)-(x1, y1), vbGreen '回归直线
End Sub
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
我用过后发现如果点1到点5的x坐标是递增的,那么画出的线是正确的.
但是如果我修改点的坐标,x值随意变化,y值递增或递减(也就是画竖线),发现有问题.
我没有上过高中,所以对数学知道很少,看上面的算法感觉是根据斜率/截距/x 来算y的坐标,但是x取值应该是多少?当画横线的时候,x取所有点的x值的最大与最小值就可以了,但是画竖线的时候哪?


请<<狼行天下>>大哥,或其它高手帮忙!!!
谢谢!!