



 
Hi Nikos, There are many possible choices for the third vertex. Suppose that the line segment AB is the base of the triangle and the distance from A to B is b. Suppose the given area of the triangle is T square units then
where h is the height of the triangle. You know b and T so any point C that is h units from the line through AB forms a triangle ABC with area T square units. Here is a construction technique which will produce two such points C. Use the distance formula to find b, the distance from A to B. Find the slope m of the line through A and B. Any line perpendicular to the line through a and B has slope 1/m. Write the equation of the line through A with slope 1/m. Let C(x_{3}, y_{3}) be a point on this line and solve for the coordinates of C so that the distance from A to C is h = 2T/b. If you have difficulty finding the coordinates of C tell us what you have done and where you are stuck and we will try to help. Harley
 


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