Given that the area of triangle ABC is 5, a (1,2), B (4,6), find the trajectory coordinates of vertex C

|Ab | = 5, triangle ABC area = | ab | D / 2 = 5, d = 2
Point C is on a line parallel to line AB at a distance of 2. There are two such lines
The equation of AB is: 4x-3y + 2 = 0
Let C (x, y), then the formula of distance from point to line is as follows:
|4x-3y+2|/5=2
The trajectory equation of point C is: 4x-3y-8 = 0, or 4x-3y + 14 = 0

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