Since you know it's a square that means that every connecting line is the same distance with exactly perpendiculare lines. It will be easiest if you draw this on a sheet of graphing paper. Then you can visualize it and see how the solution works. Turn it into two isoscelese right triangles. You know that the missing vertices will be 90 degrees because it's a square. And that the remaining sides will be of equal length for the same reason. (-1,2) and (3,2) are four four units apart. There's your hypotenuse hypotenuse (side opposite the 90). Given that it then equals √2 times the other sides you know they are 4/√2. Now you use that as your hypotenuse to get the distance up, and over from the original vertices (which will be equal, again it's a square). So your distance up is the side of the new triangle you're creating outside the square. Since the hypotenuse is 4/√2 your other sides are (4/√2)/√2. Which is the same as (4/√2)*(1/√2). Multiply the denominators (√2) and you get 2. So each side, which is the amount up and is also the amount over, is 4/2 or 2. Now you know how far to go up (and down for the other vertex) as well as over for the two other vertices of the square. The first point was (-1,2), so adding 2 to the x and y axis will give you (1,4). To get the other vertex you must do the opposite, so subtracting 2 from each in your other vertex (3,2) gives you (1,0). I know how hard that must've been to follow. I really couldn't think of how to simplify it any more. I hope I haven't mispoken anywhere and confused you. Please comment if you have any questions. :)

You wrote: "I want to know how I'll do it using either distance formula, section formula or area of a triangle formula" Here's using ONLY the distance formula (this is not the easy way to do it, but you have a stupid teacher): The distance between the two vertices is sqrt( (x2-x1)^2 + (y2-y1)^2) That tells you that the distance is sqrt(4^2 + 0) = 4 Half that distance is 2. The center of the square must be half way between the opposite vertices. A point (x,y) two units from (-1,2) and two units from (3,2) satisfies the equations: sqrt((x+1)^2 + (y-2)^2) = 2 sqrt((x-3)^2 + (y-2)^2) = 2 squaring both equations and equating the left sides gets you: 2x + 1 = -6x + 9 So x = 1 And substituting gets you y=2 The center of the square is (1,2) Now I will let x,y stand instead for one of the other vertices. You know that it is the same distance from (-1,2) as it is from (3,2) as these are sides of a square. So sqrt((x+1)^2 + (y-2)^2) = sqrt((x-3)^2 + (y-2)^2) Square this to get: (x+1)^2 = (x-3)^2 2x + 1 = -6x + 9 so x = 1 again. You also know that it is 2 units from the center of the square: sqrt((x-1)^2 + (y-2)^2) = 2 so (y-2)^2 = 4 y-2 = plus or minus 2 y = 2+2 or 2-2