Is This a Square?Date: 01/30/2001 at 12:48:25 From: Greg Subject: Given points on graph, determine if it's a square Given four points on a graph, what can I do to verify this is a square? The distance of a line is: sqrt[(x1 - x2)^2 + (y1 - y2)^2] Do I pick one (x,y) reference point and calculate the distance to each of the three other points, verifying they are all equal in length? Greg Date: 01/30/2001 at 16:34:53 From: Doctor Rick Subject: Re: Given points on graph, determine if it's a square Hi, Greg. No. If you did so, you'd be comparing two sides of the quadrilateral and one diagonal. You need to find the length of each side (the distance between two adjacent vertices) and show that all four are equal. You also must show that each side is perpendicular to the two adjacent sides. This is true if the product of the slopes of the two lines is -1. You don't have to do every one of these calculations. For instance, once you've shown that the four sides are of equal length, it's enough to show that a single pair of adjacent sides is perpendicular. A rhombus with one right angle is a square. You can also find the center of one diagonal, (xc, yc), and show that the four vertices follow this pattern, with some constants a and b: (xc+a, yc+b); (xc+b, yc-a); (xc-a, yc-b); (xc-b, yc+a) The center (xc, yc) is found by xc = (x3 - x1)/2 yc = (y3 - y1)/2 where (x1, y1) and (x3, y3) are either pair of opposite vertices of the quadrilateral. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 01/30/2001 at 16:39:12 From: Doctor Greenie Subject: Re: Given points on graph, determine if its' a square Hi, Greg -- Here is a rough picture of what you are working with... B W +-------------------*--------+ X | * * | | * * | | * * | | * * | A * * | |* * | | * * | | * *| | * * C | * * | | * * | | * * | | * * | Z +--------*-------------------+ Y D You are given points A, B, C, and D and are asked to determine if they form a square. The method you suggest won't work. If you pick point A as your reference point, then the distances to points B and D should be the same (the length of the side of the square), but the distance to point C is the length of a diagonal of the square, which is different from the length of the sides. For purposes of my discussion, I have created the rectangle WXYZ from vertical segments WZ and XY, containing points A and C, respectively, and horizontal segments WX and YZ, containing points B and D, respectively. (Actually, if ABCD is a square, then WXYZ is not just a rectangle but a square....) The sides of this rectangle are thus parallel to the x- and y-axes (the actual location of the x- and y-axes is irrelevant to the discussion). To show that ABCD is a square, you need to show that AB, BC, CD, and DA are all the same length; and you need to show that the angles at A, B, C, and D are all right angles. You could use the distance formula to show that AB, BC, CD, and DA are all the same length. And if you remember the idea of slope from algebra, you could show that each angle is a right angle by showing that the slopes of the two lines meeting at each angle are negative reciprocals of each other (for example, the slopes are -2 and +1/2). But in fact you don't need any of those sophisticated mathematical methods. You don't need to know about slopes, and you don't need to know the distance formula. All you need to do to show that ABCD is a square is to show that the lengths AW, BX, CY, and DZ are all equal and that the lengths WB, XC, YD, and ZA are all equal. If these conditions hold, then the distance formula will show the lengths AB, BC, CD, and DA to be equal; and the slope formula will show that the line segments at each corner meet at right angles - so you will be done. It is easy to measure the lengths of all these segments, because they are all either vertical or horizontal. So to show the figure is a square, you could, for example, measure the lengths AW and WB; then the figure is a square if and only if (1) BX, CY, and DZ are all the same length as AW and (2) XC, YD, and ZA are all the same length as WB. I hope this helps. Write back if my words have confused you. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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