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Take difference between points:

x2 - x1 = 6

Since you want the point of trisection nearer to 'B', take 2/3 rd of the difference. 2/3rd of difference = 6*(2/3) = 4

Similarly for difference in y: y2 - y1 = 3

2/3rd of (y2-y1) = 3*(2/3) = 2

Now the trisecting point near 'B' will be: [ Starting point + adjusted difference ]

= (1,2) + (4,2) = (5,4)

Therefore, (5,4) is the point about which the line is rotated, and thus our final line will also pass through (5,4).

Original slope of line(m) = y2-y1/(x2-x1) = 1/2. Thus angle made by original line = tan^{-1}(1/2).

Rotated by 45 deg in anti-clockwise, New angle made with x-axis = 45 + tan^{-1}(1/2)

Therefore, slope of new line = tan(45 + tan^{-1}(1/2) )

Using property: tan(A+B) = (tanA + tanB) / (1 - tanAtanB) and the property tan(tan^{-1}x) = x, we will get the slope equal to 3.

Now we have the slope(m) = 3 and the point (5,4) of the resultant line, thus calculating the line equation as:

y-y1=m(x-x1)

y - 4 = 3(x - 5)

y = 3x -11