Knowing two points of a rectangle, how can I figure out the other two?

Question

Hey there guys, I\'m learning processing.js, and I\'ve come across a mathematical problem, which I can\'t seem to solve with my limited geometry and trigonometry knowledge o

Answer 1

If you know the midpoint for the top, and the length of the top, then you know that the y will stay the same for both top corners, and the x will be the midpoint plus/minus the width of the rectangle. This will also be true for the bottom.

Once you have the four corners, there is no need to worry about the side lengths, as their points are the same as those used for the top and bottom.

                         midpoint
     x,10                 10,10                   x,10
      *--------------------------------------------*
                         width = 30

    mx = midpoint x.
    top left corner = (w/2) - mx  or 15 - 10
    top left corner coords = -5,10

    mx = midpoint x.
    top right corner = (w/2) + mx  or 15 + 10
    top left corner coords = 25,10

Answer 2

If you know your quadrilateral is a rectangle, then you can use some simple vector maths to find the coordinates of the corners. The knowns are:

• (x1,y1) - the coordinate of the midpoint on the top line
• (x2,y2) - the coordinate of the midpoint on the bottom line
• l1 - the length of the top and bottom lines
• l2 - the length of the other two lines

First, we find the vector between the two known points. This vector is parallel to the side lines:

(vx, vy) = (x2 - x1, y2 - y1)

We need to normalize this vector (i.e. make it length 1) so we can use it later as a basis to find our coordinates.

vlen = sqrt(vx*vx + vy*vy)

(v1x, v1y) = (vx / vlen, vy / vlen)

Next, we rotate this vector anticlockwise by 90 degrees. The rotated vector will be parallel to the top and bottom lines. 90 degree rotation turns out to just be swapping the coordinates and negating one of them. You can see this just by trying it out on paper. Or take at look at the equations for 2D rotations and substitute in 90 degrees.

(u1x, u1y) = (-v1y, v1x)

Now we have enough information to find the 'top-left' corner. We simply start at our point (x1, y1) and move back along that side by half the side length:

(p1x, p1y) = (x1 - u1x * l1 / 2, y1 - u1y * l1 / 2)

From here we can find the remaining points just by adding the appropriate multiples of our basis vectors. When implementing this you can obviously speed it up by only calculating each unique multiplication a single time:

(p2x, p2y) = (p1x + u1x * l1, p1y + u1y * l1)

(p3x, p3y) = (p1x + v1x * l2, p1y + v1y * l2)

(p4x, p4y) = (p3x + u1x * l1, p3y + u1y * l1)

Answer 3

If this quadrilateral is a rectangle (all four angles are 90 degrees), then it can be solved. (if it could be any quadrilateral, then it is not solvable)

if the points are (x1,y1), and (x2, y2), and if the two points are not perfectly vertical (x1 = x2) or horizontal (y1 = y2), then the slope of one edge of the rectangle is

m1 = (y2-y1) / (x2-x1) 

and the slope of the other edge is:

m2 = - 1 / m1

If you know the lengths of the sides, and the midpoints of two opposite sides, then the corrner points are easily determined by adding dx, dy to the midpoints: (if L is length of the sides that the midpoints are on)

dx = Sqrt( L^2 / (1 + m2^2) ) / 2

and

dy = m2 * dx

NOTE: if the points are vertically or horizontally aligned, this technique will not work, although the obvious solution for those degenerative cases is much simpler.

Answer 4

  function getFirstPoint(x1,y1,x2,y2,l1,l2)
    distanceV = {x2 - x1, y2 - y1}
    vlen = math.sqrt(distanceV[1]^2 + distanceV[2]^2)
    normalized = {distanceV[1] / vlen, distanceV[2] / vlen}
    rotated = {-normalized[2], normalized[1]}
    p1 = {x1 - rotated[1] * l1 / 2, y1 - rotated[2] * l1 / 2}
    p2 = {p1[1] + rotated[1] * l1, p1[2] + rotated[2] * l1}
    p3 = {p1[1] + normalized[1] * l2, p1[2] + normalized[2] * l2}
    p4 = {p3[1] + rotated[1] * l1, p3[2] + rotated[2] * l1}
    points = { p1 , p2 , p3 , p4}
    return p1
end

Answer 5

There's a difference between a "quadrilateral" and a "rectangle".

If you have the midpoint of the top and bottom, and the sides lengths, the rest is simple.

Given:

(x1, y1) -- (top_middle_x, top_middle_y) -- (x2, y1)

(x1, y2) -- (btm_middle_x, btm_middle_y) -- (x2, y2)

and top/bottom length along with right/left length.

x1 = top_middle_x - top/bottom_length / 2; x2 = x1 + top/bottom_length;

y1 = top_middle_y y2 = bottom_middle_y

Obviously, that's the simplest case and assuming that the line of (tmx, tmy) (bmx, bmy) is solely along the Y axis.

We'll call that line the "mid line".

The next trick is to take the mid line, and calculate it's rotational offset off the Y axis.

Now, my trig is super rusty.

dx = tmx - bmx, dy = tmy - bmy.

So, the tangent of the angle is dy / dx. The arctangent(dy / dx) is the angle of the line.

From that you can get your orientation.

(mind, there's some games with quadrants, and signs, and stuff to get this right -- but this is the gist of it.)

Once you have the orientation, you can "rotate" the line back to the Y axis. Look up 2D graphics for the math, it's straight forward.

That gets you your normal orientation. Then calculate the rectangles points, in this new normal form, and finally, rotate them back.

Viola. Rectangle.

Other things you can do is "rotate" a line that's half the length of the "top" line to where it's 90 deg of the mid line. So, say you have a mid line that's 45 degrees. You would start this line at tmx, tmy, and rotate this line 135 degrees (90 + 45). That point would be your "top left" corner. Rotate it -45 (45 - 90) to get the "top right" point. Then do something similar with the lower points.

Answer 6

Calculate the angle of the line joining the two midpoints using an arc-tangent function applied to the vector you get between them.

Subtract 90 degrees from that angle to get the direction of the top edge

Starting from the top-center point, move relative (1/2 top width x sin(angle), 1/2 top width x cos(angle)) - that gets the top right corner point.

Continue around the rectangle using the sin and cos of the angles and widths as appropriate

As a test: Check you made it back to the starting point