在日常生活和科学实验中，人们会经常发现因变量y和自变量x之间存在一定线性关系设一组数据为：

则y与x的关系可以用线性方程表示：

    按最小二乘法可得：

        线性关系的程度可以用相关系数r表示

        所以，如果想在图象框中根据已知的多个存在线性关系的点描出相应的离所有的点最靠近的直线，应该利用以上一元线性回归的方法，代码如下：

Private Sub Command1_Click()Picture1.Scale (0, 20)-(12, 0) '设置坐标范围Dim p(4, 1) As Double, i As IntegerFor i = 0 To 4p(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 SubSub drawline(ByVal pic As PictureBox, ByRef p() As Double)Dim sigmax As Double, sigmay As Double, sigmaxx As Double, sigmaxy As Double, n As IntegerDim i As LongDim 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 = vbBluePicture1.Print "(" & p(i, 0) & ","; p(i, 1) & ")" '数据标志sigmax = sigmax + p(i, 0) 'Σxsigmay = sigmay + p(i, 1) 'Σysigmaxx = sigmaxx + p(i, 0) ^ 2 'Σx^2sigmaxy = sigmaxy + p(i, 0) * p(i, 1) 'Σx*yNext

a = (sigmaxx * sigmay - sigmax * sigmaxy) / (n * sigmaxx - sigmax ^ 2) '截距b = (n * sigmaxy - sigmax * sigmay) / (n * sigmaxx - sigmax ^ 2) '斜率x0 = Picture1.ScaleLefty0 = a + b * x0 '左端点x1 = Picture1.ScaleLeft + Picture1.ScaleWidthy1 = a + b * x1 '右端点Picture1.Line (x0, y0)-(x1, y1), vbGreen '回归直线End Sub

结果如下图所示：